# Filtered Rings

Definition.

Let A be a ring. A filtration on A is a sequence of additive subgroups

$A = A_0 \supseteq A_1 \supseteq A_2 \supseteq \ldots$

such that $A_i A_j \subseteq A_{i+j}$ for any $i, j\ge 0$. A filtered ring is a ring with a designated filtration.

Note

Since $A\cdot A_i = A_0 \cdot A_i \subseteq A_i$, in fact each $A_i$ is an ideal of A.

Examples

1. If $A = \oplus_{i=0}^\infty A_i$ is a grading, we can form a filtration by taking $n \mapsto A_n \oplus A_{n+1} \oplus \ldots$ (where $A_n, A_{n+1}, \ldots$ refers to the grading).

2. Let $\mathfrak a\subseteq A$ be an ideal. The $\mathfrak a$adic filtration is given by $A_i = \mathfrak a^i$. E.g. we can take $A = \mathbb Z$ and $\mathfrak a = 2\mathbb Z$, then $A_i$ is the set of integers divisible by $2^i$. More generally, in a dvr with uniformizer $\pi$, we can take $A_i = (\pi^i)$.

Definition.

Suppose A is a filtered ring. A filtration on an A-module M is a sequence of additive subgroups

$M = M_0 \supseteq M_1 \supseteq M_2 \supseteq \ldots$

such that $A_i M_j \subseteq M_{i+j}$ for any $i, j\ge 0$. A filtered module is a module with a designated filtration.

Note

Since $A\cdot M_i = A_0 M_i \subseteq M_i$, each $M_i$ is an A-submodule of M.

Also, we need a fixed filtration on the base ring A before we can talk about filtrations on A-modules.

Example

Again, for an ideal $\mathfrak a \subseteq A$, we obtain the $\mathfrak a$adic filtration of M, where $M_i = \mathfrak a^i M$.

# Induced Filtrations

Definition.

Let M, N be filtered A-modules. A linear map $f:M\to N$ is said to be filtered if $f(M_i) \subseteq N_i$ for each i.

We also have:

Definition.

Let $f:M\to N$ be a linear map of A-modules.

• If $(M_i)$ is a filtration of M, the induced filtration on N via f is given by $N_i = f(M_i)$.
• If $(N_i)$ is a filtration of N, the induced filtration on M via f is given by $M_i = f^{-1}(N_i)$.

Note

Let us show that the induced filtrations are legitimate. In the first case,

$A_i N_j = A_i f(M_j) = f(A_i M_j) \subseteq f(M_{i+j}) = N_{i+j}$.

And in the second,

$f(A_i M_j) = A_i\cdot f(M_j) \subseteq A_i N_j \subseteq N_{i+j} \implies A_i M_j \subseteq M_{i+j}.$

In particular, if M is a filtered A-module and $N\subseteq M$ is a submodule, the induced filtrations on N and M/N are given by:

$N_i = M_i \cap N, \quad (M/N)_i = (M_i + N)/N.$

So far everything is natural, but beneath all this a danger lurks.

If $f:M\to N$ is a filtered A-linear map, then $M/\mathrm{ker} f$ and $\mathrm{im }f$ have induced filtrations via quotient module of M and submodule of N. Although $M/\mathrm{ker } f \cong \mathrm{im }f$ as A-modules, it is not an isomorphism in the category of filtered A-modules!

Let us write everything out explicitly. The filtration on the LHS and RHS are given respectively by

$(M / \mathrm{ker } f)_i = (M_i + \mathrm{ker } f)/\mathrm{ker } f, \quad (\mathrm{im } f)_i = (\mathrm{im} f) \cap N_i$

which gives

$(M_i + \mathrm{ker } f)/\mathrm{ker } f \cong M_i / (M_i \cap \mathrm{ker } f) = M_i / \mathrm{ker} (f|_{M_i}) \cong \mathrm{im} (f|_{M_i})$

which is not isomorphic to $(\mathrm{im} f) \cap N_i$ in general.

# Completion

Definition.

Let M be a filtered A-module, the completion of M is the following (inverse) limit of A-modules:

$\hat M := \varprojlim M/M_n.$

By the description of inverse limits in the category of A-modules, $\hat M$ comprises of the set of all $(\ldots, x_n, \ldots, x_2, x_1) \in \prod_{i=1}^\infty (M/M_i)$ such that for each i, $x_{i+1}$ maps to $x_i$ under the canonical map $M/M_{i+1}\to M/M_i$.

The canonical maps $M\to M/M_n$ induce an A-linear map $M \to \hat M$ by the universal property of inverse limits. This can be described as follows: for each $m\in M$, let $m_i$ be its image in $M/M_i$; then the map takes $(m \in M) \mapsto (\ldots, m_n, \ldots, m_2, m_1) \in \hat M$. From this we see that:

Lemma 1.

The map $i:M\to \hat M$ is injective if and only if $\cap_n M_n = 0$, in which case we say the filtration is Hausdorff.

Next, by setting MA, we also have $\hat A = \varprojlim A/A_n$. Although we defined $\hat A$ as an A-module, one sees by the explicit construction that $\hat A$ has a ring structure. To be specific,

$(a_n)_{n=1}^\infty, (b_n)_{n=1}^\infty \in \hat A \implies (a_n) \times (b_n) := (a_n b_n)_{n=1}^\infty.$

Exercise A

1. Prove that $\hat M$ has a canonical structure as an $\hat A$-module.

2. Prove that $\hat A$ is the inverse limit of $A/A_n$ in the category of rings. In particular, the canonical map $i : A\to \hat A$ is a ring homomorphism.

3. Prove that if the filtration on M is Hausdorff, we can define a metric on M via:

$d(x, y) := 2^{-|x-y|}$, where $|z| := \sup\{ n : z\in M_n\},$

such that the collection of all cosets $\{ m + M_n : m\in M, n \ge 0\}$ forms a basis for the resulting topology. In fact, d is an ultrametric.

# Limits in Completion

The completion $\hat M$ enables us to take limits of infinite sequences and sums of infinite series in modules.

Definition.

Let $m_1, m_2, \ldots \in M$ be a sequence in a filtered module M. We say the sequence is Cauchy if: for any i, there exists n such that

$m_n \equiv m_{n+1} \equiv m_{n+2} \equiv \ldots \pmod {M_i}$.

The astute reader would note that if the filtration is Hausdorff, then $(m_n)$ is Cauchy here if and only if it is Cauchy with respect to the metric of exercise A.3.

Definition.

The limit of a Cauchy sequence $m_1, m_2, \ldots \in M$ is defined as follows. For each i, pick n such that $m_n \equiv m_{n+1} \equiv \ldots$ modulo $M_i$, with image $y_i \in M/M_i$. Now define:

$\lim_{n\to\infty} m_n := (\ldots, y_n, \ldots, y_2, y_1) \in \hat M$.

From the definition, it is clear that if $(m_n), (m_n')$ (resp. $(a_n)$) are Cauchy sequences in M (resp. in A), then $(m_n + m_n')$ and $(a_n m_n)$ are also Cauchy and we have

\begin{aligned}\lim_{n\to\infty} (m_n + m_n') &= (\lim_{n\to\infty} m_n) + (\lim_{n\to\infty} m_n') \in \hat M, \\ \lim_{n\to\infty} (a_n m_n) &= (\lim_{n\to\infty} a_n) (\lim_{n\to\infty} m_n) \in \hat M.\end{aligned}

Pick a ring A with maximal ideal $\mathfrak m$; we will take the $\mathfrak m$-adic filtration $A_n = \mathfrak m^n$. Given a sequence $x_i \in A_i$ for $i=0, 1, \ldots$, let

$y_i := x_0 + \ldots + x_{i-1} \pmod {\mathfrak m^i}$, an element of $A/A_i$.

Then $(y_i)$ is a Cauchy sequence in A and we write:

$\sum_{i=0}^\infty x_i := \lim_{n\to\infty} y_n \in \hat A$.

Arithmetic Example.

Let $A = \mathbb Z$ and $A_n = p^n \mathbb Z$ for a prime p. The completion $\hat A$ is called the ring of p-adic integers and denoted by $\mathbb Z_p$. Let us take p = 2 and the element:

$y = 1 + 2^2 + 2^4 + 2^6 + \ldots \in \mathbb Z_2.$

Then $3y = 3 + 3(2^2) + 3(2^4) + \ldots$ is congruent to -1 modulo any $2^n$. Thus $y = -\frac 1 3$. In general, an element of $\mathbb Z_p$ can be regarded as having an infinite base-p expansion. Thus the above $y \in \mathbb Z_2$ would have base-2 expansion $(\ldots 1010101)_2$. One easily checks that three times this value is $(\ldots 1111)_2$, which is -1.

Geometric Example

Let A be a ring, $B = A[X, Y]$ and $B_n = \mathfrak m^n$ where $\mathfrak m = (X, Y)$. Again, we can take the infinite sum $\alpha = 1 + XY + (XY)^2 + \ldots$ and check that $\alpha(1-XY) = 1$. Note that $\hat B \cong A[[X,Y]]$, the ring of formal power series with coefficients in A.

Definition.

Let A be any ring. A formal power series in X with coefficients in A is an expression

$f(X) = a_0 + a_1 X + a_2 X^2 + \ldots$, where $a_i \in A$.

Unlike polynomials, we allow infinitely many $a_i$ to be non-zero. Addition and multiplication of formal power series are defined as follows. For $f(X) = \sum_{i=0}^\infty a_n X^n$ and $g(X) = \sum_{j=0}^\infty b_n X^m$,

\begin{aligned} f(X) + g(X) &=(a_0 + b_0) + (a_1 + b_1)X + (a_2 + b_2)X^2 + \ldots \\ f(X)\times g(X) &= (a_0 b_0) + (a_0 b_1 + a_1 b_0)X + (a_0 b_2 + a_1 b_1 + a_2 b_0)X^2 + \ldots \end{aligned}

This gives a ring structure on the set $A[[X]]$ of formal power series. To define rings of formal power series in multiple variables, we set recursively

$A[[X_1, \ldots, X_n]] := (A[[X_1, \ldots, X_{n-1}]])[[X_n]]].$

As another example, let us take the $\mathfrak m$-adic completion for $A = \mathbb C[X, Y]/(Y^2 - X^3 + X)$ and $\mathfrak m = (X, Y)$. We will prove later that

$\hat A \cong \mathbb C[[X, Y]]/(Y^2 - X^3 + X)$

and that the map $\mathbb C[[Y]] \to \hat A$ is an isomorphism. Geometrically, this means when we project $E: Y^2 = X^3 - X$ to the Y-axis, the map is locally invertible at the origin.

[ Image edited from GeoGebra plot. ]

Functorially, the ring $A[[X, Y]]$ behaves quite differently from $A[X, Y]$, because as an A-module, it is a countably infinite direct product of copies of A, unlike $A[X, Y]$ which is a direct sum. If we follow the earlier guideline, it is (for example) generally false that $B \otimes_A A[[X, Y]] \cong B[[X, Y]]$ for any A-algebra B.

Exercise B

1. Prove that the canonical map $\mathbb C[[X]] \to \mathbb C[[X, Y]]/(Y^2 - X^3 + X)$ is not an isomorphism.

2. Let A be a filtered ring. Prove that if $f(X) \in A[[X]]$ is $a_0 + a_1 X + a_2 X^2 + \ldots$, then f defines a map

$A_1 \to \hat A, \quad f(\alpha) := \sum_{n=0}^\infty a_n \alpha^n \in \hat A.$

Prove that if we fix $\alpha \in A_1$, we get a ring homomorphism $A[[X]] \to \hat A$, $f\mapsto f(\alpha)$.

## Commutative Algebra 53

Definition.

A grading on a ring A is a collection of additive subgroups $A_0, A_1, \ldots \subseteq A$ such that

$A = A_0 \oplus A_1 \oplus A_2 \oplus \ldots$

as abelian groups, and $A_i A_j \subseteq A_{i+j}$ for any $i, j\ge 0$, i.e..

$a\in A_i, b\in A_j \implies ab \in A_{i+j}.$

Note

The notation $A = \oplus_i A_i$ means every $a\in A$ can be uniquely written as a finite sum $a_0 + a_1 + \ldots + a_n$ for some $a_i \in A_i$ (uniqueness holds up to appending or removal of $0\in A_m$). Then $a_d$ is called the degree-d component of $a$.

An $a\in A$ is said to be homogeneous of degree d if $a \in A_d$; we write $\deg a = d$. Note that d is unique if $a\ne 0$.

Example

The standard example of a graded ring is the ring of polynomials $B = A[X_1, \ldots, X_n]$ over some ring A, where $B_d$ has A-basis given by the set of monomials $X_1^{m_1} \ldots X_n^{m_n}$ satisfying $m_i \ge 0$ and $\sum_i m_i = d$.

Lemma 1.

$A_0$ is a subring of A and each $A_i$ is a module over $A_0$.

Proof

For the first statement, since $A_0 A_0 \subseteq A_0$ it suffices to show $1 \in A_0$; we may assume $1\ne 0$. Write $1 = a_0 + \ldots + a_n$ with $a_i \in A_i$ and $a_n \ne 0$. Suppose n > 0. For any $b_m \in A_m$ we have

$b_m = b_m\cdot 1 = b_m(a_0 + \ldots + a_n) = b_m a_0 + \ldots + b_m a_n.$

Since the LHS is homogeneous of degree m, we have $b_m a_n = 0$. Thus $a_n A_m = 0$ for any m so we have $a_n A = 0$. This gives $a_n = 0$, a contradiction.

The second statement is clear. ♦

By definition if $a, b\in A$ are homogeneous of degrees m and n, then $ab$ is homogeneous of degree m+n. We also have:

Lemma 2.

Suppose A is an integral domain with grading. If $a,b\in A$ are non-zero elements such that $ab$ is homogeneous, then so are a and b.

Proof

Write $a = \sum_m a_m$ and $b = \sum_n b_n$ as sums of their components. Let m (resp. m’) be the minimum (resp. maximum) degree for which $a_m \ne 0$ (resp $a_{m'} \ne 0$). Similarly, let n (resp. n’) be the minimum (resp. maximum) degree for which $b_n \ne 0$ (resp $b_{n'} \ne 0$). By definition $a_m b_n, a_{m'}b_{n'} \ne 0$. Since ab is homogeneous we have $m+n = m'+n'$ and thus $m = m', n = n'$. So a and b are homogeneous. ♦

Exercise A

Find a graded ring A, with $a,b \in A-\{0\}$ such that ab is homogeneous but a and b are not. [ Hint: come back to this exercise after finishing the whole article. ]

Definition.

A grading on an A-module M is a collection of additive subgroups $M_0, M_1, \ldots \subseteq M$ such that

$M = M_0 \oplus M_1 \oplus M_2 \oplus \ldots$

as additive groups, and $A_i M_j \subseteq M_{i+j}$ for any $i,j \ge 0$.

Note

We need a fixed grading on the base ring A before we can talk about grading on A-modules.

• As before if we write $m\in M$ as a sum $m_0 + m_1 + \ldots + m_n$ with $m_i \in M_i$, then $m_d$ is called the degree-d component of m.
• Also $m\in M$ is homogeneous of degree d if $m \in M_d$; again write $\deg m = d$.

The following result is quite important for grading of submodules.

Proposition 1.

Let M be a graded A-module and $N\subseteq M$ be a submodule. The following are equivalent.

1. There is a generating set for N comprising of homogeneous elements.
2. If $n\in N$, all homogeneous components of n lie in N.
3. We have $N = (N \cap M_0) \oplus (N \cap M_1) \oplus \ldots$.

Proof

(1⇒3) It suffices to show $N = \sum_i (N \cap M_i)$. Pick a generating set S for N comprising of non-zero homogeneous elements. For each $n\in N$, write $n = \sum_{i=1}^k a_i m_i$ with $a_i \in A$ and $m_i \in S$ with $\deg m_i = d_i$. Fix $d\ge 0$ and let n’ be the degree-d component of n. Then n’ is the sum of the degree-d components of $a_i m_i$. Hence

$n' = \sum_{i=1}^k a_i' m_i$, where $a_i'$ is the degree-$(d-d_i)$ component of $a_i$

so $n'\in N \cap M_d$.

(3⇒2) Let $n\in N$; we can write $n = n_0 + n_1 + \ldots + n_d$ where $n_i \in N \cap M_i$. Since $n_i \in M_i$, it is homogeneous of degree i; thus $n_i$ is the degree-i component of n and it lies in N.

(2⇒1) Pick any generating set S of N, then take the homogeneous components of all $m\in S$ to obtain a homogeneous generating set.

Definition.

Let M be a graded A-module. A submodule $N\subseteq M$ is said to be graded if it satisfies the conditions of proposition 1.

An ideal $\mathfrak a\subseteq A$ is said to be graded if it is graded as a submodule.

Proposition 1 immediately gives the following.

Corollary 1.

Given a graded module M, with collection of graded submodules $(N_i)$, graded submodule N, and graded ideal $\mathfrak a\subseteq A$,

$\sum_i N_i, \quad \cap_i N_i, \quad \mathfrak a N$

are all graded submodules of M.

Proof

Since each $N_i$ is generated by homogeneous elements, so is $\sum N_i$; thus $\sum N_i$ is graded. Suppose $n\in \cap_i N_i$, if n’ is a homogeneous component of n, then for each i, $n \in N_i \implies n' \in N_i$ and hence $n' \in \cap N_i$. Thus $\cap N_i$ is graded. Finally, if S (resp. T) is a generating set of $\mathfrak a$ (resp. N) comprising of homogeneous elements, then $\{an : a\in S, n\in T\}$ is a generating set of $\mathfrak aN$ comprising of homogeneous elements. ♦

Exercise B

Decide if each statement is true.

• If $\mathfrak a\subseteq A$ is a graded ideal of A, then $r(\mathfrak a)$ is graded.
• If $N, P \subseteq M$ are graded submodules, then $(N : P) = \{a \in A: aP \subseteq N\}$ is a graded ideal.

# Quotient

Proposition 2.

Let N be a graded submodule of a graded module M. Then M/N has a canonical grading given by

$(M/N)_i := M_i / (N\cap M_i) \hookrightarrow M/N$.

Proof

Note that we have $A_i (M/N)_j \subseteq (M/N)_{i+j}$. It remains to show M/N is a direct sum of $(M/N)_i$.

Let $m\in M$. Write $m = \sum_i m_i$ with $m_i \in M_i$. Then $m+N \in M/N$ is the sum of images of $m_i + (N \cap M_i) \in M/(N\cap M_i)$ in $M/N$. So $M/N = \sum_i (M/N)_i$.

Suppose $m + N$ can be written as $\sum_i [m_i + (N\cap M_i)]$ and $\sum_i [m_i' + (N\cap M_i)]$. Then

$m+N = (\sum_i m_i) + N = (\sum_i m_i') + N \implies \sum_i (m_i - m_i') \in N.$

By condition 2 of proposition 1, each $m_i - m_i' \in N$ so $m_i + (N\cap M_i) = m_i' + (N\cap M_i)$. ♦

Corollary 2.

If $\mathfrak a\subseteq A$ is a graded ideal, then $A/\mathfrak a$ is a graded ring under the above grading.

Proof

Since $A_i A_j \subseteq A_{i+j}$, multiplication gives

$A_i / (\mathfrak a \cap A_i) \times A_j / (\mathfrak a \cap A_j) \longrightarrow A_{i+j} / (\mathfrak a \cap A_{i+j}).$

Example

1. Suppose A is the coordinate ring of a variety $V\subseteq \mathbb A^n$, so that $A \cong k[X_1, \ldots, X_n]/I(V)$. If I(V) is homogeneous, then A is a graded ring. In particular, the group of units in A is $k^*$, the multiplicative group $k - \{0\}$. This does not work for non-homogeneous I(V), e.g. the unit group of $k[X, Y]/(XY - 1)$ contains $\{ c X^n : n \in \mathbb Z, c \in k^*\}$ (does equality hold?).

2. We will do exercise C here, i.e. show that $A = \mathbb C[X, Y, Z]/(Z^2 - X^2 - Y^2)$ is not a UFD. Indeed we have the equality $Z \cdot Z = (X+iY)(X - iY)$ in A; we claim that $Z, X+iY, X-iY \in A$ are all irreducible. By lemma 2, since Z is homogeneous of degree 1, we can only factor it as a product of a degree-0 and a degree-1 element. But all non-zero degree-0 elements of A are units (see example 1). Similarly $X+iY, X-iY$ are irreducible and $Z, X+iY, X-iY$ are clearly not associates.

3. By the same reasoning, $\mathbb C[W, X, Y, Z]/(Z^2 - W^2 - X^2 - Y^2)$ is not a UFD.

Note: however $\mathbb C[X_1, \ldots, X_n]/(X_1^2 + \ldots + X_n^2)$ is a UFD for all $n\ge 5$, as we will see later.

Definition.

Let M and N be graded A-modules, and $f:M\to N$ be A-linear. We say f is graded if $f(M_i) \subseteq N_i$ for each i.

Immediately we have:

Lemma 3.

If $f:M\to N$ is a graded map of graded A-modules, then $\mathrm{ker} f$ is a graded submodule of M and $\mathrm{im} f$ is a graded submodule of N.

Proof

If $m \in M$ and $m = m_0 + \ldots + m_d$ with $m_i \in M_i$, then $f(m_i) \in N_i$ and hence $f(m) = f(m_0) + \ldots + f(m_d)$ is the unique decomposition of f(m) into its homogeneous components. The rest is an easy exercise. ♦

Proposition 3 (First Isomorphism Theorem).

For a graded map $f:M\to N$ of graded modules, we have an isomorphism

$g : M/\mathrm{ker} f \longrightarrow \mathrm{im } f, \quad m + \mathrm{ker } f \mapsto f(m).$

in the category of graded A-modules.

Proof

Let us show that g is graded. The grading on the LHS is given by $i\mapsto M_i / (\mathrm{ker} f \cap M_i)$. Since f is graded, g takes the LHS into $f(M_i) = N_i$.

Since g is bijective, it remains to show that $g^{-1}$ is also graded. For $n\in N_i$, let $m = g^{-1}(n)$. Write $m = m_0 + \ldots + m_d$ as a sum of homogeneous components; since $g(m_i)$ is the degree-i homogeneous component of $g(m)$ we have $g(m) = g(m_i)$ so $m = m_i$. ♦

Exercise C

State and prove the remaining two isomorphism theorems.

# Direct Limits of Rings

Let $((A_i), (\beta_{ij}))$ be a directed system of rings. Regard them as a directed system of abelian groups (i.e. ℤ-modules) and take the direct limit A.

Proposition 1.

The abelian group A has a natural structure of a commutative ring.

Note

General philosophy of the direct limit: “if something happens at index i and another thing happens at index j, then by picking k greater than i and j, both things may be assumed to happen at the same index”. The cautious reader is advised to fill in the gaps in the following proof.

Sketch of Proof

For each $i\in J$, let $\epsilon_i : A_i \to A$ be the canonical map.

Given $a,b \in A$, by proposition 2 here there exists $j\in J$ and $a', b' \in A_j$ such that $a = \epsilon_j(a')$, $b = \epsilon_j(b')$. Now define multiplication in A by $a\times b := \epsilon_j(a'b')$. This does not depend on our choice of j and $a',b' \in A_j$. Indeed, if we have another index $i\in J$ and $a_1, b_1 \in A_i$ such that $a = \epsilon_i(a_1)$, $b = \epsilon_i(b_1)$, by proposition 2 here again pick $k\in J$ greater than i and j such that

$\beta_{ik}(a_1) = \beta_{jk}(a'), \ \beta_{ik}(b_1) = \beta_{jk}(b') \implies \beta_{ik}(a_1 b_1) = \beta_{jk}(a'b').$

This gives us the desired equality

$\epsilon_i(a_1 b_1) = \epsilon_k(\beta_{ik}(a_1 b_1)) = \epsilon_k (\beta_{jk}(a'b')) = \epsilon_j(a'b').$

Clearly product in A is commutative. To show associativity, given $a,b,c\in A$, pick $j\in J$ and $a',b',c' \in A_j$ such that $a = \epsilon_j(a')$, $b = \epsilon_j(b')$ and $c = \epsilon_j(c')$. Then $a(bc) = \epsilon_j(a'(b'c')) = \epsilon_j((a'b')c') = (ab)c$.

To define $1\in A$, we pick any index $i\in J$ and set $1_A := \epsilon_i(1_{A_i})$. ♦

Exercise A

1. Prove that $1_A$ in the above proof is well-defined, and $1_A \times a = a$ for all $a\in A$.

2. Prove that the resulting ring A with the canonical $\epsilon_i : A_i \to A$ gives the direct limit of $A_i$ in the category of rings.

3. Prove that if the direct limit of rings $A_i$ is zero, then $A_i = 0$ for some i. [ Hint: a ring is zero if and only if 1 = 0. ]

4. Suppose $(B_i)_{i\in I}$ is an arbitrary collection of A-algebras. For each finite subset $L = \{i_1, \ldots, i_n\} \subseteq I$, define $B_L := B_{i_1} \otimes_A \ldots \otimes_A B_{i_n}$. Define a directed system of $B_L$ over the directed set of all finite subsets of I, ordered by inclusion $L\subseteq L'$. The tensor product of $B_i$ over A is defined to be the direct limit of this system. Prove that this gives the coproduct of $(B_i)_{i\in I}$ in the category A-algebras.

Note

Since direct limits are denoted by $\varinjlim$, we will write $\varprojlim$ for the earlier limits and call them inverse limits. For most cases of interest, inverse limits will be taken over J such that $J^{op}$ is directed.

Even over directed sets, taking the inverse limit is not exact. A useful criterion for determining exactness is given by the Mittag-Lefler condition, which we will not cover (for now).

# Taking Stock

We have seen many constructions which commute and some which do not. In the following examples, $(M_i)$ is an arbitrary collection of modules; M and N are modules, B is an A-algebra and $S\subseteq A$ is a multiplicative subset.

Case 1 (proposition 1 here): $S^{-1} (\oplus_i M_i) = \oplus_i S^{-1} M_i$ but $S^{-1}(\prod_i M_i) \ne \prod_i S^{-1}M_i$ in general.

Case 2 (exercise A here): $\mathfrak a (\oplus_i M_i) = \oplus_i \mathfrak a M_i$ but $\prod_i \mathfrak a M_i \ne \mathfrak a (\prod_i M_i)$.

Case 3 (corollary 1 here): a direct sum of projective modules is projective. A direct product of projective modules is not projective in general, but a counter-example is not too easy to construct.

Case 4 (exercise B here): for any collection $(N_i)$ of submodules of M, we have $S^{-1}(\sum_i N_i) = \sum_i S^{-1}N_i$ but $S^{-1}(\cap_i N_i) \ne \cap_i S^{-1} N_i$ in general.

Case 5 (exercise A here): if M is a flat A-module, then $M^B$ is a flat B-module.

Case 6 (proposition 1 here): we have $(\oplus_i M_i) \otimes_A N \cong \oplus_i (M_i \otimes_A N)$.

Case 7: more generally, we have $(\mathrm{colim}_{i\in J} M_i) \otimes_A N \cong \mathrm{colim}_{i\in J} (M_i \otimes_A N)$ if $(M_i)_{i\in J}$ is a diagram of A-modules of type J.

Case 8: hence we have an isomorphism $(\mathrm{colim}_{i\in J} M_i)^B \cong \mathrm{colim}_{i\in J}(M_i)^B$ of B-modules; in particular $S^{-1}(\mathrm{colim}_{i\in J} M_i) \cong \mathrm{colim}_{i\in J} (S^{-1} M_i)$.

Case 9 (proposition 3 here): if $f: M_1 \to M_2$ is surjective, then $f\otimes_A 1_N : M_1 \otimes_A N \to M_2 \otimes_A N$ is surjective; however, if f is injective $f\otimes_A 1_N$ is not injective in general.

Exercise B

Prove that there is always a canonical map between $(\prod_i M_i) \otimes_A N$ and $\prod_i (M_i \otimes_A N)$. Find an example where the map is not an isomorphism.

If M is a projective A-module, must $M^B = B\otimes_A M$ be a projective B-module?

# Duality Principle

Remembering all the above relations may seem like a pain: in general if we have n constructions, we have about $O(n^2)$ relations to learn. It turns out most of these constructions can be classified as either “left-adjoint-like” or “right-adjoint-like”, which saves us a whole lot of effort in remembering them.

In the following table, constructions on the same side tend to commute or have consistent properties. Constructions on different sides may commute under specific additional conditions (e.g. finiteness, noetherianness).

 Left-adjoint-like Right-adjoint-like Sum of submodules Intersection of submodules Coproducts Products Right-exact functors Left-exact functors Pushouts Pullbacks, fibre products Direct sum of modules Direct product of modules Tensor products of modules Hom modules HomA(-, M) HomA(M, -) Colimits / Direct limits Limits / Inverse limits Injective maps Surjective maps Quotient modules Submodules Induced modules $M\mapsto M^B$. (Coinduced modules) Projective / free modules (Injective modules) Localization Multiplying ideal by module: $\mathfrak a M$

Do consider this table as a very rough guide. For example, if $0 \to N \to M \to P \to 0$ is a short exact sequence of A-modules, we do not get a right-exact sequence $\mathfrak a N \to \mathfrak a M \to \mathfrak a P \to 0$. [ Take $0\to 2\mathbb Z \to \mathbb Z \to \mathbb Z / 2\mathbb Z \to 0$ and $\mathfrak a = 2\mathbb Z$. ]

Also note that the terms in brackets have not been defined yet.

### Further Examples

1. We have $\mathrm{Hom}_A(\oplus_i N_i, M) \cong \prod_i \mathrm{Hom}_A(N_i, M)$.

2. The functor $\mathrm{Hom}_A(-, M)$ takes a right-exact sequence to a left-exact sequence.

3. Recall that the colimit of a diagram of A-modules was constructed by taking a quotient of the direct sum of these modules. Dually, its limit can be constructed by taking a submodule of the direct product.

4. The tensor product was constructed by taking a quotient of a free (hence projective) module.

Exercise C (Coinduced Modules)

1. Let B be an A-algebra. Prove that for an A-module M, $M_B := \mathrm{Hom}_A(B, M)$, the set of all A-linear maps $B\to M$, has a natural structure of a B-module.

2. Prove that we get a functor

$F :A\text{-}\mathbf{Mod} \longrightarrow B\text{-}\mathbf{Mod}, \quad M \mapsto M_B$

such that there is a natural bijection

$\mathrm{Hom}_A(N, M) \cong \mathrm{Hom}_B(N, M_B).$

for any B-module N. Thus $F = \mathrm{Hom}_A(B, -)$ is right-adjoint to the forgetful functor $B\text{-}\mathbf{Mod} \to A\text{-}\mathbf{Mod}$.

We call $M_B$ the coinduced B-module from M.

# Limits Are Left-Exact

By example 6 and proposition 2 in the previous article, one is inclined to conclude that taking the colimit in $\mathcal C = A\text{-}\mathbf{Mod}$ is a right-exact functor, but there is a rather huge issue here: the functors are between $\mathcal C$ and $\mathcal D = \mathcal C^J$, the category of diagrams in $\mathcal C$ while we only defined exactness of functors between categories of modules. The proper way to do this is to introduce the framework of abelian categories and extend our concept of additive functors and exact functors there. However, doing this will take us too far afield so we will prove it directly (which is, admittedly, a bit of a cop out).

Proposition 1.

Let J be an index category, and $D', D, D'' : J\to A\text{-}\mathbf{Mod}$ be diagrams of type J. For concreteness, write these diagrams as

$((N_i), (\beta_e^N : N_i \to N_j)), \ ((M_i), (\beta_e^M : M_i \to M_j)),\ ((P_i), (\beta_e^P : P_i \to P_j))$

where $i\in J$ and $e:i\to j$. Let $D'\to D \to D''$ be morphisms, written as a collection of $N_i\stackrel {\phi_i} \to M_i \stackrel {\psi_i} \to P_i$ over $i\in J$. Then

\left( \begin{aligned}0\to N_i\stackrel {\phi_i} \to M_i \stackrel {\psi_i} \to P_i\\ \text{exact for each } i\in J\end{aligned}\right) \implies 0 \to \lim N_i \to \lim M_i \to \lim P_i \text{ exact.}

\left( \begin{aligned}N_i\stackrel {\phi_i} \to M_i \stackrel {\psi_i} \to P_i \to 0\\ \text{exact for each } i\in J\end{aligned}\right) \implies \mathrm{colim} N_i \to \mathrm{colim} M_i \to \mathrm{colim} P_i \to 0\text{ exact.}

Note

In summary, taking the limit is left-exact while taking the colimit is right-exact.

Proof

We prove the second claim, leaving the first as an exercise. By proposition 1 here, $\mathrm{colim} M_i$ is concretely described as follows. Take the quotient of $\oplus_{i\in J} M_i$ by all $m_i - \beta_e^M(m_i)$, where $e:i\to j$ is an arrow in J, $m_i \in M_i$ and $\beta_e^M(m_i)\in M_j$ are identified with their images in $\oplus_i M_i$.

With this description, clearly $\mathrm{colim} M_i \to \mathrm{colim} P_i$ is surjective. Also, composing $\mathrm{colim} N_i \to \mathrm{colim} M_i \to \mathrm{colim} P_i$ is the zero map so $\mathrm{im}(\mathrm{colim} \phi_i) \subseteq \mathrm{ker}(\mathrm{colim} \psi_i)$. Now write $\phi : \oplus_i N_i \to \oplus_i M_i$ for $\oplus \phi_i$ and $\psi : \oplus_i M_i \to \oplus_i P_i$ for $\oplus \psi_i$.

Conversely, let $x\in \oplus_i M_i$ represent an element in the kernel of $\mathrm{colim} \psi_i$. Thus $\psi(x) \in\oplus_i P_i$ is a finite sum of $p_i - \beta_e^P(p_i)$. Since $\psi_i$ is surjective, we can write such a term as

$\psi_i(m_i') - \beta_e^P(\psi_i(m_i')) = \psi_i(m_i') - \psi_j(\beta_e^M(m_i')) = \psi(m_i' - \beta_e^M(m_i'))$

for some $m_i'\in M_i$. Since $\psi(x)$ is a finite sum of $\psi(m_i' - \beta_e^M(m_i'))$, we can replace x by another representative such that $\psi(x) = 0$. Then $x = \phi(y)$ for some $y\in \oplus N_i$. ♦

Neither the limit nor the colimit functor is exact in general. For the colimit case, consider the following commutative diagram of A-modules

where all maps $A\to A$ are identities. The rows are short exact sequences and the squares all commute, but taking the colimit of the columns gives

$0 \longrightarrow A^2 \longrightarrow A \longrightarrow 0 \longrightarrow 0$

which is not exact.

Exercise A

Find an example for the case of limits.

# Direct Limits

We will describe a special case where taking the colimit is exact.

Given a poset $(S, \le)$, we recall the category $\mathcal C(S)$ whose objects are elements of S, and between any $x,y\in S$, $|\mathrm{hom}(x, y)| \le 1$ with equality if and only if $x \le y$. Composition is the obvious one.

Definition.

A poset $(S, \le)$ is called a directed set if for any $a,b\in S$, there is a $c\in S$ such that $a\le c$ and $b\le c$.

In other words, a poset is directed if every finite set has an upper bound.

Definition.

If J is an index category obtained from $\mathcal C(S)$ for some directed set S, then a diagram in $\mathcal C$ of type J is called a directed system. The colimit of $(A_i)_{i\in J}$ is called the direct limit and denoted by

$\varinjlim_{i\in J} A_i$.

In other words, direct limit = colimit over directed set. We will abuse notation a little and regard J as the directed set itself.

To avoid set-theoretic difficulties, the directed set J is always assumed to be non-empty.

Example

In exercise C.3 here, for a multiplicative $S\subseteq A$ and A-module M, we have an isomorphism of A-modules

$\mathrm{colim}_{f\in S} M_f \cong M_S$

where $f\le g$ if g is a multiple of f. Since S is multiplicative, any {fg} has an upper bound fg. Hence $M_S$ is the direct limit of $M_f$ over $f\in S$:

$\varinjlim_{f\in S} M_f \cong M_S$.

Similarly, we have the following direct limit in the category of rings:

$\varinjlim_{f\in S} A_f \cong A_S$.

Next we will discuss the general direct limit in the categories A-Mod and Ring.

# Direct Limit of Modules

Let A be a fixed ring; the following holds for direct limits in the category of A-modules.

Proposition 2.

Suppose $((M_i)_{i\in J}, (\beta_{ij})_{i\le j})$ is a directed system of A-modules over a directed set J. Let

$M = \varinjlim_{i\in J} M_i$, with canonical $\epsilon_i : M_i \to M$ for each $i\in J$.

Then for each $m\in M$, there exists an $i\in J$ and $m_i \in M_i$ such that $\epsilon_i(m_i) = m$.

Also if $m_i\in M_i$ satisfies $\epsilon_i(m_i) = 0$, then there exists $j\ge i$ such that $\beta_{ij}(m_i) = 0 \in M_j$.

Note

The philosophy is that “whatever happens in the direct limit happens in $M_j$ for some sufficiently large index j“.

Proof

By proposition 1 here, the colimit M is described concretely by taking the quotient of $P = \oplus_{i\in J} M_i$ (with canonical $\nu_i : M_i \to P$) by relations of the form

$\nu_k(m_k) - \nu_l\beta_{kl}(m_k),\ m_k \in M_i,\ k\le l\ (k, l\in J).$

Hence any $m\in M$ can be written as $\epsilon_{i_1}(m_1) + \ldots + \epsilon_{i_N}(m_N)$ for $m_1 \in M_{i_1}, \ldots, m_N \in M_{i_N}$. But J is a directed set, so we can pick index $j\in J$ such that $j\ge i_1, \ldots, i_N$; then

$m = \epsilon_j(m_j)$, where $m_j = \beta_{i_1 k}(m_1) + \ldots + \beta_{i_N k}(m_N)$,

proving the first claim.

For the second claim, if $\epsilon_i(m_i) = 0$ then $\nu_i(m_i) \in \oplus_i M_i$ is a finite sum of the above relations. Pick an index $j\in J$ larger than i and all indices kl in the sum; then $\beta_{ij}(m_i)$ is the sum of the images of these relations in $M_k$. But each such relation has image $\beta_{kj}(m_k) - \beta_{lj}\beta_{kl}(m_k) = 0$ in $M_k$, so $\beta_{ij}(m_i) = 0$ as desired. ♦

Corollary 1.

If $((M_i), (\beta_{ij}))$ is a directed system of A-modules such that $\beta_{ij}$ are all injective, then

$\epsilon_j : M_j \longrightarrow \varinjlim_i M_i$

is also injective for each $j\in J$.

Finally we have:

Proposition 3.

Let $((M_i), (\beta_{ij}))$ and $((N_i), (\gamma_{ij}))$ be directed systems of A-modules and $\phi_i : M_i \to N_i$ be a morphism of the directed systems, i.e. for any $i\le j$, we have $\gamma_{ij} \circ \phi_i = \phi_j \circ \beta_{ij} : M_i \to N_j$.

If each $\phi_i$ is injective, so is $\phi : \varinjlim M_i \to \varinjlim N_i$.

Since taking the colimit is right-exact by proposition 1, we see that taking the direct limit is exact.

Proof

Write $\epsilon_i^M : M_i \to \varinjlim M_i$ and $\epsilon_i^N : N_i \to \varinjlim N_i$ for the canonical maps.

Suppose $\phi(m) = 0$ for $m \in \varinjlim M_i$. By proposition 2, we have $m = \epsilon_i^M(m_i)$ for some $m_i \in M_i$; then

$0 = \phi(m) = \phi(\epsilon_i^M(m_i)) = \epsilon_i^N(\phi_i(m_i))$

so by proposition 2 again, there exists $j\ge i$ such that $\gamma_{ij}(\phi_i(m_i)) = 0$, so $\phi_j(\beta_{ij}(m_i)) = 0$. Since $\phi_j$ is injective we have $\beta_{ij}(m_i) = 0$ so $m = \epsilon_i^M(m_i) = \epsilon_j^M\beta_{ij}(m_i) = 0$. ♦

Exercise B

Describe the direct limit of sets $(S_i)$ over J. State and prove an analogue of proposition 2.

## Commutative Algebra 50

Adjoint functors are a general construct often used for describing universal properties (among other things).

Take two categories $\mathcal C$ and $\mathcal D$.

Definition.

Covariant functors $F:\mathcal D\to \mathcal C$ and $G: \mathcal C \to \mathcal D$ are said to be adjoint if we have isomorphisms

$A \in \mathcal C, B \in \mathcal D \implies \mathrm{hom}_{\mathcal C}(F(B), A) \cong \mathrm{hom}_{\mathcal D}(B, G(A))$

which are natural in A and B, when we regard both sides as functors $\mathcal D^{\text{op}} \times \mathcal C \to \mathbf{Set}$.

We also say F is left adjoint to G and G is right adjoint to F.

Unwinding the definition, for any $B\in \mathcal D$ and $A\in \mathcal C$, we have a bijection

$T(B, A) : \mathrm{hom}_{\mathcal C}(F(B), A) \stackrel\cong\longrightarrow \mathrm{hom}_{\mathcal D}(B, G(A))$

such that the following diagram commutes for any $g:B' \to B$ and $f:A\to A'$.

Note

Suppose F and G are adjoint. For any $B\in \mathcal D$, the covariant functor

$\mathcal C \longrightarrow \mathbf{Set}, \quad A \mapsto \mathrm{hom}_{\mathcal D}(B, G(A))$

is representable since it is isomorphic to $\mathrm{hom}_{\mathcal C}(F(B), -)$. Similarly for each $A\in\mathcal C$, the contravariant functor $\mathcal D \to \mathbf{Set}$ taking $B\mapsto \mathrm{hom}_{\mathcal C}(-, G(A))$ is representable.

There are various equivalent ways of defining adjoint functors but we will not delve into those here. Instead, let us contend ourselves with some examples before restricting to the case of modules.

# Examples Galore

“Adjoint functors arise everywhere.” – Saunders Mac Lane, Categories for the Working Mathematician.

### Example 1

Let $F: \mathbf{Set} \to \mathbf{Gp}$ take a set S to the free group F(S) on S and $U: \mathbf{Gp} \to \mathbf{Set}$ be the forgetful functor. By the universal property of free groups, we have a natural bijection

$\mathrm{hom}_{\mathbf{Gp}}(F(S), G) \cong \mathrm{hom}_{\mathbf{Set}}(S, U(G))$

for any set S and group G.

### Example 2

Let $F : \mathbf{Set} \to \mathbf{Ring}$ take a set S to the (commutative) ring $\mathbb \mathbb Z[S] := Z[x_s : s\in S]$ and $U :\mathbf{Ring} \to \mathbf{Set}$ be the forgetful functor. Again we get

$\mathrm{hom}_{\mathbf{Ring}}(\mathbb Z[S], R) \cong \mathrm{hom}_{\mathbf{Set}}(S, U(R))$

for any set S and ring R.

In general, free objects are left adjoint to forgetful functors.

### Example 3

Let B be an A-algebra. Recall that for each A-module M, we have an induced B-module $M^B = B\otimes_A M$; this gives a functor $F: A\text{-}\mathbf{Mod} \to B\text{-}\mathbf{Mod}$. On the other hand, we have the forgetful functor $U : B\text{-}\mathbf{Mod} \to A\text{-}\mathbf{Mod}$ from the canonical ring homomorphism $A\to B$. By the universal property of induced B-modules

$\mathrm{hom}_{B\text{-}\mathbf{Mod}}(M^B, N) \cong \mathrm{hom}_{A\text{-}\mathbf{Mod}}(M, U(N)).$

Thus the induced module construction is also left adjoint to a forgetful functor.

### Example 4

Let $\mathcal C = A\text{-}\mathbf{Mod}$ and $N \in \mathcal C$. Take functors $F_N, G_N : \mathcal C \to \mathcal C$ where $F_N(M) = M\otimes_A N$ and $G_N(P) = \mathrm{Hom}_A(N, P)$. In proposition 2 here we obtained a natural isomorphism $\mathrm{Hom}_A(M\otimes N, P) \cong \mathrm{Hom}_A(M, \mathrm{Hom}_A(N, P))$. This translates to a natural isomorphism

$\mathrm{hom}_{\mathcal C}(F_N(M), P) \cong \mathrm{hom}_{\mathcal C}(M, G_N(P)).$

In short $- \otimes_A N$ is left adjoint to $\mathrm{Hom}_A(N, -)$.

### Example 5

Let $\mathcal D = \mathcal C \times \mathcal C$. Suppose the coproduct of any two objects in $\mathcal C$ exist. Let $\Sigma:\mathcal D \to \mathcal C$ take $(A, A')\mapsto A \amalg A'$ and $\Delta : \mathcal C \to \mathcal D$ take $A\mapsto (A, A)$. By definition of coproduct in categories, we have

\begin{aligned} \mathrm{hom}_{\mathcal D}((A, A'), \Delta(B)) &= \mathrm{hom}_{\mathcal C}(A, B) \times \mathrm{hom}_{\mathcal C}(A', B)\\ &\cong \mathrm{hom}_{\mathcal C}(A \amalg A', B)\\ &= \mathrm{hom}_{\mathcal C}(\Sigma(A, A'), B)\end{aligned}

for any $(A, A') \in \mathcal D$ and $B\in \mathcal C$. Hence the coproduct functor is left adjoint to the diagonal functor. Dually, the product is right adjoint to the diagonal functor.

### Example 6

More generally, let J be an index category and $\mathcal D = \mathcal C^J$ be the category of all diagrams in $\mathcal C$ of type J. Assuming all colimits of type J exist in $\mathcal C$, let $\mathrm{colim}_J : \mathcal D \to \mathcal C$ be this colimit functor. On the other hand, take the diagonal embedding $\mathcal C \to \mathcal D$ which takes an object A to the diagram where all objects are A and all morphisms are $1_A$. Then

$\mathrm{hom}_{\mathcal D}(D, \Delta(B)) \cong \mathrm{hom}_{\mathcal C}(\mathrm{colim} D, B)$

for any diagram $D\in \mathcal D$ and object $B \in \mathcal C$.

Exercise A

Let $U : \mathbf{Top} \to \mathbf{Set}$ be the forgetful functor. Find a left adjoint and a right adjoint to U (these are two different functors of course).

Let $U : \mathbf{AbGp} \to \mathbf{Gp}$ be the inclusion functor, where $\mathbf{AbGp}$ is the category of abelian groups. Find a left adjoint functor to U.

# Properties

Suppose $F:\mathcal D \to \mathcal C$ is left adjoint to $G : \mathcal C \to \mathcal D$. If coproducts exist in both categories, then $F(B) \amalg F(B') \cong F(B\amalg B')$ because we have natural isomorphisms

\begin{aligned} A \in\mathcal C \implies \mathrm{hom}_{\mathcal C}(F(B\amalg B'), A) &\cong \mathrm{hom}_{\mathcal D}(B \amalg B', G(A)) \\ &\cong \mathrm{hom} _{\mathcal D}(B, G(A)) \times \mathrm{hom}_{\mathcal D}(B', G(A)) \\ & \cong \mathrm{hom}_{\mathcal C}(F(B), A) \times \mathrm{hom}_{\mathcal C}(F(B'), A) \\ &\cong \mathrm{hom}_{\mathcal D}(F(B) \amalg F(B'), A).\end{aligned}

More generally, we have:

Proposition 1.

Let J be an index category; assume that colimits of diagrams of type J all exist $\mathcal C$ and $\mathcal D$. Then for any diagram $(B_i)_{i\in J}$ in $\mathcal D$ of type J,

$\mathrm{colim }_{i\in J} F(B_i) \cong F(\mathrm{colim}_{i\in J} B_i)$.

Proof

By the universal property of colimits and limits, we have, for any diagram $(A_i)_{i\in J}$ in $\mathcal C$ of type J, and any object $A\in \mathcal C$,

$\mathrm{hom}_{\mathcal C}(\mathrm{colim}_{i\in J} A_i, A) \cong \mathrm{lim}_{i\in J^{\text{op}}} (\mathrm{hom}_{\mathcal C}(A_i, A)).$

Now we apply this universal property twice to obtain:

\begin{aligned}A \in \mathcal C \implies \mathrm{hom}_{\mathcal C}(F(\mathrm{colim}_{i \in J} B_i), A) &\cong \mathrm{hom}_{\mathcal D}(\mathrm{colim}_{i \in J} B_i, G(A))\\ &\cong \mathrm{lim}_{i \in J^{\text{op}}} (\mathrm{hom}_{\mathcal D}(B_i, G(A))) \\ &\cong \mathrm{lim}_{i\in J^{\text{op}}} ( \mathrm{hom}_{\mathcal C} (F(B_i), A))\\ &\cong \mathrm{hom}_{\mathcal C} (\mathrm{colim}_{i\in J} F(B_i), A)\end{aligned}

and we are done. ♦

Exercise B

1. State and prove the dual of the above properties.

2. Prove that the forgetful functor $U:\mathbf{Gp} \to \mathbf{Set}$ has no right adjoint.

3. Prove that the inclusion functor $U:\mathbf{AbGp} \to \mathbf{Gp}$ has no right adjoint.

4. Prove that for any diagram of A-modules $(M_i)_{i \in J}$ and A-module N, we have

$\mathrm{colim}_{i \in J} (M_i \otimes_A N) \cong (\mathrm{colim}_{i\in J} M_i) \otimes_A N.$

As a special case, this implies tensor product is distributive over direct sums (already proved in proposition 1 here).

$F : B\text{-}\mathbf{Mod} \to A\text{-}\mathbf{Mod}, \quad G : A \text{-}\mathbf{Mod} \to B\text{-}\mathbf{Mod}$

such that F is left adjoint to G. Further, we assume that the natural $\mathrm{Hom}_B(F(N), M) \cong \mathrm{Hom}_A(N, G(M))$, for A-module M and B-module N, is an isomorphism of additive groups

Note

In fact, one can show that if F and G are adjoint functors as above, then they must be additive and $\mathrm{Hom}_B(F(N), M) \cong \mathrm{Hom}_A(N, G(M))$ must preserve the additive structure.

Proposition 2.

The functors F and G are right-exact and left-exact respectively.

In summary, left adjoint functors are right-exact and vice versa.

Proof

Suppose $M_1 \to M_2 \to M_3 \to 0$ is an exact sequence of A-modules. To prove $F(M_1) \to F(M_2) \to F(M_3) \to 0$ is exact, by proposition 4 here it suffices to show:

$0 \longrightarrow \mathrm{Hom}_B(F(M_3), N) \longrightarrow \mathrm{Hom}_B(F(M_2), N) \longrightarrow \mathrm{Hom}_B(F(M_1), N)$

is exact for every B-module N. But by adjointness, the above is naturally isomorphic to

$0 \longrightarrow \mathrm{Hom}_A(M_3, G(N)) \longrightarrow \mathrm{Hom}_B(M_2, G(N)) \longrightarrow \mathrm{Hom}_B(M_1, G(N))$

which is exact since $\mathrm{Hom}_A(-, G(N))$ is a left-exact functor by proposition 3 here. The case for G is similar. ♦

### Examples

1. By example 4, we see that for any A-module M, the functor $M\otimes -$ is right-exact.

2. By example 5, for any A-algebra B, taking the induced module $M\mapsto M^B$ is right-exact.

# Morphism of Diagrams

Throughout this article $\mathcal C$ denotes a category and J is an index category.

Definition

Given diagrams $D, D' : J\to \mathcal C$, a morphism $D \to D'$ is a natural transformation $T : D\Rightarrow D'$.

Thus we have the category of all diagrams in $\mathcal C$ of type J, which we will denote by $\mathcal C^J$.

For example if we write D and D’ as tuples:

$\left((A_i)_{i\in J}, (\beta_{e}: A_i \to A_j)_{(e : i\to j)}\right), \quad \left((A'_i)_{i\in J}, (\beta'_{e}: A'_i \to A'_j)_{(e : i\to j)}\right)$

a morphism $D\to D'$ is a collection of morphisms $\gamma_i : A_i \to A_i'$ such that

$(e:i\to j) \implies \beta'_e \circ \gamma_i = \gamma_j \circ \beta_e : A_i \to A_j'.$

In diagram form, we have the following, where all “rectangles” commute.

Exercise A

Suppose any diagram of type J in $\mathcal C$ has a colimit. Prove that we get a functor $\mathrm{colim} : \mathcal C^J \longrightarrow \mathcal C$ which takes a diagram in $\mathcal C$ to its colimit. In other words show that a morphism of two diagrams of same type in $\mathcal C$ induces a morphism of their colimits.

# Category of Modules

Proposition 1.

Colimits always exist in the category of A-modules.

Proof

Suppose $\left((M_i)_{i\in J}, (\beta_{e}: M_i \to M_j)_{(e : i\to j)}\right)$ is a diagram of type J. Let $P = \oplus_{i\in J} M_i$ with canonical embeddings $\nu_i : M_i \to P$. Let $Q\subseteq P$ be the submodule generated by all elements of the form $\nu_i(m_i) - \nu_j(\beta_e(m_i))$ over all $m_i \in M_i$ and $e : i\to j$ in J. We claim that $P/Q$ satisfies our desired universal properties. Define

$\epsilon_i : M_i \longrightarrow P/Q, \qquad M_i \stackrel {\nu_i}\longrightarrow \oplus_{i\in J} M_i = P \stackrel \pi \longrightarrow P/Q.$

By definition $\epsilon_j \circ \beta_e = \pi\circ \nu_j \circ \beta_e = \pi\circ \nu_i = \epsilon_i$ for all $e:i \to j$.

Now suppose we have a module N with linear maps $(\alpha_i : M_i \to N)_{i\in J}$ such that for any $e:i\to j$ we have $\alpha_j \circ \beta_e = \alpha_i$. The collection of $\alpha_i$ induce, by definition of direct sum, a unique map $g : \oplus_{i\in J} M_i \to N$ such that $g\circ\nu_i = \alpha_i$ for each $i\in J$. Hence

$e:i\to j \implies g\circ (\nu_j \circ \beta_e - \nu_i) = g\circ \nu_j \circ \beta_e - g\circ \nu_i = \alpha_j \circ \beta_e - \alpha_i = 0$

so $g$ factors through $f:P/Q \to N$ such that $f\circ \pi = g$. Thus for each $i\in J$ we have $f\circ \epsilon_i = f\circ \pi\circ \nu_i = g\circ \nu_i = \alpha_i$. ♦

Exercise B

Prove that colimits exist in the categories $\mathbf{Set}$, $\mathbf{Top}$ and $\mathbf{Gp}$.

# Another Functoriality

Definition.

Suppose $F: J_0 \to J$ is a morphism of index categories. Composition then gives:

$(D : J \to \mathcal C) \mapsto (D\circ F : J_0 \to \mathcal C)$.

Thus a diagram of type J gives us a diagram of type $J_0$. If $J_0$ is a subcategory of J, this is just the restriction of D to $J_0$, denoted by $D|_{J_0}$.

In fact we get a functor $F: \mathcal C^{J} \to \mathcal C^{J_0}$. Indeed, a morphism between diagrams $D_1, D_2 : J \to\mathcal C$ is a natural transformation $T : D_1\Rightarrow D_2$. We let F take this T to

$F(T) = 1_F * T : (D_1 \circ F) \Rightarrow (D_2\circ F)$,

where * is a form of “horizontal composition” of natural transformation (see the optional exercise here).

Although the abstract definition looks harrowing, the underlying concept is quite easy when $J_0$ is a subcategory of J, so it helps to keep this special case in mind. We denote the diagrams $D_1, D_2$ by the following tuples

$((A_i)_{i\in J}, (\beta_e : A_i \to A_j)_{e:i\to j}), \quad ((A'_i)_{i\in J}, (\beta'_e : A'_i \to A'_j)_{e:i\to j})$

so that a morphism $D_1 \to D_2$ is of a collection of morphisms $\gamma_i : A_i \to A_i'$ in $\mathcal C$ such that for any $e:i\to j$ we have $\beta'_e\circ \gamma_i = \gamma_j \circ \beta_e$. Now the new diagrams $D_1|_{J_0}$ and $D_2|_{J_0}$ are the same tuples but with $i\in J_0$ and $e:i\to j$ running through morphisms in $J_0$. Hence, $F(T)$ is given by the same collection of $\gamma_i$, except now i runs through $i \in J_0$.

Proposition 2.

Let $F : J_0 \to J$ be a morphism of index categories. Then for any diagram in $\mathcal C$ of type J, denoted by the pair $(A_i)$ and $(\beta_e)_{e:i\to j}$, we have an induced

$f: \mathrm{colim}_{i_0 \in J_0} A_{F(i_0)} \longrightarrow \mathrm{colim}_{i\in J} A_i$

assuming both colimits exist.

Proof

By definition $A := \mathrm{colim}_{i\in J} A_i$ comes with a collection of morphisms $(\epsilon_i : A_i \to A)_{i\in J}$ such that $\epsilon_j \circ \beta_e = \epsilon_i$ for all $e:i\to j$ in J.

Similarly $A_0 := \mathrm{colim}_{i_0 \in J_0} A_{F(i_0)}$ comes with a collection of morphisms $(\epsilon_{i_0} : A_{F(i_0)} \to A_0)_{i_0 \in J_0}$ such that $\epsilon_{j_0} \circ \beta_{F(e_0)} = \epsilon_{i_0}$ for all $e_0 : i_0 \to j_0$ in $J_0$.

From restricting the first colimit, we get a collection $(\epsilon_{F(i_0)} : A_{F(i_0)} \to A)_{i_0 \in J_0}$ such that $\epsilon_{F(j_0)} \circ \beta_{F(e_0)} = \epsilon_{F(i_0)}$ for all $e_0 : i_0 \to j_0$ in $J_0$.

By universal property of the colimit $A_0$, this induces a unique morphism $f:A_0 \to A$ such that $f\circ \epsilon_{i_0} = \epsilon_{F(i_0)}$ . ♦

Example

By restricting the following diagram

we obtain a morphism $B \coprod B' \to B \coprod_A B'$, assuming both objects exist. More generally we have $\coprod_{i \in J} A_i \to \mathrm{colim}_{i\in J} A_i$.

In fact, the proof of proposition 1 gives us a clue on how to construct general colimits. First take the coproduct, which corresponds to colimit over a diagram of vertices. Next we “add the arrow relations” by taking the coequalizer for each arrow.

# Limits

Limits are the dual of colimits.

Definition.

Take a diagram in $\mathcal C$ of type J, written as

$((A_i)_{i\in I}, (\beta_e : A_i \to A_j)_{(e:i\to j)}$

The limit of the diagram comprises of the following data:

$(A, (\pi_i : A \to A_i)_{i\in J})$

where $A = \lim_{i \in J} A_i \in \mathcal C$ is an object, $\pi_i : A \to A_i$ is a morphism in $\mathcal C$ for each $i\in J$, such that for any arrow $e:i\to j$, we have $\beta_e\circ\pi_i = \pi_j$.

We require the following universal property. For any tuple

$(B, (\alpha_i : B \to A_i)_{i\in J})$

where $B\in \mathcal C$ is an object, $\alpha_i$ is a morphism for each $i\in J$, such that for any arrow $e:i\to j$, we have $\beta_e \circ \alpha_i = \alpha_j$, there is a unique morphism $f : B \to A$ such that

$\pi_i\circ f = \alpha_i \text{ for each } i\in J.$

As before, we have the following special cases.

### Example 1: Products

If J is obtained from an index set I, the limit is the product $\prod_{i\in I} A_i$.

### Example 2: Pullbacks (Fiber Products)

If J is the following, the resulting limit is the pullback.

### Example 3: Equalizers

Definition.

The equalizer of $\beta_1, \beta_2 : A\to B$ in a category $\mathcal C$ is the limit of the following diagram.

This is a pair $(C, \pi : C\to A)$ such that $\beta_1\circ \pi = \beta_2\circ \pi$ and, for any pair $(D, \alpha : D\to A)$ such that $\beta_1 \circ \alpha = \beta_2 \circ \alpha$, there is a unique $f:D\to C$ such that $\pi\circ f = \alpha$.

Exercise C

Prove that limits always exist in the category of A-modules.

# Initial and Terminal Objects

Definition.

An object $A\in \mathcal C$ is said to be initial (resp. terminal) if for any object $B\in \mathcal C$, there is a unique morphism $A\to B$ (resp. $B\to A$).

Note

• If A is initial or terminal, there is a unique morphism $A\to A$, i.e. the identity.
• A is initial if and only if it is a (colimit / limit) of the empty diagram. [ Exercise: pick the right option and write the dual statement. ]

Exercise D

Prove that if A and A’ are initial, there is a unique isomorphism $A\to A'$. Dually, the same holds for terminal objects. In summary, initial (resp. terminal) objects are unique up to unique isomorphism.

The phrase “unique up to unique isomorphism” has been used multiple times while looking at universal properties. This is not a coincidence, for initial and terminal objects can be used to describe universal properties of various constructions. Here is an example.

Lemma 1.

Let M, N be A-modules. Consider the category $\mathcal C(M, N)$, whose objects are pairs

$(P, B : M\times N \to P)$

where P is an A-module, B is an A-bilinear map. The morphisms

$(B : M\times N \to P) \longrightarrow (B' : M\times N \to P')$

are A-linear maps $f:P\to P'$ such that $f\circ B = B'$. Then $(M\otimes_A N, (m, n)\mapsto m\otimes n)$ is an (initial / terminal) object in $\mathcal C$. [ Exercise: pick the right option. ]

Proof

Follows directly from the definition. ♦

# Introduction

For the next few articles we are back to discussing category theory to develop even more concepts. First we will look at limits and colimits, which greatly generalize the concept of products and coproducts and cover loads of interesting cases.

As a starting example, recall that for A-algebras B and C, we have $B\otimes_A C$ which is the coproduct of B and C in the category of A-algebras. But the category of A-algebras corresponds to the coslice category $A\downarrow \mathcal C$ whose objects are morphisms $A\to B$ (as B runs through objects of $\mathcal C$), and morphisms are just morphisms $B\to B'$ in $\mathcal C$ making the diagram commute. If we unwind the definition, coproduct in the coslice category means the following.

Definition.

Let $\beta: A\to B$, $\beta':A \to B'$ be morphisms in the category $\mathcal C$. The pushout of $\beta$ and $\beta'$ is a triplet:

$(C, \epsilon : B \to C, \epsilon' : B'\to C)$,

where $C := B \coprod_A B'\in \mathcal C$ is an object, $\epsilon$, $\epsilon'$ are morphisms in $\mathcal C$ satisfying $\epsilon\circ \beta = \epsilon' \circ \beta'$ such that for any triplet:

$(D, \alpha : B \to D, \alpha' : B'\to D)$

of object $D\in \mathcal C$ and morphisms $\alpha$, $\alpha'$ satisfying $\alpha \circ \beta = \alpha' \circ \beta'$, there is a unique morphism $f : C\to D$ such that $f\circ \epsilon = \alpha$ and $f\circ \epsilon' = \alpha'$.

Pictorially, the pushout gives a correspondence as follows.

The idea is that $B\coprod_A B'$ classifies all morphisms “from” the diagram in red.

The pushout may not exist; if it does, it is unique up to unique isomorphism.

## Examples

1. We already saw that in the category of rings, the pushout of $A\to B$ and $A\to B'$ is $B\otimes_A B'$.

2. Take $\beta : S\to T$ and $\beta' : S\to T$ in the category $\mathbf{Set}$. Then:

$T \coprod_S T' = \{T \coprod T' \} / (\beta(s) \sim \beta'(s), s\in S)$

where $T\coprod T'$ is the disjoint union of T and T’.

3. Let $\beta : H\to G$ and $\beta' : H\to G'$ be injective group homomorphisms in $\mathbf{Gp}$. The pushout is called the amalgamation of G and G’ over H and can be concretely described in terms of words in H, coset representatives of G/H and those of G’/H. A description is given in Trees by Jean-Pierre Serre.

Exercise A

1. Prove example 2.

2. Find the pushforward for:

• $\alpha : X\to Y$ and $\alpha' :X \to Y'$ in the category $\mathbf{Top}$ of topological spaces;
• $\alpha : M\to N$ and $\alpha' : M\to N'$ in the category $A\text{-}\mathbf{Mod}$.

# Pullbacks (Fibre Products)

Dually, we can define the following.

Definition.

Let $\beta :B\to A$, $\beta' :B' \to A$ be morphisms in $\mathcal C$. The pullback (or fibre product) of $\beta$ and $\beta'$ corresponds to the pushforward of $\beta^{\text{op}}$ and $\beta'^{\text{op}}$ in the opposite category $\mathcal C^{\text{op}}$.

The underlying object of the pullback is denoted by $B\times_A B'$.

Easy exercise

Write out the details of the definition (see below diagram).

## Examples

1. Let $\gamma : T \to S$ and $\gamma' : T' \to S$ be functions in $\mathbf{Set}$. The pullback is given by

$T\times_S T' = \{(t, t') \in T\times T' : \gamma(t) = \gamma'(t') \in S\}$

with projection maps $T\times_S T' \to T$, $T\times_S T' \to T'$ taking $(t,t')$ to $t, t'$ respectively.

2. As a special case suppose $T\subseteq S$ and $\gamma : T\to S$ is the inclusion map. Then $T\times_S T' = \gamma'^{-1}(T)$, i.e. the fibre space of $\gamma' : T' \to S$ over the subset T.

3. As a special case of special case, suppose $T, T'\subseteq S$ and $\gamma, \gamma'$ are both inclusions. Then $T\times_S T' = T\cap T'$.

Exercise B

Do the same constructions work for $\mathbf{Gp}$, $\mathbf{Ring}$, $\mathbf{Top}$, $A\text{-}\mathbf{Mod}$, $A\text{-}\mathbf{Alg}$?

Explain how the three constructions at the end of Chapter 30 are all fibre products in the category of k-schemes (or pushouts, if we take the dual category of finitely generated k-algebras).

# Colimits

Now we will generalize the above constructions, by taking limits and colimits over a diagram in a category.

Definition.

An index category J is a category such that its class of objects is a set.

If you are concerned about the difference between classes and sets, please read the note at the end.

In summary, an index category consists of a set of vertices and arrows between them such that we can compose arrows. For convenience, we will employ this terminology for an index category: objects are called vertices and edges are called arrows.

Now an index set I may be regarded as an index category J, where we pick a vertex for each $i\in I$ and the only morphisms are the identities. E.g. the index set {1,2,3} has 3 vertices and one identity arrow for each object.

Definition.

Let $J$ be an index category; a diagram of type $J$ in the category $\mathcal C$ is a covariant functor $D: J \to \mathcal C$.

Thus to each vertex $i \in J$ we assign an object $A_i \in \mathcal C$ and to each arrow $e : i \to j$ in J, we assign a morphism $\beta_{e} : A_i \to A_j$.

[ Note: when representing a diagram in pictorial form, we often remove the arrows which are implied, e.g. the arrow corresponding to $\beta_{35}\circ \beta_{13}$ is not drawn. In particular, identity maps are not drawn. ]

In particular, if J is obtained from an index set I, a diagram of type J is just a collection of objects $(A_i)$ indexed by $i\in I$ (with the identity arrows mapped to the identity morphisms).

Definition.

Let $D : J\to \mathcal C$ be a diagram, written as

$\left((A_i)_{i\in J}, (\beta_{e}: A_i \to A_j)_{(e : i\to j)}\right)$.

The colimit of this diagram comprises of the tuple

$(A, (\epsilon_i : A_i\to A)_{i\in J})$

where $A = \mathrm{colim }_{i \in J} A_i \in \mathcal C$ is an object, $\epsilon_i$ is a morphism for each $i\in J$, such that for any arrow $e: i\to j$, we have $\epsilon_j \circ \beta_e = \epsilon_i$ as morphisms $A \to A_j$.

We require the following universal property to hold: for any tuple

$(B, (\alpha_i : A_i \to B)_{i\in I})$

where $B\in \mathcal C$ is an object, $\alpha_i$ is a morphism for each $i\in J$, such that for any arrow $e:i\to j$ in $J$ we have $\alpha_j \circ\beta_e = \alpha_i$, there is a unique morphism $f:A\to B$ in $\mathcal C$ such that

$f\circ \epsilon_i = \alpha_i \text{ for each } i\in J$.

Clearly, the colimit is unique up to unique isomorphism if it exists.

### Example 1: Coproducts

If J is obtained from an index set I, the resulting colimit is the coproduct $\coprod_{i\in I} A_i$.

### Example 2: Pushout

If J is the following index category, the resulting colimit is the pushout.

### Example 3: Coequalizers

Definition.

The coequalizer of $\beta_1, \beta_2 :A\to B$ in a category $\mathcal C$ is the colimit of the following diagram.

This is a pair $(C, \epsilon : B\to C)$ where $C\in \mathcal C$ is an object such that $\epsilon \circ \beta_1 = \epsilon\circ \beta_2$ and, for any pair $(D, \alpha : B\to D)$ satisfying $\alpha\circ \beta_1 = \alpha\circ \beta_2$, there exists a unique $f : C \to D$ such that $f\circ \epsilon = \alpha$.

Exercise C

1. Describe the colimits for each of the following two diagrams.

2. Compute the coequalizer in the categories $\mathbf{Set}$, $\mathbf{Top}$ and $\mathbf{Gp}$.

3. Let $S\subseteq A$ be a multiplicative subset and M an A-module. Prove that we have an isomorphism of A-modules:

$\mathrm{colim}_{f\in S} M_f \cong M_S$,

with morphisms $M_f \to M_g$ defined as follows. If g = af we take the canonical map $M_f \to M_g$ via $\frac m f \mapsto \frac{am}g$. Otherwisee there is no map between $M_f$ and $M_g$. Prove also that we get a ring isomorphism $\mathrm{colim}_{f\in S} A_f \cong A_S$.

In particular for any prime $\mathfrak p \subset A$ we have

$\mathrm{colim}_{f\in A-\mathfrak p} M_f \cong M_{\mathfrak p}, \quad \mathrm{colim}_{f\in A-\mathfrak p} A_f \stackrel{\text{rings}}\cong A_S$.

## Note

For readers who are unfamiliar with the set-theoretic notion of class and set, try not to fret too much over this. Roughly, a set is a class which is “not too huge”. If we allow arbitrary collections to be sets, then Russell’s paradox occurs (where one can take the set of all sets not containing themselves). A common way out of this paradox is to restrict the type of collections which can be considered as sets.

As a very rough guide, in Zermelo-Fraenkel set theory, a collection is a set if it can be “built from the set of natural numbers ℕ”. Thus the collection of real numbers ℝ is a set because its cardinality is the same as the power set of ℕ. Similarly, the set of open subsets of ℝ forms a set because it is a subset of the power set of ℝ, so we can do topology on ℝ. On the other hand, the collection of all groups does not form a set.

# Minkowski Theory: Introduction

Suppose $K/\mathbb Q$ is a finite extension and $\mathcal O_K$ is the integral closure of $\mathbb Z$ in K.

In algebraic number theory, there is a classical method by Minkowski to compute the Picard group of $\mathcal O_K$ (note: in texts on algebraic number theory, this is often called the divisor class group; there is a slight difference between the two but for Dedekind domains they are identical).

We will only consider the simplest cases here to give readers a sample of the theory.

Minkowski’s Lemma.

If a measurable region $X\subseteq \mathbb R^2$ has area > 1, then there exist distinct $x, y\in X$ such that $x-y\in \mathbb Z^2$.

Proof

Use the following picture:

Since the area > 1, there exist two points which overlap in the unit square on the right. The corresponding $x,y\in X$ then give $x-y\in \mathbb Z^2$. ♦

Minkowski’s Theorem.

Let $X\subseteq \mathbb R^2$ be a measurable region which is convex, symmetric about the origin and has area > 4. Then X has a lattice point other than the origin.

Proof

Take $\{\frac 1 2 x : x\in X\}$, which has area > 1. By Minkowski’s lemma, there exist distinct $x, y\in X$ such that $\frac 1 2 x - \frac 1 2 y \in \mathbb Z^2$. Since X is symmetric about the origin replace y by –y (so $x\ne -y$) to give $\frac 1 2 x + \frac 1 2 y \in \mathbb Z^2$. And since X is convex, $x,y\in X \implies \frac 1 2 x + \frac 1 2 y \in X$. Finally since $x \ne -y$, $\frac 1 2 x + \frac 1 2 y$ is not the origin. ♦

By applying a linear transform to $\mathbb R^2$ we obtain the more useful version of Minkowski’s theorem.

Minkowski’s Theorem B.

Take a full lattice $\mathbb Z^2 \cong L \subset \mathbb R^2$ (i.e. discrete subgroup which spans $\mathbb R^2$). Taking a basis $(v_1, v_2)$ of L, we obtain a fundamental domain

$\{ \alpha_1 v_1 + \alpha_2 v_2 : 0 \le \alpha_1, \alpha_2 < 1\}$

of area D. If $X\subseteq \mathbb R^2$ is convex, symmetric about the origin, and has area > 4D, then X has a non-zero point of L.

# Picard Group of Number Rings

Now we use this to compute $\mathrm{Pic} A$ for $A =\mathcal O_{\mathbb Q(\sqrt{-5})} = \mathbb Z[\sqrt{-5}]$.

First note that any non-zero ideal $\mathfrak a \subseteq A$ has finite index since if $x\in \mathfrak a - \{0\}$, then $N(x) \in \mathfrak a$ is a non-zero integer in $\mathfrak a$ so $N(x)A \subseteq \mathfrak a$, where N is the norm function. We let

$N(\mathfrak a) := [A : \mathfrak a]$.

Recall that if $\mathfrak a = \mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n}$, by proposition 4 here, the composition factors of $A/\mathfrak a$ comprise of exactly $k_i$ copies of $A/\mathfrak m_i$ for each i, so

$N(\mathfrak a) = N(\mathfrak m_1)^{k_1} \ldots N(\mathfrak m_n)^{k_n}.$

In particular, $N(\mathfrak {ab}) = N(\mathfrak a)N(\mathfrak b)$ for any non-zero ideals $\mathfrak a, \mathfrak b \subseteq A$, and we can extend the norm function to the set of all fractional ideals of A.

Exercise

Prove that if $\alpha = x + y\sqrt{-5} \in A - \{0\}$ then

$N(\alpha A) = x^2 + 5y^2 = N(\alpha)$

so we can consider $N(\mathfrak a)$ as an extension of the norm function to the set of ideals.

To apply Minkowski’s theorem B, we identify $x+y\sqrt{-5} \in A$ with $(x,y) \in \mathbb R^2$. Let $\mathfrak a\subseteq A$ be any non-zero ideal with norm N, considered as a full lattice in $\mathbb R^2$. We take

$S = \{(x,y) \in \mathbb R^2 : x^2 + 5y^2 \le t\}$

where t will be decided later. Note that S is convex, symmetric about the origin and has area $\frac{\pi t}{\sqrt 5}$. If $t = \frac {4\sqrt{5}}\pi N + \epsilon$ where $\epsilon > 0$ then Minkowski’s theorem B assures us there exists $a\in \mathfrak a - \{0\}$ with $N(a) \le t$. Since this holds for all $\epsilon > 0$ we have:

$N(a) \le \frac {4\sqrt 5}\pi N(\mathfrak a), \quad a\in \mathfrak a - \{0\}.$

Now $\mathfrak b := a\mathfrak a^{-1} \subseteq A$ has norm $\le {4\sqrt 5}\pi$, i.e. $\le 2$. Hence every element of Pic A can be represented by an ideal $\mathfrak a$ of norm 1 or 2. Since A has norm 1 and $\mathfrak m = (2, 1 + \sqrt{-5})$ has norm 2, we have proven:

$\mathrm{Pic} (\mathbb Z[\sqrt{-5}]) = \mathbb Z / 2\mathbb Z$.

More generally, one can show the following.

Theorem.

Let $K/\mathbb Q$ be a finite extension. Then the Picard group of $\mathcal O_K$ is finite; its cardinality is called the class number of K.

Exercise

Prove that $K = \mathbb Q(\sqrt{-23})$ has class number 3. Note that $\mathcal O_K = \mathbb Z[ \frac{1 + \sqrt{-23}}2]$.

Prove that $K = \mathbb Q(\sqrt{-163})$ has class number 1.

Prove that $K = \mathbb Q(\sqrt{10})$ has class number 2.

[ Hint: identify $a + b\sqrt{10} \in \mathbb Z[\sqrt{10}]$ with $(a + b\sqrt{10}, a - b\sqrt{10}) \in \mathbb R^2$. Pick the square $\{ (x, y) \in \mathbb R^2 : |x| + |y| \le t \}$ for a suitable t. You could also pick $\{(x, y) \in \mathbb R^2 : |x|, |y| \le t\}$ but it is not as efficient. ]

# Geometric Example: Elliptic Curve Group

Take the elliptic curve E over $\mathbb C$ given by $\{(x,y) : y^2 = x^3 - x\}$ and let $A = \mathbb C[E]$ be its coordinate ring. In Exercise B.1 here, we showed A is a normal domain. Clearly it is noetherian. By Noether normalization theorem, $\dim A = 1$. Hence, A is a Dedekind domain.

We will show how computation of Pic A leads to point addition on the elliptic curve. For each maximal ideal $\mathfrak m \subset A$, write $[\mathfrak m]$ for its image in Pic A. Recall that points $P\in E$ correspond bijectively to maximal ideals $\mathfrak m_P \subset A$.

Lemma 1.

Suppose $f\in A$ is not a unit. Then taking $A/(f)$ as a complex vector space, $\dim_{\mathbb C} A/(f) \le 2$ if and only if $f$ can be represented by a linear function in X, in which case

$A/(f) \cong \mathbb C[Y]/(Y^2 - \beta)$

for some $\beta\in \mathbb C$.

Note

The intuition is that the curve $f(X, Y) = 0$ and the elliptic curve cannot have less than 3 intersection points (with multiplicity) unless we take a vertical line.

Proof

Since $Y^2 = X^3 - X$ in the ring A, without loss of generality we can write

$f(X, Y) = Y\cdot g(X) + h(X)$

for $g(X), h(X) \in \mathbb C[X]$. The condition $\dim_{\mathbb C} A/(f) \le 2$ implies $Y^2 - X^3 + X$ and $Y\cdot g(X) + h(X)$ have at most two intersection points. Solving gives us

$g(X)^2 (X^3 - X) = h(X)^2$.

If $g(X) \ne 0$, the LHS has odd degree while the RHS has even degree; hence the equation has at least 3 roots and each corresponds to at least one point on the elliptic curve. If $g(X) = 0$ and $\deg h(X) = m$, then $\dim_{\mathbb C} A/(f) = 2m$ so $h(X)$ is linear in X, and we are done. ♦

Exercise

If we replace $\mathbb C$ by a general algebraically closed field k, would the proof still work? What additional conditions (if any) need to be imposed?

Corollary 1.

No maximal ideal of A is principal.

Proof

If $\mathfrak m \subset A$ is generated by f then $A/(f) \cong \mathbb C$, which is impossible by lemma 1. ♦

Corollary 2.

For any points $P = (\alpha_1, \beta_1)$ and $Q = (\alpha_2, \beta_2)$ on E, $\mathfrak m_P \mathfrak m_Q$ is principal if and only if

$\alpha_1 = \alpha_2, \beta_1 = -\beta_2.$

When that happens, we write $P = -Q$.

Proof

(⇐) If $\alpha_1 = \alpha_2, \beta_1 = -\beta_2$ then setting $f = X - \alpha_1$ gives

$A/(f) \cong \mathbb C[X, Y]/(Y^2 - X^3 + X, X - \alpha_1) \cong \mathbb C[Y]/(Y^2 - (\overbrace{\alpha_1^3 - \alpha_1}^{\beta_1^2})).$

If $\beta_1 \ne 0$, this ring has exactly two maximal ideals, corresponding to maximal ideals $\mathfrak m_P$ and $\mathfrak m_Q$ of A. If $\beta_1 = 0$, it has exactly one maximal ideal so we still have $(f) = \mathfrak m_P^2 = \mathfrak m_P \mathfrak m_Q$.

(⇒) If $\mathfrak m_P \mathfrak m_Q = (f)$ is principal, then $\dim_{\mathbb C} A/(f) = 2$ so by lemma 1, f can be represented by a linear function in X so we must have P and Q as described. ♦

Corollary 3.

If $P,Q\in E$ satisfy $[\mathfrak m_P] = [\mathfrak m_Q]$, then $P=Q$.

Proof

Let $R = -P$ as in corollary 2. Then $[\mathfrak m_P][\mathfrak m_{R}] = 1$. By the given condition $[\mathfrak m_Q][\mathfrak m_R] = 1$ so by corollary 2 again we have $R = -Q$ and hence $P = Q$. ♦

Lemma 2.

For any $P, Q\in E$ with $P\ne -Q$, there is a unique $R\in E$ such that

$[\mathfrak m_P]\cdot [\mathfrak m_Q]\cdot [\mathfrak m_R] = 1.$

Proof

First suppose $P\ne \pm Q$ so P and Q have different x-coordinates. Let $f = Y - cX - d$ be the equation of PQ. We get:

$A/(f) \cong \mathbb C[X,Y]/(Y^2 - X^3 + X, Y - cX - d) \cong \mathbb C[X]/((cX+d)^2 - X^3 + X),$

which has complex dimension 3. Since $(f)$ is divisible by $\mathfrak m_P \mathfrak m_Q$ we have $(f) = \mathfrak m_P \mathfrak m_Q \mathfrak m_R$ for some $R\in E$. Geometrically R is the third point of intersection of PQ with E, which can be equal to P or Q.

If $P = Q = (\alpha, \beta)$ with $\beta \ne 0$, we can similarly pick a line through P of gradient $c = \frac{3\alpha^2 - 1}{2\beta}$. Then as above $A/(f) \cong \mathbb C[X]/((cX+d)^2 - X^3 + X)$ where $(cX + d)^2 - X^3 + X$ has a double root for $X = \alpha$ (this requires some algebraic computation). Hence $(f) = \mathfrak m_P^2 \mathfrak m_R$ for some $R\in E$. ♦

Summary.

The Picard group of A is given by

$\{ [\mathfrak m] : \mathfrak m \subset A \text{ maximal} \} \cup \{1\}$.

In particular it is infinite.

# Properties of Dedekind Domains

Proposition 1.

Every fractional ideal of A can be written as

$\mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n}, \quad k_1, \ldots, k_n \in \mathbb Z - \{0\}$

where each $\mathfrak m_i$ is a maximal ideal. The expression is unique up to permutation of terms.

Note

In particular, the maximal ideals of A generate its Picard group.

Proof

Existence. First suppose $\mathfrak a \subseteq A$ is a non-zero ideal. If $\mathfrak a = (1)$ we are done. Otherwise it is contained in some maximal ideal $\mathfrak m$, necessarily invertible. Now $\mathfrak a \ne \mathfrak a\mathfrak m^{-1}$, because if equality holds multiplying both sides by $\mathfrak {ma}^{-1}$ gives $\mathfrak m= A$, so $\mathfrak a \subsetneq \mathfrak a\mathfrak m^{-1} \subseteq A$. We replace $\mathfrak a$ by $\mathfrak a \mathfrak m^{-1}$ and repeat the process; this cannot continue indefinitely or we would have

$\mathfrak a \subsetneq \mathfrak a \mathfrak m_1^{-1}\subsetneq \mathfrak a \mathfrak m_1^{-1} \mathfrak m_2^{-1} \subsetneq \ldots$

Hence $\mathfrak a$ is a finite product of maximal ideals. Now a general fractional ideal is of the form $\frac 1 x \mathfrak a$ for a non-zero ideal $\mathfrak a \subseteq A$ and $x\in A-\{0\}$. Write both $\mathfrak a$ and $xA$ as a product of maximal ideals and we are done.

Uniqueness. Suppose $\mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n} = \mathfrak m_1^{j_1} \ldots \mathfrak m_n^{j_n}$, where $j_i, k_i \in \mathbb Z$ (some possibly zero). If $(k_1, \ldots, k_n) \ne (j_1, \ldots, j_n)$, after cancelling terms and reordering, we obtain a relation of the form:

$\mathfrak m_1^{k_1} \ldots \mathfrak m_s^{k_s} = \mathfrak m_{s+1}^{k_{s+1}} \ldots \mathfrak m_t^{k_t}, \quad k_i \in \mathbb Z_{>0}.$

This is an equality of ideals. We get a contradiction because the LHS is contained in $\mathfrak m_1$ but the RHS is not. ♦

Exercise A

Prove that an integral domain is a PID if and only if it is a Dedekind domain and a UFD. Hence we have the following Venn diagram.

Now we have the following.

Proposition 2.

Let M, N be non-zero fractional ideals of A. Write, as in proposition 1,

$M = \mathfrak m_1^{d_1} \mathfrak m_2^{d_2} \ldots \mathfrak m_k^{d_k}, \quad N = \mathfrak m_1^{e_1} \mathfrak m_2^{e_2} \ldots \mathfrak m_k^{e_k}$

where each $\mathfrak m_i \subset A$ is maximal and $d_i, e_i \in \mathbb Z$. Then

\begin{aligned} M+N &= \mathfrak m_1^{\min(d_1, e_1)} \ldots \mathfrak m_k^{\min(d_k, e_k)}\\ M\cap N &= \mathfrak m_1^{\max(d_1, e_1)} \ldots \mathfrak m_k^{\max(d_k, e_k)}\\ MN &= \mathfrak m_1^{d_1 + e_1} \ldots \mathfrak m_k^{d_k + e_k},\\ (M:N) &= \mathfrak m_1^{d_1 - e_1} \ldots \mathfrak m_k^{d_k - e_k}\end{aligned}

Proof

Exercise. [ Localize at each $\mathfrak m_i$; reduce to the case where A is a dvr. ]

# Constructing Dedekind Domains

How do we find Dedekind domains? For starters, a PID is a Dedekind domain so we have $\mathbb Z$ and $k[X]$ (for any field k), but these are not too interesting.

Here is a standard way to construct new Dedekind domains from an existing one A. Let K be its field of fractions and L be a finite extension of K; we let B be the integral closure of A in L.

Exercise B

Prove that L is the field of fractions of B.

Now B has the following properties:

• it is a normal domain by construction;
• dim B = 1 since B is an integral extension of A, and dim A = 1 (by main theorem here).

Hence, if we can show that B is noetherian, then it is a Dedekind domain. But this result, called the Krull-Akizuki theorem, is surprisingly hard to prove. In most cases of interest, B will turn out to be a finitely generated A-module; thus B is noetherian as an A-module and hence as a module over itself.

Examples

1. Let $K/\mathbb Q$ be a finite extension and $\mathcal O_K$ be the integral closure of $\mathbb Z$ in K. One can show that $\mathcal O_K$ is a discrete subgroup of $K$, when we topologize $K$ by identifying it with $\mathbb Q^n$. Hence $\mathcal O_K$ is a finitely generated $\mathbb Z$-module; in particular it is a noetherian ring and hence a Dedekind domain. A more general result of this nature will be proven later.

2. Let k be any field such that $\mathrm{char} k \ne 2$. The integral closure of $k[X]$ in $L = k(X)[Y]/(Y^2 - X^3 + X)$ is $A = k[X, Y]/(Y^2 - X^3 +X)$. [ Proof: exercise. ] Since A is clearly noetherian, it is a Dedekind domain.

Exercise C

Let $m\in \mathbb Z - \{\pm 1\}$ be a square-free integer and $K = \mathbb Q(\sqrt m)$, which is quadratic over $\mathbb Q$. Prove that

$\mathcal O_K = \begin{cases} \mathbb Z[\frac{1 + \sqrt{m}}2], \quad &\text{if } m\equiv 1 \pmod 4, \\ \mathbb Z[\sqrt{m}], \quad &\text{if } m\equiv 2,3 \pmod 4.\end{cases}$

# Valuation

Definition.

Let $K = \mathrm{Frac} A$. Each maximal $\mathfrak m \subset A$ gives a function

$\nu_{\mathfrak m} : K - \{0\} \longrightarrow \mathbb Z$,

which takes $x$ to the exponent of $\mathfrak m$ in the above factorization of $xA$. This is called the valuation of $x$ at $\mathfrak m$.

Exercise D

1. Prove the following properties of $\nu = \nu_{\mathfrak m}$: for any $x,y\in K - \{0\}$ we have:

$\nu(1) = 0, \quad \nu(xy) = \nu(x) + \nu(y), \quad \nu(x+y) \ge \min(\nu(x), \nu(y)) \text{ if } x+y\ne 0$.

Verify that if we set $\nu(0) = \infty$ these properties hold for all $x,y\in K$.

2. Define a distance function on K by $d(x, y) = e^{-\nu(x-y)}$. Prove that (Kd) is a metric space, where the metric satisfies the strong triangular inequality.

$x,y, z\in K \implies d(x, z) \le \max(d(x, y), d(y, z)).$

A metric which satisfies this inequality is called an ultrametric. Interpret and prove the following statement: in an ultrametric space, every triangle is isoceles.

## Valuation in Geometry

Suppose $A = \mathbb C[X, Y]/(Y^2 - X^3 + X)$, a Dedekind domain. Recall that points P on the curve $Y^2 = X^3 - X$ correspond bijectively to maximal ideals $\mathfrak m_P \subset A$. Then $\nu_{\mathfrak m_P}(f)$ is just the order of vanishing of at P.

For example, let $\mathfrak m = (X, Y) = \mathfrak m_P$ where $P = (0, 0)$; for convenience write $\nu := \nu_{\mathfrak m}$. Let us compute $\nu(f)$ for $f = X^2 - Y^4 \in A$:

$X^2 - Y^4 = X^2 - (X^3 - X)^2 = 2X^4 - X^6 = X^4 ( 2 - X^2).$

Now $\nu(2 - X^2) = 0$ since f does not vanish on P. From an earlier example, we have $\mathfrak m_P^2 = (X)$ so $\nu(X) = 2$ and $\nu(X^2 - Y^4) = 8$.

As a simple exercise, compute $\nu(Y)$.

# Composition Series

Recall in the last example here that for distinct maximal ideals $\mathfrak m_1, \ldots, \mathfrak m_n$ of any ring A and $k_1, \ldots, k_n \ge 1$ we have a ring isomorphism

$A/(\mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n}) \cong A/\mathfrak m_1^{k_1} \times \ldots \times A/\mathfrak m_n^{k_n}.$

This gives the following.

Proposition 3.

If A is a Dedekind domain with finitely many maximal ideals $\mathfrak m_1, \ldots, \mathfrak m_n$, then it is a PID.

Proof

It suffices to prove that each $\mathfrak m_i$ is principal. Let us consider $\mathfrak m_1$. Since $\mathfrak m_i^2 \subsetneq \mathfrak m_i$ we can pick $y \in \mathfrak m_1 - \mathfrak m_1^2$. From the isomorphism

$A/(\mathfrak m_1 \ldots\mathfrak m_n) \cong A/\mathfrak m_1 \times \ldots \times A/\mathfrak m_n$,

we can find $x\in A$ such that

$x\equiv y \pmod {\mathfrak m_1}, \ x \equiv 1 \pmod {\mathfrak m_2}, \ \ldots, \ x \equiv 1 \pmod {\mathfrak m_n}.$

Then x is divisible by $\mathfrak m_1$ but not its square, and not divisible by $\mathfrak m_2, \ldots, \mathfrak m_n$. Thus $\mathfrak m_1 = (x)$. ♦

Proposition 4.

Let $\mathfrak a \subseteq A$ be a non-zero ideal; write

$\mathfrak a = \mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n}.$

Then the composition factors of $A/\mathfrak a$ as an A-module comprise of $k_i$ copies of $A/\mathfrak m_i$ for $1 \le i \le n$. In particular

$l_A(A/\mathfrak a) = \sum_{i=1}^n k_i$.

Note

Recall that the composition factors of $A/\mathfrak a$ as an A-module are identical to those as a module over itself.

Proof

The above ring isomorphism

$A/(\mathfrak m_1 \ldots\mathfrak m_n) \cong A/\mathfrak m_1 \times \ldots \times A/\mathfrak m_n$

is in fact an isomorphism of A-algebras. Hence $l_A(A/\mathfrak a) = \sum_i l_A(A/\mathfrak m_i^{n_i})$. We obtain a series of submodules of $A/\mathfrak m^n$

$0 \subsetneq \mathfrak m^{n-1}/\mathfrak m^n \subsetneq \mathfrak m^{n-2}/\mathfrak m^n \subsetneq \ldots \subsetneq \mathfrak m/\mathfrak m^n \subsetneq A/\mathfrak m^n$

with successive quotients $\mathfrak m^i/\mathfrak m^{i+1}$. It suffices to show each of these is of dimension 1 over $A/\mathfrak m$. To prove that, we recall that $A_{\mathfrak m}$ is a dvr so $\mathfrak n := \mathfrak m A_{\mathfrak m}$ is a principal ideal. Since $\mathfrak m^i / \mathfrak m^{i+1}$ is already a module over $A_{\mathfrak m}$ we have:

$\dim_{A/\mathfrak m} \mathfrak m^i/\mathfrak m^{i+1} = \dim_{A_{\mathfrak m}/\mathfrak n} \mathfrak n^i / \mathfrak n^{i+1} = 1$

as desired. ♦

Note

Consider the special case where $A = k[V]$ is the coordinate ring of a variety V over an algebraically closed field k. If A is a Dedekind domain, then since $A/\mathfrak m \cong k$ for each maximal ideal $\mathfrak m \subset A$ we have

$l_A(A/\mathfrak a) = \dim_k A/\mathfrak a$.

# Invertibility is Local

In this article, we again let A be an integral domain and K its field of fractions. We continue our discussion of invertible fractional ideals of A.

Proposition 1.

A fractional ideal M of A is invertible if and only if the following hold.

• M is a finitely generated A-module.
• For any maximal ideal $\mathfrak m \subset A$, $M_{\mathfrak m}$ is an invertible fractional ideal of $A_{\mathfrak m}$.

Proof

(⇒) This was proven earlier (propositions 2 and 5).

(⇐) It suffices to show $(A:M) M = A$. By definition $(A:M) M\subseteq A$. Let $\mathfrak m \subset A$ be any maximal ideal. Since M is finitely generated and $M_{\mathfrak m}$ is invertible, we have (by proposition 3 here):

$A_{\mathfrak m} = (A_{\mathfrak m} : M_{\mathfrak m})M_{\mathfrak m} = (A:M)_{\mathfrak m} M_{\mathfrak m} = [(A:M)M]_{\mathfrak m}.$

If the A-module $(A:M) M$ is contained in any maximal $\mathfrak m$ then $[(A:M)M]_{\mathfrak m} \subseteq \mathfrak m A_{\mathfrak m}$, a contradiction. Hence $(A:M)M = A$. ♦

Complementing the above result, we have:

Proposition 2.

Let $(A,\mathfrak m)$ be a local domain. A fractional ideal M of A is invertible if and only if it is principal and non-zero. Thus $\mathrm{Pic} A = 0$.

Proof

Suppose M is invertible; write MNA. Then $1 = x_1 y_1 + \ldots + x_n y_n$ for $x_i \in M$ and $y_i \in N$. Since all $x_i y_i \in A$ and sum to 1, not all $x_i y_i$ lie in $\mathfrak m$. Thus $x_i y_i \in A$ is invertible for some i. Multiplying $x_i$ by a unit we may assume $x_i y_i = 1$.

It remains to show $M = Ax_i$. Clearly since $x_i \in M$, we have $M\supseteq A x_i$. Conversely if $m\in M$ then $m = x_i (my_i) \in Ax_i$ since $MN\subseteq A$. ♦

Thus we have:

Corollary 1.

A non-zero fractional ideal M of A is invertible if and only if:

• it is finitely generated, and
• for each maximal $\mathfrak m \subset A$, $M_{\mathfrak m}$ is a principal fractional ideal of $A_{\mathfrak m}$.

Example

Let $A = \mathbb Z[2\sqrt 2] = \{a + 2b\sqrt 2 : a,b \in \mathbb Z\}$ with maximal ideal $\mathfrak m = \{2a + 2b\sqrt 2: a,b\in \mathbb Z\}$. We claim that $\mathfrak m$ is not invertible; it suffices to show that $\mathfrak m A_{\mathfrak m}$ is not a principal ideal of $A_{\mathfrak m}.$

For that we apply Nakayama’s lemma to compute $\dim_{A/\mathfrak m} \mathfrak m /\mathfrak m^2$. Let $t = 2\sqrt 2$; we can check that $\mathfrak m = 2A + tA$ so $\mathfrak m^2 = 4A + 2tA = 2\mathfrak m$ since $t^2 = 8$. Since $\mathfrak m/\mathfrak m^2$ has 4 elements, we see that $\dim_{A/\mathfrak m} \mathfrak m /\mathfrak m^2 = 2$. Hence $\mathfrak m A_{\mathfrak m}$ is not principal.

Exercise

Let $A = \mathbb C[X, Y]/(Y^2 - X^3)$ with $\mathfrak m = (X, Y)$. Is $\mathfrak m$ an invertible ideal?

# Dedekind Domains

Definition.

Dedekind domain A is an integral domain in which every fractional ideal is invertible (or equivalently, every ideal is invertible).

We obtain some immediate facts about Dedekind domains.

• Since invertible ideals are finitely generated, a Dedekind domain is noetherian.
• If A is a Dedekind domain so is its localization $S^{-1}A$ if $0\not\in S$; this follows from proposition 3 here: every ideal of $S^{-1}A$ is of the form $\mathfrak a(S^{-1}A)$ for some ideal $\mathfrak a \subseteq A$.

Proposition 3.

In a Dedekind domain A, every non-zero prime ideal is maximal. Thus its Krull dimension is at most 1.

Hence Spec A is as follows:

Proof

If A is a field we are done. Otherwise, we will show that every maximal ideal $\mathfrak m\subset A$ has height 1. Since $\mathfrak m$ is invertible, $\mathfrak m A_{\mathfrak m}\subset A_{\mathfrak m}$ is principal by proposition 2. By exercise A.2 here, $\mathfrak m A_{\mathfrak m}$ is a minimal non-zero prime so $\mathrm{ht} \mathfrak m = 1$. ♦

Proposition 4.

A Dedekind domain A is normal.

Proof

Since normality is a local property, we may assume $(A, \mathfrak m)$ is local. And since $\mathfrak m$ is invertible, it is principal by proposition 2. Thus it suffices to prove the following.

• If $(A,\mathfrak m)$ is a noetherian local domain such that $\mathfrak m$ is principal, then A is normal.

The whole of the next section is devoted to such rings.

# Discrete Valuation Rings

Definition.

discrete valuation ring (dvr) is a noetherian local domain $(A,\mathfrak m)$ such that $\mathfrak m \ne 0$ is principal. If $\mathfrak m = (\pi)$, we call $\pi$ a uniformizer of the dvr.

Note

As we saw earlier, $(\pi)$ has height 1, so the spectrum of A is easy to describe:

Here is an easy way to construct such rings.

Lemma.

Let A be a noetherian integral domain with a prime element $\pi \in A$. If $\mathfrak p = (\pi)$, then $A_{\mathfrak p}$ is a dvr.

Proof

Trivial. ♦

In particular, we can take any irreducible element π of a noetherian UFD A; then π is a prime element so $A_{(\mathfrak \pi)}$ is a dvr. Common examples include:

$\mathbb Z_{(2)} = \{\frac a b \in \mathbb Q : b \text{ odd}\},\quad k[X]_{(X)} = \{ \frac {f(X)}{g(X)} : g(0) \ne 0\},$

where k is a field.

Proposition 5.

Let A be a dvr with uniformizer $\pi$. Then every non-zero ideal of A is uniquely of the form $(\pi^n)$ for some $n\ge 0$.

Proof

We have $A \supsetneq (\pi) \supsetneq (\pi^2) \supsetneq \ldots$.

First we show that $\cap_n (\pi^n) = 0$. Indeed if $x \in \cap_n (\pi^n), x\ne 0$ then we can write $x = \pi^n u_n$ for $u_1, u_2, \ldots \in A$. Then $u_n = \pi u_{n+1}$ for each n so $(u_1) \subsetneq (u_2) \subsetneq \ldots$ which is impossible since A is noetherian.

Now for each $x\in A-\{0\}$, let

$\nu(x) = \max(n\ge 0 : x \in (\pi^n))$, where $(\pi^0) = A$.

We leave the remaining as an exercise.

• Prove that $(x) = (\pi^k)$ where $k = \nu(x)$.
• Hence show that any non-zero ideal of A is of the form $(\pi^k)$ for some k. ♦

Thus a dvr is a special type of UFD with exactly one prime element (its uniformizer). This makes prime factoring in the ring rather trivial: every non-zero element is a unit times a power of the uniformizer.

Corollary 2.

A dvr is a PID, hence a UFD so it is a normal domain.

To recap, we have shown that Dedekind domains are noetherian, of Krull dimension at most 1, and normal. Next we will see that the converse is true.

# Conditions for DVR

dvr ⟹ PID ⟹ UFD ⟹ normal.

The following is a converse statement.

Proposition 6.

Let $(A,\mathfrak m)$ be a noetherian 1-dimensional local domain (i.e. A has exactly two prime ideals: 0 and $\mathfrak m$).

If A is normal then it is a dvr.

Proof

Pick $x\in \mathfrak m- \{0\}$ so that $A/(x)$ is a noetherian ring with exactly one prime, i.e. it is a local artinian ring, so $\mathfrak m/(x) \subset A/(x)$ is nilpotent. Thus $\mathfrak m^n \subseteq (x) \subseteq \mathfrak m$ for some n > 0. We may assume $\mathfrak m^{n-1} \not\subseteq (x)$.

Let $\mathfrak a = \{a\in A: a\mathfrak m\subseteq (x)\}$, an ideal of A. Fix an $a\in \mathfrak a$ and set $\alpha := \frac a x \in\mathrm{Frac} A$. Then $\alpha \mathfrak m = \frac a x \mathfrak m \subseteq A$ is an ideal of A. If $\alpha \mathfrak m = A$, then $\mathfrak m = \frac 1 {\alpha} A$ is principal.

Otherwise $\alpha \mathfrak m \subseteq \mathfrak m$. We claim that $\alpha$ is integral over A. To prove this, pick a finite generating set $y_1, \ldots, y_n$ for the ideal $\mathfrak m$; each $\alpha y_i$ can be written as an A-linear combination of $y_1, \ldots, y_n$ so

$\alpha (y_1, \ldots, y_n)^t = M (y_1, \ldots, y_n)^t$

where M is an $n \times n$ matrix with entries in A. Hence $(\alpha I - M)(y_1, \ldots, y_n)^t = 0$ where equality holds in $\mathrm{Frac} A$. Multiplying by the adjugate matrix, we get

$\det(\alpha I - M) \cdot (y_1, \ldots, y_n)^t = 0$.

Since $\mathfrak m \ne 0$, we get $\det(\alpha I - M) = 0$; expanding gives a monic polynomial relation in $\alpha$ with coefficients in A.

Hence $\alpha \in \mathrm{Frac} A$ is integral over A; since A is normal, $\alpha \in A$. Thus $\frac a x \in A$ for all $a\in \mathfrak a$ so $\mathfrak a \subseteq (x)$. Since $x \in \mathfrak a$ we have $\mathfrak a = (x)$. But this means $\mathfrak m^{n-1} \subseteq \mathfrak a = (x)$, a contradiction. ♦

Corollary 3.

Let A be a noetherian 1-dimensional domain. If A is normal, then it is a Dedekind domain.

Proof

Let $\mathfrak a \subseteq A$ be any ideal; it is finitely generated since A is noetherian. To prove it is invertible by proposition 1, it suffices to show $\mathfrak aA_{\mathfrak m}$ is principal for each maximal ideal $\mathfrak m$. But $A_{\mathfrak m}$ is a local 1-dimensional noetherian domain; since it is normal, by proposition 6 it is a dvr. Hence $\mathfrak aA_{\mathfrak m}$ is principal. ♦

Summary.

Let A be an integral domain. Every non-zero ideal of A is invertible if and only if it is noetherian, normal and of Krull dimension at most 1.

If $A = k[V]$ is the coordinate ring of a variety V, geometrically the above conditions translate as follows:

• Krull dimension = 1 ⟺ V is a curve;
• A is a domain ⟺ V is irreducible;
• A is normal ⟺ V is “smooth”.

Hence philosophically, a Dedekind domain is “like a smooth curve”.

We put the term “smooth” in quotes because we based the concept on geometric intuition rather than rigour. E.g. in our example above, the curve $Y^2 = X^3$ is not smooth at the origin.