Commutative Algebra 17

Posted on April 4, 2020 by limsup

Field of Fractions

Throughout this article, A denotes an integral domain (which may not be a UFD).

Definition.

The field of fractions of A is an embedding of A into a field K,

$i : A\hookrightarrow K$

such that every element of K can be expressed as $\frac a b$ , where $a\in A$ and $b\in A-\{0\}$ .

Exercise A

Prove that the field of fractions of A is unique up to isomorphism.

Examples

1. $\mathbb Q$ is a field of fractions of $\mathbb Z$ .

2. $\mathbb Q[\sqrt{-1}]$ is a field of fractions of $\mathbb Z[\sqrt{-1}]$ .

3. A field of fractions of $k[X]$ (k = field) is given by the set of all $\frac{f(X)}{g(X)}$ where $f(X), g(X)\in k[X]$ and $g(X) \ne 0$ .

The third example strongly hints at a general construction.

Proposition 1.

Every domain A has a field of fractions.

The trick is to “formally” invert elements. Although the proof is long, most of it is tedious mechanical work and not particularly illuminating.

Proof

We take the set of all pairs $(a,b) \in A\times (A - \{0\})$ , together with the relation:

$(a,b) \sim (a',b') \iff ab' = a'b.$

This is an equivalence relation: to check transitivity, if $ab' = a'b$ and $a'b'' = a''b'$ where $b, b', b'' \ne 0$ , then $(ab'' - a''b)b' = (ab')b'' - (a''b')b = (a'b)b'' - (a'b'')b = 0$ and since $b'\ne 0$ we have $ab'' = a''b$ .

Write $\frac a b$ for the equivalence class of $(a,b)$ and let K be the set of equivalence classes. We define addition and product on K as follows.

$\frac a b + \frac {c}{d} = \frac{ad + bc}{bd}, \qquad \frac a b \times \frac {c}{d} = \frac{ac}{bd}.$

Now show that these operations are well-defined, e.g. for addition if $\frac {c}{d} = \frac{c'}{d'}$ then $\frac{ad + bc}{bd} = \frac{ad' + bc'}{bd'}$ , which follows from

$(ad+bc)(bd') = abdd' + b^2(cd') = abdd' + b^2(c'd) = (ad' + bc')(bd).$

We leave it to the reader to verify that K, with the above operations, forms a commutative ring with 1. It is a field since if $\frac a b \ne 0$ , then $a\ne 0$ so $\frac a b \times \frac b a = 1$ . ♦

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Field of Fractions of UFD

Let A be a UFD and K its field of fractions. We say that $x,y\in K^*$ are associates if $\frac x y$ is a unit in A.

Now fix a set of primes $P \subset A$ modulo associates. For example, in ℤ, we can take $P = \{2, -3, -5, 7, 11, 13, \ldots\}$ .

Proposition 2.

Any $x\in K^*$ can be uniquely written as

$x = u\pi_1^{e_1} \pi_2^{e_2} \ldots \pi_n^{e_n},$

where $u\in U(A)$ , $\pi_i \in P$ and $e_i \in \mathbb Z - \{0\}$ . We call this the prime factorization of $x\in K^*$ .

Proof

Write $x = \frac a b$ where $a,b\in A-\{0\}$ . Each of a and b can be written as $u \pi_1^{e_1} \ldots \pi_k^{e_k}$ where u is a unit, $\pi_i \in P$ and $e_i \in \mathbb Z_{\ge 0}$ . Expanding $x = \frac a b$ and removing terms with exponent 0 gives us an expression as above.

For uniqueness, suppose $u \pi_1^{e_1} \ldots\pi_k^{e_k} = u' \pi_1^{f_1} \ldots \pi_k^{f_k}$ where $e_i, f_i \in \mathbb Z$ . Without loss of generality assume $e_1 \ge f_1$ . If $e_1 > f_1$ , then $u \pi_1^{e_1 - f_1} \pi_2^{e_2} \ldots \pi_k^{e_k} = \pi_2^{f_2} \ldots \pi_k^{f_k}$ . The LHS is a multiple of $\pi_1$ in A but the RHS is not; this gives a contradiction. Hence $e_1 = f_1$ . This holds for the remaining terms as well. ♦

Now we can extend the following notions for $x,y \in K^*$ . Let

$x = u\pi_1^{e_1} \ldots \pi_k^{e_k}, \ \ y = u'\pi_1^{f_1} \ldots \pi_k^{f_k}, \quad u, u' \in U(A), \pi_i \in P, e_i, f_i \in \mathbb Z$

be their prime factorizations.

Divisibility : we write $x|y$ if $\frac y x \in A$ , equivalently $e_i \le f_i$ for each $1\le i \le k$ .
Gcd : write $\gcd(x,y) = \pi_1^{\min(e_1, f_1)} \ldots \pi_k^{\min(e_k, f_k)}$ .
Lcm : write $\mathrm{lcm}(x,y) = \pi_1^{\max(e_1, f_1)} \ldots \pi_k^{\max(e_k, f_k)}$ .

As before if $\gcd(x_1, \ldots, x_n) = g$ , then $y|x_i$ for each i if and only if $y|g$ . Similarly, if $\mathrm{lcm}(x_1, \ldots, x_n) = h$ , then $x_i|y$ for each i if and only if $h|y$ .

Example

In $\mathbb Q$ we have

$\gcd(\frac{20}3, \frac{50}9, \frac{8}5) = \frac{2}{45}, \quad \mathrm{lcm}(\frac{20}3, \frac{50}9, \frac{8}5) = 200.$

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Gauss’s Lemma

Here is our theorem of the day.

Theorem (Gauss).

If $A$ is a UFD, so is its ring of polynomials $A[X]$ .

Proof

We say $f \in A[X] - \{0\}$ is primitive if its coefficients have gcd 1.

Step 0. Any prime $\pi \in A$ is also prime in $A[X]$ .

Because $A[X]/(\pi) \cong (A/\pi)[X]$ is an integral domain.

Step 1. The product of two primitive polynomials is primitive.

Let $f = a_0 + a_1 X + \ldots \in A[X]$ and $g = b_0 + b_1 X + \ldots \in A[X]$ be primitive.

Let $\pi \in A$ be any prime; it suffices to show there is a coefficient of fg not divisible by $\pi$ .
Since f is primitive there is a maximum d such that $a_d \not\in (\pi)$ ; similarly, there is a maximum e such that $b_e \not\in (\pi)$ .
The coefficient of $X^{d+e}$ in fg is not a multiple of $\pi$ since it is:

$a_0 \overbrace{b_{d+e}}^{\in (\pi)} + a_1 \overbrace{b_{d+e-1}}^{\in (\pi)} + \ldots + \overbrace{a_d b_e}^{\not\in (\pi)} + \ldots + \overbrace{a_{d+e}}^{\in (\pi)} b_0.$

Step 2. Extend A into a field.

Let K be the field of fractions of A. For any $g \in K[X] - \{0\}$ , let $c(g)$ be the gcd of the coefficients of $g$ and $p(g) := c(g)^{-1} g$ . Note that $p(g) \in A[X]$ since $c(g)\in K^*$ divides every coefficient of $g$ . Also, the gcd of all coefficients of $p(g)$ is 1 so we have:

$g = c(g)p(g), \quad c(g)\in K^*, p(g) \in A[X] \text{ primitive.}$

This expression is clearly unique: if $g = c'p$ where $c'\in K^*$ and $p\in A[X]$ is primitive, we have $c' = c(g)$ and $p = p(g)$ .

Step 3. Prove that c and p are multiplicative.

To show that $c(fg) = c(f)c(g)$ and $p(fg) = p(f)p(g)$ , write

$f = c(f)p(f), \ g = c(g)p(g), \quad c(f), c(g)\in K^*, p(f), p(g) \in A[X] \text{ primitive.}$

Then $c(fg) p(fg) = fg = c(f)c(g)p(f)p(g)$ . By step 1, $p(f)p(g)$ is primitive; hence $c(fg) = c(f)c(g)$ and $p(fg) = p(f)p(g)$ .

Step 4. Every irreducible non-constant f in A[X] is irreducible in K[X].

Let $f \in A[X]$ be non-constant and irreducible. Note that it must be primitive since a prime element of $A$ is also prime in $A[X]$ (step 0). If $f$ is reducible in $K[X]$ we have $f = gh$ where $g, h\in K[X]$ are not constant. Then $f = p(f) = p(g)p(h)$ gives a factorization in $A[X]$ as well, which is a contradiction.

Step 5. Every irreducible f in A[X] is prime.

First suppose $f\in A[X]$ is non-constant and irreducible.

Suppose $gh = fh'$ where $g, h, h' \in A[X]$ . Then $p(g)p(h) = p(f)p(h') = f\cdot p(h')$ since $f$ is primitive. Hence we may assume $g, h, h'$ are primitive. Since $K[X]$ is a PID, one of $g, h$ is a multiple of $f$ in $K[X]$ (by step 4). Without loss of generality write $g = fg'$ for some $g' \in K[X]$ . But now $g = p(g) = p(f)p(g') = f\cdot p(g')$ so in fact $g$ is a multiple of $f$ in $A[X]$ .

Finally if $f\in A[X] - \{0\}$ is constant and irreducible, it is irreducible in A. ♦

Exercise B

Complete the proof by showing that $A[X]$ satisfies a.c.c. on principal ideals.

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Consequences

While proving the above theorem, we also showed

Proposition 3.

Let A be a UFD and K its field of fractions. Then $f\in A[X] - \{0\}$ is irreducible in $A[X]$ if and only if either

$f$ is a prime element of A, or

$\deg f > 0$ , $f$ is primitive, and irreducible in $K[X]$ .

Thus $k[X_1, \ldots, X_n]$ and $\mathbb Z[X_1, \ldots, X_n]$ are UFDs for any field k. Looking at the spectra of these rings, we see a striking difference between PIDs and general UFDs.

Proposition 4.

Every non-zero prime ideal of a PID $A$ is maximal.

Proof

Suppose we have ideals $\mathfrak p \subseteq \mathfrak a$ of A with $\mathfrak p$ a non-zero prime. Since A is a PID we write $\mathfrak p = (\pi)$ and $\mathfrak a = (\alpha)$ where $\pi \ne 0$ . Then $\pi = \alpha\beta$ for some $\beta\in A$ . Since $\pi$ is irreducible either $\alpha$ or $\beta$ is a unit. In the former case, $\mathfrak a = (1)$ ; in the latter case $\mathfrak a = \mathfrak p$ . Hence $\mathfrak p$ is maximal. ♦

In particular, if A is a PID and not a field, its Krull dimension is 1.

spec_pid_and_ufd

In contrast, each $k[X_1, \ldots, X_n]$ is a UFD of Krull dimension at least n. Thus UFDs are structurally much more complicated than PIDs. They can be of arbitrarily high dimension.

Exercise C

Prove that $\mathbb C[X, Y, Z]/(Z^2 - X^2 - Y^2)$ is not a UFD.

[Hint: define some norm function to an easier ring. ]

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Posted in Advanced Algebra | Tagged field of fractions, gauss lemma, krull dimension, primes, principal ideal domains, UFDs, unique factorisation | 6 Comments

Commutative Algebra 16

Posted on April 3, 2020 by limsup

Gcd and Lcm

We assume A is an integral domain throughout this article.

If A is a UFD, we can define the gcd (greatest common divisor) and lcm (lowest common multiple) of two elements as follows. For $a,b\in A-\{0\}$ , we can write the ideals (a) and (b) uniquely as a product of primes:

$\begin{aligned} (a) &= (\pi_1)^{e_1} (\pi_2)^{e_2} \ldots (\pi_k)^{e_k}, \\ (b) &= (\pi_1)^{f_1} (\pi_2)^{f_2} \ldots (\pi_k)^{f_k}.\end{aligned}$

where $\pi_i \in A$ are prime elements and $e_i, f_i \ge 0$ are integers.

Lemma 1.

We have $a|b$ if and only if $e_i \le f_i$ for each $1 \le i \le k$ .

Proof

Only (⇒) is non-trivial. Write $b = ac$ for $c\in A$ . If, say, $e_1 > f_1$ , then we have

$c\pi_1^{e_1 - f_1} \pi_2^{e_2} \ldots \pi_n^{e_n} = \pi_2^{f_2} \ldots \pi_n^{f_n}.$

The prime element $\pi_1$ occurs in the LHS but not the RHS, which violates unique factorizability. ♦

Definition.

The gcd and lcm of a and b are respectively defined by elements of A satisfying:

$\begin{aligned}(\gcd(a, b)) &:= (\pi_1)^{\min(e_1, f_1)} \ldots (\pi_k)^{\min(e_k, f_k)}, \\ (\mathrm{lcm}(a,b)) &:= (\pi_1)^{\max(e_1, f_1)} \ldots (\pi_k)^{\max(e_k, f_k)}.\end{aligned}$

Note

The gcd and lcm are well-defined only up to associates. We also define the gcd and lcm of $x_1, \ldots, x_n$ recursively via

$\begin{aligned}\gcd(x_1, \ldots, x_n) &= \gcd(\gcd(x_1, \ldots, x_{n-1}), x_n),\\ \mathrm{lcm}(x_1, \ldots, x_n) &= \mathrm{lcm}(\mathrm{lcm}(x_1, \ldots, x_{n-1}), x_n).\end{aligned}$

The following properties are important and easily derived from lemma 1.

Let $g = \gcd(x_1, \ldots, x_n)$ ; if $y\in A$ satisfies $y|x_i$ for each $1\le i \le n$ then $y | g$ .
Let $h = \mathrm{lcm}(x_1, \ldots, x_n)$ ; if $y\in A$ satisfies $x_i|y$ for each $1 \le i \le n$ then $h|y$ .

warning We often say elements $x,y\in A$ are coprime if their gcd is 1. This potentially conflicts with our earlier definition of coprime ideals, for if $x,y\in A$ are coprime here, it does not mean $(x,y) = (1)$ . E.g. in the next article we will show that $k[X, Y]$ is a UFD but (X, Y) is a proper ideal.

Hence we will refrain from using the term coprime in this case.

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Principal Ideal Domains

Definition.

A principal ideal domain (PID) is a domain in which every ideal is principal.

The key property we want to show is:

Theorem.

Let A be a PID. Then A satisfies a.c.c. on principal ideals and A is a UFD.

Proof

For the first claim, let $(x_1) \subseteq (x_2) \subseteq \ldots$ be a chain of principal ideals of A. Now the union of an ascending chain of ideals is an ideal (see proof of proposition 2 here), so $\cup_i (x_i)$ is an ideal, necessarily principal. Write $\cup_i (x_i) = (y)$ . Then $y\in (x_n)$ for some n. So

$(y) \subseteq (x_n) \subseteq \cup_i (x_i) = (y) \implies (y) = (x_n)$

and $(y) = (x_n) = (x_{n+1}) = \ldots$ . This proves the first claim

For the second, let $x\in A$ be irreducible and suppose $yz \in (x)$ ; we need to show y or z lies in (x). Then $(x, y)$ is principal so we write $(x, y) = (y')$ . Thus x is a multiple of y’; since x is irreducible this means either (x) = (y’) or y’ is a unit.

Former case gives us $y \in (y') = (x)$ .
Latter case gives us $ax + by = 1$ for some $a, b\in A$ so $z = (ax + by)z = axz + byz \in (x)$ since $yz \in (x)$ . ♦

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Euclidean Domains

Finally, we have an effective method for identifying some PIDs.

Definition.

An Euclidean function on A is a function $N : A - \{0\} \to \mathbb Z_{\ge 0}$ such that

for any $a, b \in A$ where $b\ne 0$ , we can write $a = bq + r$ for some $q, r\in A$ such that $r=0$ or $N(r) < N(b)$ .

If A has an Euclidean function on it, we call A an Euclidean domain.

Note

The Euclidean function gives the “size” of an element $a\in A$ , such that we can always divide b by a to obtain a remainder r which is either zero, or strictly smaller. Repeating this process gives us the gcd of two elements via the Euclidean algorithm.

Theorem.

An Euclidean domain A is a PID.

Proof

Fix an Euclidean function N on A. Let $\mathfrak a\subseteq A$ be an ideal. If $\mathfrak a = 0$ it is clearly principal. Otherwise, pick $x\in \mathfrak a - \{0\}$ such that $N(x)$ is minimum. We claim that $\mathfrak a = (x).$ Clearly we only need to show ⊆.

Suppose $y\in \mathfrak a$ . Thus we can write $y = qx + r$ where $r=0$ or $N(r) < N(x)$ . In the former case we have $y \in (x)$ . The latter case cannot happen since $r \in \mathfrak a$ , $N(r) < N(x)$ but $N(x)$ is already minimum among elements of $\mathfrak a$ . Thus $\mathfrak a \subseteq (x)$ . ♦

In a nutshell we have:

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Example: k[X]

Let $A = k[X]$ , where k is a field. We take the degree of a polynomial

$\deg : A- \{0\} \to \mathbb Z_{\ge 0}.$

For any $a(X)\in A, b(X) \in A-\{0\}$ , the remainder theorem gives $a(X) = b(X) q(X) + r(X)$ where $r(X) = 0$ or $\deg r < \deg b$ . This is exactly the condition for an Euclidean function.

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Case Study: Gaussian Integers

We shall prove that the norm function N on $A = \mathbb Z[\sqrt{-1}]$ given by $N(a+bi) = a^2 + b^2$ is Euclidean. Note that N extends to

$N : \mathbb Q[\sqrt{-1}] = \{a + bi : a,b\in \mathbb Q\} \longrightarrow \mathbb Q, \quad N(a+bi) = a^2 + b^2$

which is still multiplicative. Now to prove that N is Euclidean, let $a,b\in \mathbb Z[\sqrt{-1}]$ with $b\ne 0$ . Define $x = \frac a b \in \mathbb Q[\sqrt{-1}]$ . By rounding off the real and imaginary parts of x, we obtain $q\in \mathbb Z[\sqrt{-1}]$ such that

$x - q = \alpha + \beta i, \quad -\frac 1 2 \le \alpha, \beta <\frac 1 2.$

It follows that $N(x-q) \le \frac 1 2 < 1$ . Hence we have:

$N(a - bq) = N(b) N(\frac a b - q) = N(b) N(x-q) < N(b)$ .

In conclusion, we have shown:

Theorem.

The ring $\mathbb Z[\sqrt{-1}]$ is an Euclidean domain, hence a PID and UFD.

Since the norm is effectively computable, we can write a program to perform the Euclidean algorithm on this ring.

Factoring in Gaussian Integers

Here, we consider how an integer factors in $A = \mathbb Z[\sqrt{-1}]$ .

Proposition 1.

Let $p\in \mathbb Z$ be prime..

If $p = 2$ we have $2 = (-i)(1 + i)^2$ , where $1+i \in A$ is prime.

If $p\equiv 3\pmod 4$ , then p is still prime in A.

If $p \equiv 1 \pmod 4$ , then $p = \alpha\beta$ where $\alpha, \beta$ are primes which are not associates.

Proof

The case for p = 2 is obvious (1+i is prime since its norm is 2, a prime). For odd p, we have

$A/(p) \cong \mathbb Z[X]/(X^2 + 1, p) \cong \mathbb F_p[X]/(X^2 + 1).$

Now this ring is a domain if and only if $X^2 + 1 \equiv 0 \pmod p$ has no solution. From theory of quadratic residues, this holds if and only if $p\equiv 3 \pmod 4$ . For $p\equiv 1 \pmod 4$ , we have

$A/(p) \cong F_p[X]/((X - u)(X-v))$

for distinct u, v modulo p. On the RHS ring we have $(X-u)(X-v) = 0$ ; hence in the original ring A we have $(p) = (p, i-u)\times (p, i-v)$ since X corresponds to i. And since A is a PID, we have $p = \alpha\beta$ for primes $\alpha, \beta$ which are not associates. ♦

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Summary

We have thus shown the following in this article and the previous.

To ensure factoring terminates after finitely many steps, we need a.c.c. on the principal ideals of A.
Assuming a.c.c. on principal ideals, A is a UFD if and only if all irreducibles in A are prime.
If A is a PID, then it satisfies a.c.c. on principal ideals and is a UFD.
If A has an Euclidean function, then it is a PID and Euclidean algorithm allows us to compute, for any $x,y\in A$ , their gcd $z\in A$ such that $(z) = (x,y)$ .

Exercises

1. Find all solutions to $x^2 + y^2 = 5^5 \times 13^4 \times 17^3 \times 29^2$ in each of the following cases:

x, y are any integers;
x > y are coprime positive integers.

Note: since the numbers are rather large, the reader is advised to work in Python.

[Hint: $x^2 + y^2$ is the norm of $x+ yi \in \mathbb Z[\sqrt{-1}]$ .]

2. Prove that the norm function is Euclidean for the following rings:

$\mathbb Z[\sqrt{-2}], \ \ \mathbb Z[\frac{1 + \sqrt{-3}} 2], \ \ \mathbb Z[\frac{1 + \sqrt{-7}}2], \ \ \mathbb Z[\frac{1 + \sqrt{-11}}2].$

[Hint for the last case: this diagram may help.]

3. Fill in the gaps in the solution to the sample problem in the previous article.

4. (Hard) Solve $x^2 + 7 = 2^n$ for positive integers x and n.

5. Write a Python program to compute the Euclidean algorithm in $\mathbb Z[\sqrt{-1}]$ .

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Posted in Advanced Algebra | Tagged commutative rings, euclidean domains, gaussian integers, primes, principal ideal domains, principal ideals, rings, unique factorisation | Leave a comment

Commutative Algebra 15

Posted on April 2, 2020 by limsup

Unique Factorization

Through this article and the next few ones, we will explore unique factorization in rings. The inspiration, of course, comes from ℤ. Here is an application of unique factorization. Warning: not all steps may make sense to the reader at this point of time.

Sample Problem.

Find all integer solutions to $x^2 + 2 = y^3$ .

Solution (Sketch)

Factor the equation to obtain $y^3 = (x + \alpha)(x - \alpha)$ where $\alpha = \sqrt{-2}$ . Use the fact that $\mathbb Z[\alpha]$ satisfies unique factorization; deduce that $x + \alpha$ and $x-\alpha$ must be coprime (since x, y are odd). Hence $x + \alpha = a^3$ and $x - \alpha = b^3$ for some $a,b \in \mathbb Z[\alpha]$ with $ab = y$ . This gives $2\sqrt{-2} = a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ . Now we factor $2\sqrt{-2}$ in all possible ways and solve a pair of simultaneous equations in a and b. E.g.

$a-b = 2\sqrt{-2}, \ a^2 + ab + b^2 = 1\ \implies \ (a, b) = (1 + \sqrt{-2}, 1 - \sqrt{-2})$

which gives $(x,y) = (-5, 3).$ ♦

Before starting, here are some preliminary definitions.

Throughout this article, A denotes an integral domain.

Definition.

Recall that A is a unit if it is invertible under product.

Elements $x,y\in A$ are associated (written as $x\sim y$ ) if there is a unit $u\in A$ such that $y = ux$ .

For $x,y\in A$ we say x divides y (written as $x|y$ ) if there exists $z\in A$ such that $y = xz$ .

Let $x\in A - \{0\}$ , $x$ not a unit. We say it is reducible if we can find $y,z\in A$ which are non-unit such that $x=yz$ . Otherwise, $x$ is irreducible.

All the above conditions can be expressed in terms of ideals.

x and y are associated if and only if (x) = (y) as ideals of A. Thus, the relation is an equivalence relation.
x divides y if and only if $(y) \subseteq (x)$ as ideals. Thus $x|y$ and $y|x$ if and only if $x\sim y$ .
x is irreducible if and only if when (x) = (y)(z) as a product of ideals, either (x) = (y) or (x) = (z). In particular, if x is irreducible, so are all its associates.

The descriptions in terms of principal ideals are much cleaner.

Examples

1. Suppose $A = \mathbb Z[i] = \{a + b\sqrt{-1} : a,b\in \mathbb Z\}$ . The only units are {-1, +1, –i, +i}. So each non-zero element has 4 associates, e.g.

$(1+2i, -2+i, -1-2i, 2-i)$ are associates, $(1-2i, -2-i, -1+2i, 2+i)$ are associates.

Note that 1 + 2i and 1 – 2i are not associates. On the other hand, 1 + i and 1 – i are associates. (Verify this!)

2. Suppose $A = \mathbb Z[\sqrt 2] = \{a + b\sqrt 2 : a,b \in \in \mathbb Z\}$ . The unit group is infinite since it contains $\pm (1 + \sqrt 2)^n$ for all $n\in \mathbb Z$ . Thus every non-zero element has infinitely many associates.

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Factoring Into Irreducibles

Let $x\in A$ be non-zero, non-unit. If x is reducible, we can factor it as x = yz where y, z are non-zero and non-unit. Again, if either of them is reducible, we factor them further. If this process terminates, we obtain x as a factor of irreducibles.

Unfortunately, there are cases where it does not.

Example

Let $A = \mathbb R[x^{1/n} : n = 1, 2, \ldots]$ . Now $x\in A$ can be factorized indefinitely:

$x=x^{1/2} \cdot x^{1/2} = (x^{1/4} \cdot x^{1/4}) \cdot (x^{1/4}\cdot x^{1/4}) = \ldots$

Here is a condition which ensures that factorization terminates.

Proposition 1.

Partially order the principal ideals of A by reverse inclusion:

$(x) \le (y)\iff (y)\subseteq (x)$ .

If this poset is noetherian, then every non-unit $x\in A-\{0\}$ can be factored as a product of irreducibles.

Note

By proposition 1 here, the above condition is equivalent to either of the following.

Any non-empty collection of principal ideals of A has a maximal element.
In any chain of principal ideals $(x_1) \subseteq (x_2) \subseteq \ldots$ of A, there exists n such that $(x_n) = (x_{n+1}) = \ldots$ .

One also says A satisfies ascending chain condition (a.c.c.) on the set of principal ideals.

Proof

Let S be the collection of all non-unit $x\in A-\{0\}$ which cannot be factored as a product of irreducibles. If S is non-empty, the set of principal ideals (x) for $x\in S$ has a maximal element (z) with respect to ⊆.

Since z cannot be factored as a product of irreducibles, it is not irreducible. Write z = xy where x, y are non-unit and non-zero. Then $(z) \subsetneq (x)$ and $(z) \subsetneq (y)$ so by maximality of (z) we have $x,y\not\in S$ . Hence x and y can be expressed as a product of finitely many irreducibles; since z = xy so does z, a contradiction. ♦

But how does one check the above abstract condition for a ring A? There are two answers to this.

Most rings we are interested in actually satisfy a.c.c. on all ideals. Called noetherian rings, we will see later that they are everywhere.
More immediately, we will give a criterion for unique factorizability which also implies a.c.c. on principal ideals.

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UFDs

Finally, here is our main object of interest.

Definition.

An integral domain A is called a unique factorization domain (UFD) if every proper principal ideal $(z) \subsetneq A$ can be factored as a product of irreducible principal ideals.

$(z) = (x_1) (x_2) \ldots (x_k)$

uniquely up to permutation.

Here is a non-example. Suppose $A = \mathbb Z[\sqrt{-5}]$ . In this ring, we have

$6 = 2 \times 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}).$

It is easy to see that no two of these elements are associates. Thus to see that unique factorization fails, it remains to show all of them are irreducible. For that, take the norm function $N : A \to \mathbb Z$ where $N(a+b\sqrt{-5}) = a^2 + 5b^2.$ An easy calculation shows:

$N(xy) = N(x) N(y)$ ,
$N(x) \ge 0$ for all x,
$N(x) = 1 \iff x = \pm 1$ .

Now the above elements have norms

$N(2) = 4, \ N(3) = 9, \ N(1+\sqrt{-5}) = N(1 - \sqrt{-5}) = 6.$

If any of them had a proper factor, such a factor would have norm 2 or 3. But it is easy to see no element of norm 2 or 3 exists.

Exercise A

Prove that $\mathbb Z[\sqrt{10}]$ is not a UFD.

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Prime Elements

We end this section by stating a necessary and sufficient condition for unique factorization to hold.

Definition.

Let $x\in A - \{0\}$ . We say x is prime if $(x)$ is a prime ideal of A.

Thus x is prime if and only if:

whenever yz is a multiple of x, either y or z is a multiple of x.

Lemma 1.

A prime element $x\in A - \{0\}$ is irreducible.

Proof

Suppose x = yz, where y, z are non-zero. Then yz is a multiple of x so either y or z is a multiple of x. Assume the former, so y is a multiple of x, but since x = yz we have x is a multiple of y also. Thus x and y are associates and z is a unit. ♦

Finally, the main result we want is as follows.

Theorem 1.

Let A be a domain which satisfies a.c.c. on principal ideals.

Then A is a unique factorization domain if and only if all irreducible elements are prime.

Proof

(⇒) Let A be a UFD and $x\in A$ be irreducible. To show x is prime, suppose $yz \in (x)$ and $y\not\in (x)$ ; we need to show $z\in (x)$ . Write $yz = ax$ for some $a\in A$ . Write y, z and a as a product of irreducibles. By unique factorization, since x does not occur in the factoring of y, it must occur in the factoring of z. Thus $z\in (x)$ .

(⇐) Suppose all irreducibles are prime. For any $a\in A$ , suppose we have factorizations $a = x_1 x_2 \ldots x_k = y_1 y_2 \ldots y_l,$ where $x_i, y_j \in A$ are all irreducible, and hence prime. Write this in terms of principal ideals:

$(x_1) (x_2) \ldots (x_k) = (y_1) (y_2) \ldots (y_l).$

Now since $y_1 \ldots y_l \in (x_1)$ , one of the terms, say $y_1$ must be in $(x_1)$ . Hence $y_1$ is a multiple of $x_1$ ; by irreducibility we have $(x_1) = (y_1)$ . So $x_1$ and $y_1$ are associates and upon division we have

$u x_2 \ldots x_k = y_2 \ldots y_l\ (u = \text{unit}) \implies (x_2)\ldots (x_k) = (y_2)\ldots (y_l).$

Repeat until we run out of terms on one side so we get (1). The other side must also be empty so we have established unique factorization. ♦

In the next article, we will find many examples of UFDs based on this criterion.

Example

Let us show that $A = \mathbb Z[\sqrt {-5}]$ is not a UFD by finding an irreducible element which is not prime: 2. By the norm argument above, 2 is irreducible but it is not prime because

$A/(2) = \mathbb Z[X]/(X^2 + 5, 2) = \mathbb F_2[X]/(X^2 + 5) = \mathbb F_2[X]/(X^2 + 1)$

is not an integral domain.

Exercise B

Find all prime ideals of $A = \mathbb Z[\sqrt{-5}]$ which contain 2 or 3.

[Hint (highlight to read): prime ideals of A containing 2 correspond to prime ideals of A/(2).]

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Posted in Advanced Algebra | Tagged integral domains, irreducibles, noetherian, primes, UFDs, unique factorisation | Leave a comment

Commutative Algebra 14

Posted on April 1, 2020 by limsup

Basic Open Sets

For $f\in A$ , let $D(f) := \mathrm{Spec} A - V(f)$ , an open subset of Spec A. Note that

$D(f) = \{\mathfrak p \subset A \text{ prime}: \mathfrak p \not\ni f\}$ .

Proposition 1.

The collection of $D(f)$ over all $f\in A$ forms a basis for the topology of $\mathrm{Spec} A$ .

Proof

Let $U = \mathrm{Spec} A - V(\mathfrak a)$ be an open subset of Spec A. Suppose $\mathfrak p \in U$ so that $\mathfrak p\not\supseteq \mathfrak a$ . Pick an $f \in \mathfrak a - \mathfrak p$ . Since $f\not\in \mathfrak p$ we have $\mathfrak p \in D(f)$ . It remains to show $D(f) \subseteq U$ . Indeed if $\mathfrak q\not\ni f$ then certainly $\mathfrak q \not\supseteq \mathfrak a$ so we have $\mathfrak q \in U$ . ♦

With this we have:

Proposition 2.

$\mathrm{Spec} A$ is quasi-compact, in the sense that every open cover has a finite subcover.

Note

The modern topological term for this is just compact. However, the term quasi-compact is firmly entrenched in the literature for commutative algebra and algebraic geometry so we will respect the tradition here.

Proof

To prove quasi-compactness, it suffices to check covering via basic open sets. Thus suppose $\mathrm{Spec} A = \cup_i D(f_i)$ is a basic open covering; we need to show there is a finite subcover. Now $\emptyset = \cap_i V(f_i) = V(\mathfrak a)$ where $\mathfrak a = (f_i)$ , i.e. the ideal generated by the $f_i$ . This can only happen if $\mathfrak a = (1)$ so we can write

$1 = g_{i_1} f_{i_1} + \ldots + g_{i_k} f_{i_k}, \quad g_{i_j} \in A$

for finitely many choices of $f_i$ . Hence $(1) = (f_{i_1}, \ldots, f_{i_k})$ so we have

$\cap_{j=1}^k V(f_{i_j}) = \emptyset \implies \cup_{j=1}^k D(f_{i_j}) = \mathrm{Spec} A.$ ♦

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More on Zariski Topology

Let $X = \mathrm{Spec} A$ for a ring A. Immediately, we see some differences between X and affine varieties. Firstly, points of X are prime ideals $\mathfrak p \subset A$ and often $\{\mathfrak p\}$ is not closed in X. More generally, we have the following.

Proposition 3.

For $\mathfrak p \in \mathrm{Spec} A$ , the closure of $\{\mathfrak p\}$ is $V(\mathfrak p)$ . In particular, $\{\mathfrak p\}$ closed if and only if $\mathfrak p$ is a maximal ideal, in which case we call it a closed point.

Proof

The closure C of $\{\mathfrak p\}$ is the intersection of all $V(\mathfrak a)$ containing $\mathfrak p$ . This includes $V(\mathfrak p)$ so we have $C \subseteq V(\mathfrak p)$ . On the other hand, any closed subset $V(\mathfrak a)$ containing $\mathfrak p$ gives $\mathfrak p \supseteq \mathfrak a$ so $V(\mathfrak p) \subseteq V(\mathfrak a)$ . Thus $C = V(\mathfrak p)$ . ♦

For example, if A is a domain, then (0) is a prime ideal so $\mathrm{Spec} A = V(0)$ is the closure of the point (0). In a topological space X, a generic point of X is a $p\in X$ such that the closure of $\{p\}$ is X.

Note

In topological lingo, this says that Spec A is not a T1 space. It is, however, a T0 space. Compare this with the case of affine varieties which are T1 (i.e. all points are closed), but not T2. [For example, $V = \mathbb A^1$ has the cofinite topology.]

Elements of Spec ℤ

spec_z_diagram

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Nilradical and Jacobson Radical

Definition.

The intersection of all prime ideals of A is called the nilradical of A, denoted $n(A)$ . From the previous article,

$n(A) = \{x \in A: x^n = 0 \text{ for some } n > 0\}.$

Correspondingly, the intersection of all maximal ideals of A is called the Jacobson radical of A, denoted by $J(A)$ .

We have the following classification for Jacobson radical.

Proposition 4.

We have $J(A) = \{ x \in A: 1 + xA \subseteq U(A)\}$ .

Proof

(⊇) If $x\not\in\mathfrak m$ for maximal ideal $\mathfrak m$ then since $\mathfrak m + xA \supsetneq \mathfrak m$ we have $\mathfrak m + xA = (1)$ so there exists $y\in A, z\in \mathfrak m$ such that $z + xy = 1$ so $1 + x(-y)$ is not a unit, i.e. $1 + xA \not\subseteq U(A)$ .

(⊆) Suppose $x\in J(A)$ ; let $y\in A$ . If (1 + xy) is not a unit, it is contained in some maximal ideal $\mathfrak m$ . Then $x\not \in \mathfrak m$ , or else we would have $1 \in \mathfrak m$ . ♦

Philosophy

Following affine varieties, we now look at rings from a new point of view.

“Points” correspond to prime ideals $\mathfrak p \subset A$ .
“Functions on points” correspond to elements $f\in A$ .
“Function vanishes on point” corresponds to $f\in \mathfrak p$ .

Thus the nilradical (resp. Jacobson radical) is the set of all functions vanishing on all points (resp. closed points). In a way, non-closed points act as “irreducible closed sets” of closed points, so one naturally asks when the nilradical is equal to the Jacobson radical. When this holds, the ring in question is called a Jacobson ring.

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Connected Components

Proposition 5.

Let $A = B\times C$ be a product of rings. Then $\mathrm{Spec} A$ is the topological disjoint union of $\mathrm{Spec} B$ and $\mathrm{Spec} C$ .

Conversely suppose $\mathrm{Spec A} = V \cup W$ where V, W are disjoint closed subsets, then

$A \cong B \times C, \quad V \cong \mathrm{Spec} B, W \cong \mathrm{Spec} C$ .

Proof

The projection maps $A\to B$ gives continuous map $\mathrm{Spec} B \to \mathrm{Spec} A$ which takes $\mathfrak q \mapsto \mathfrak q \times C$ . This map is injective, continuous and takes closed set $V(\mathfrak b)$ to closed set $V(\mathfrak b \times C)$ . Hence it is a subspace embedding.

Similarly, we have a subspace embedding $\mathrm{Spec} C\to \mathrm{Spec} A$ ; by exercise A here, the two images form a disjoint union so the first claim is done.

For the second claim, write $V = V(\mathfrak a)$ and $W = V(\mathfrak b)$ so

$\mathrm{Spec} A = V(\mathfrak a) \cup V(\mathfrak b) = V(\mathfrak a \cap \mathfrak b), \quad \emptyset = V(\mathfrak a) \cap V(\mathfrak b) = V(\mathfrak a + \mathfrak b).$

Thus $\mathfrak a + \mathfrak b = (1)$ ; pick $x\in \mathfrak a$ , $y\in \mathfrak b$ such that $x+y=1$ so (x) and (y) are coprime ideals. Now $xy\in \mathfrak a\cap \mathfrak b$ which is contained in the nilradical since it is contained in all prime ideals. Thus $(xy)^n = 0$ for some n > 0 so $(x^n), (y^n)$ are coprime ideals with product zero. By Chinese Remainder Theorem this gives $A \cong A/(x^n) \times A/(y^n)$ .

Since $x^n \in \mathfrak a$ we have $V \subseteq V(x^n)$ ; similarly $W \subseteq V(y^n)$ . Since $V(x^n) \cong \mathrm{Spec} A/(x^n)$ and $V(y^n) \cong \mathrm{Spec} A/(y^n)$ also form a disjoint union of Spec A by the first part, we are done. ♦

Pictorially we have

spectrum_ring_product

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Irreducible Spaces

Proposition 6.

The space $\mathrm{Spec} A$ is irreducible if and only if the nilradical is a prime ideal.

Proof

(⇐) If the nilradical is prime $\mathfrak p$ , and Spec A is a union of two closed subsets, one of them must contain $\mathfrak p \in \mathrm{Spec} A$ ; hence it must contain all primes. Thus Spec A is irreducible.

(⇒) Conversely, suppose Spec A is irreducible. Pick $x,y\in A$ with $xy\in \mathfrak n(A)$ . Then

$\begin{aligned} (x)(y) \subseteq n(A) &\implies V(x) \cup V(y) \supseteq V(n(A)) = \mathrm{Spec} A\\ &\implies V(x) \cup V(y) = \mathrm{Spec} A.\end{aligned}$

By irreducibility, either V(x) or V(y) is the whole Spec A. Assume the former; then x lies in all prime ideals of A so it is nilpotent, i.e. $x\in n(A)$ . ♦

In particular, we have:

Corollary 1.

If $\mathrm{Spec A}$ is irreducible then it has a generic point.

irreducible_spec

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Prime Chains

Finally, another important aspect of Spec A are the prime chains.

Definition.

A prime chain of A is a strictly increasing sequence of prime ideals

$\mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \ldots \subsetneq \mathfrak p_d$

of A; the length of the chain is d.

For example, in the ring $A = k[X, Y]$ for a field X, we have $(0) \subset (X) \subset (X, Y)$ which is a prime chain of length 2.

Definition.

The Krull dimension (or just dimension) of a ring A is the supremum of all lengths of prime chains in A.

Examples

1. The dimension of a field is 0.

2. The dimension of ℤ is 1.

3. We have $\dim A[X] \ge \dim A + 1$ , since for any prime chain $(\mathfrak p_i)_{i=0}^d$ of A, we have the prime chain of A[X]:

$\mathfrak p_0[X] \subsetneq \mathfrak p_1[X] \subsetneq \ldots \subsetneq \mathfrak p_d[X] \subsetneq \mathfrak p_d[X] + (X).$

Note that these ideals are all prime since $A[X]/\mathfrak p_i[X] \cong (A/\mathfrak p_i)[X]$ and $A[X]/(\mathfrak p_d[X] +(X)) \cong A/\mathfrak p_d$ .

4. From the above, we have $\dim k[X_1, \ldots, X_n] \ge n$ and $\dim \mathbb Z[X_1, \ldots, X_n] \ge n+1$ . Eventually, we will see that equality holds!

5. We have $\dim (A\times B) = \sup (\dim A, \dim B)$ .

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Posted in Advanced Algebra | Tagged connected components, irreducible spaces, jacobson radical, maximal ideals, nilradical, prime ideals, rings, spectrum, zariski topology | Leave a comment

Commutative Algebra 13

Posted on March 31, 2020 by limsup

Zariski Topology for Rings

In this article, we generalize earlier results in algebraic geometry to apply to general rings. Recall that points on an affine variety V correspond to maximal ideals $\mathfrak m\subset k[V]$ . For general rings, we have to switch to taking prime ideals because of the following.

Lemma 1.

If $f : A \to B$ is a ring homomorphism and $\mathfrak q$ is a prime ideal of B, then $\mathfrak p := f^{-1}(\mathfrak q)$ is a prime ideal of A.

Proof

The composition $A \stackrel f\to B \to B/\mathfrak q$ has kernel $f^{-1}(\mathfrak q) = \mathfrak p$ . Hence this induces an injective ring homomorphism $A/\mathfrak p \hookrightarrow B/\mathfrak q$ . Since $B/\mathfrak q$ is a domain, so is $A/\mathfrak p$ so $\mathfrak p$ is prime. ♦

However it is not true that if $\mathfrak q$ is maximal in B, $\mathfrak p$ must be maximal in A. For instance consider the inclusion $i:\mathbb Z \hookrightarrow \mathbb Q$ . The zero ideal is maximal in $\mathbb Q$ but $i^{-1}((0)) = (0)$ is not maximal in $\mathbb Z$ .

Now we define our main object of interest.

Definition.

Given a ring A, the spectrum of A is its set of prime ideals $\mathrm{Spec} A$ . The Zariski topology on $\mathrm{Spec A}$ is defined as follows: $V\subseteq \mathrm{Spec A}$ is closed if and only if it is of the following form

$V(S) = \{ \mathfrak p \in \mathrm{Spec A} : \mathfrak p \supset S\},$

for some subset $S\subseteq A$ .

Note that since $V(S) = V(\mathfrak a)$ where $\mathfrak a$ is the ideal generated by S, we lose no generality by taking S to be ideals of A. If S = {f} we will write V(f) instead of V({f}) or V((f)) to reduce the clutter.

Immediately, we prove that the above gives a bona fide topology.

Proposition 1.

For any collection of ideals $(\mathfrak a_i)$ of A and ideals $\mathfrak a, \mathfrak b$ of A, we have

$V(1) = \emptyset, V(0) = \mathrm{Spec} A$ ,

$\cap_i V(\mathfrak a_i) = V(\sum_i \mathfrak a_i)$ ,

$V(\mathfrak a) \cup V(\mathfrak b) = V(\mathfrak a\cap \mathfrak b) = V(\mathfrak {ab})$ .

Proof

The first claim is obvious. For the second, $\mathfrak p$ contains $\mathfrak a_i$ for each i if and only if it contains $\sum_i \mathfrak a_i$ .

Finally, for the third claim, we have $\mathfrak {ab} \subseteq \mathfrak a \cap \mathfrak b \subseteq \mathfrak a$ and $\mathfrak a\cap \mathfrak b \subseteq \mathfrak b$ so we have

$V(\mathfrak {ab}) \supseteq V(\mathfrak a\cap \mathfrak b) \supseteq V(\mathfrak a) \cup V(\mathfrak b).$

Finally suppose $\mathfrak p \not\in V(\mathfrak a)\cup V(\mathfrak b)$ . Then $\mathfrak p \not\supseteq \mathfrak a$ and $\mathfrak p \not\supseteq \mathfrak b$ so there are $x \in \mathfrak a - \mathfrak p$ and $y \in \mathfrak b - \mathfrak p$ . Then $xy \in \mathfrak{ab} - \mathfrak p$ since $\mathfrak p$ is prime so $\mathfrak p \not\in V(\mathfrak{ab})$ . Hence if $\mathfrak p \in V(\mathfrak{ab})$ then $\mathfrak p\in V(\mathfrak a) \cup V(\mathfrak b)$ . ♦

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Structure of Spec A

Thanks to Zorn’s lemma, we have the following results on the structure of Spec A.

Proposition 2.

If A is a ring, every proper ideal $\mathfrak a\subsetneq A$ is contained in a maximal ideal.

Proof

Let $\Sigma$ be the collection of all proper ideals $\mathfrak b\subsetneq A$ containing $\mathfrak a$ , ordered partially by inclusion. We claim that every chain $\Sigma' \subseteq \Sigma$ has an upper bound. Since $\mathfrak a \in \Sigma$ we may assume without loss of generality $\Sigma' \ne\emptyset$ .

Take $\mathfrak c = \cup_{\mathfrak b \in \Sigma'} \mathfrak b$ . Let us show that $\mathfrak c$ is an ideal of A.

Clearly $0 \in \mathfrak c$ . Let $x,y\in \mathfrak c$ and $a\in A$ . Then $x\in \mathfrak b$ and $y\in \mathfrak b'$ for some $\mathfrak b, \mathfrak b' \in \Sigma'$ . Since $\Sigma'$ is totally ordered, either $\mathfrak b'\subseteq \mathfrak b$ or $\mathfrak b' \subseteq \mathfrak b'$ . Assuming the former, this gives $x, y\in \mathfrak b$ so $ax - y\in \mathfrak b \subseteq \mathfrak c$ . Thus $\mathfrak c$ is an ideal of A.

Since none of the $\mathfrak b\in \Sigma'$ contains 1, we have $1\not\in \mathfrak c$ . Hence $\mathfrak c \in \Sigma$ is an upper bound of $\Sigma'$ .

By Zorn’s lemma $\Sigma$ has a maximal element, which is precisely a maximal ideal of A containing $\mathfrak a$ ♦

We say that $\mathfrak p \in \mathrm{Spec} A$ minimal if it is a minimal element with respect to inclusion.

Proposition 3.

Any $\mathfrak p\in \mathrm{Spec} A$ contains a minimal prime $\mathfrak q$ .

Proof

Let $\Sigma$ be the collection of all prime ideals of A contained in $\mathfrak p$ , ordered partially by reverse inclusion, i.e. $\mathfrak q \le \mathfrak q'$ if $\mathfrak q \supseteq \mathfrak q'$ . To apply Zorn’s lemma we need to show that every chain $\Sigma'\subseteq \Sigma$ has a maximal element (i.e. minimal with respect to inclusion). Without loss of generality we may assume $\Sigma' \ne \emptyset$ .

Let $\mathfrak p' = \cap_{\mathfrak q \in \Sigma'} \mathfrak q$ , an ideal of A. We need to show that it is prime. Suppose $x,y\in A - \mathfrak p'$ so there exist $\mathfrak q_1, \mathfrak q_2\in \Sigma'$ such that $x\not\in \mathfrak q_1$ and $y\not\in \mathfrak q_2$ . Since $\Sigma'$ is a chain either $\mathfrak q_1 \subseteq \mathfrak q_2$ or $\mathfrak q_2 \subseteq \mathfrak q_1$ ; assuming the former we have $x,y\not\in \mathfrak q_1$ . Since $\mathfrak q_1$ is prime we have $xy\not\in \mathfrak q_1$ so $xy\not\in \mathfrak p'$ .

Now apply Zorn’s lemma to obtain a maximal element, which is a minimal prime contained in $\mathfrak p$ . ♦

Corollary.

If A is non-trivial, $\mathrm{Spec} A$ must contain a maximal ideal and a minimal prime ideal.

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Homomorphisms

Suppose $f:A\to B$ is a ring homomorphism. We saw that this induces a map

$f^* : \mathrm{Spec} B \to \mathrm{Spec} A,\quad \mathfrak q \mapsto f^{-1}(\mathfrak q)$ .

Proposition 4.

The map $f^*$ is continuous with respect to the Zariski topology; in fact

$(f^*)^{-1}(V(\mathfrak a)) = V(f(\mathfrak a)).$

Note that $f(\mathfrak a)$ is not an ideal of B in general, but that is okay since we defined V on arbitrary subsets.

Proof

Let $\mathfrak q \in \mathrm{Spec} B$ . This lies in the LHS if and only if $f^*(\mathfrak q) = f^{-1}(\mathfrak q)$ contains $\mathfrak a$ , if and only if $\mathfrak q$ contains $f(\mathfrak a)$ . ♦

Clearly the identity ring homomorphism on A induces the identity map on $\mathrm{Spec A}$ , also for ring homomorphisms $f:A\to B$ and $g:B\to C$ we have

$(g\circ f)^* = f^* \circ g^* : \mathrm{Spec} C \to \mathrm{Spec} A.$

Exercise A

Prove that for an ideal $\mathfrak a\subseteq A$ , the canonical map $\pi : A \to A/\mathfrak a$ induces

$\pi^* : \mathrm{Spec}(A / \mathfrak a) \to \mathrm{Spec} A$

which is a subspace embedding onto the closed subset $V(\mathfrak a)$ .

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Comparison with Algebraic Geometry

Proposition 5.

Let $V = V(\mathfrak a)$ be a closed subset of $\mathrm{Spec} A$ . Then

$\bigcap \{ \mathfrak p : \mathfrak p \in V\} = r(\mathfrak a)$ ,

the radical of $\mathfrak a$ .

Proof

(⊇) Suppose $x\in A$ , $x^n \in \mathfrak a$ for some n > 0. Now each $\mathfrak p \in V$ contains $\mathfrak a$ and hence contains $x^n$ . Since $\mathfrak p$ is prime we have $x\in \mathfrak p$ .

(⊆) Suppose $x\in A$ , $\mathfrak a$ does not contain any power of x. We will prove the existence of $\mathfrak p \supseteq \mathfrak a$ not containing x.

Let $\Sigma$ be the set of all ideals $\mathfrak b\supseteq \mathfrak a$ not containing any power of x; thus $\mathfrak a \in \Sigma$ . Order $\Sigma$ by inclusion. We will show that any chain $\Sigma' \subseteq \Sigma$ has an upper bound. Indeed, let $\mathfrak c = \cup_{\mathfrak b \in \Sigma'} \mathfrak b$ . As before, since $\Sigma'$ is a chain of ideals ordered by inclusion, $\mathfrak c$ is an ideal of A. It does not include any power of x so $\mathfrak c\in \Sigma$ .

Hence by Zorn’s lemma, $\Sigma$ has a maximal element $\mathfrak p$ ; by construction $\mathfrak p$ contains $\mathfrak a$ but not any power of x. It remains to show $\mathfrak p$ is prime: suppose not so there are $y, z\in A - \mathfrak p$ with $yz \in \mathfrak p$ . Now $\mathfrak p + (y)$ and $\mathfrak p + (z)$ strictly contain $\mathfrak p$ , so by maximality of $\mathfrak p$ , we have $\mathfrak p + (y), \mathfrak p + (z) \not\in\Sigma$ . Hence

$\left. \begin{aligned} \exists m > 0,\ x^m \in \mathfrak p+(y) \\ \exists n>0, \ x^n \in \mathfrak p + (z)\end{aligned} \right\} \implies x^{m+n} \in (\mathfrak p + (y))(\mathfrak p + (z)) \subseteq \mathfrak p + (yz) \subseteq \mathfrak p.$

a contradiction. Thus $\mathfrak p$ is prime. ♦

Exercise B

From proposition 5, prove that we have a bijection

zariski_correspondence_rings

The correspondence reverses inclusion. For $\cap V_i$ and $V_1 \cup V_2$ in Spec A, write down the corresponding operations for radical ideals of A.

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Posted in Advanced Algebra | Tagged homomorphism, maximal ideals, prime ideals, rings, spectrum, topology, zariski topology | Leave a comment

Commutative Algebra 12

Posted on March 30, 2020 by limsup

Some Results on Posets

In this article we have two goals in mind:

to introduce the idea of noetherian posets, and
to state Zorn’s lemma and give some examples.

The latter is of utmost importance in diverse areas of mathematics.

Definition.

A partial ordering on a set S is a relation ≤ on S × S, satisfying the following.

(Reflexive) For any $x\in S$ , we have $x\le x$ .

(Transitive) For any $x, y, z\in S$ , if $x\le y$ and $y\le z$ , then $x\le z$ .

(Anti-symmetric) For any $x, y\in S$ , if $x\le y$ and $y\le x$ , then $x=y$ .

A total ordering is a partial ordering such that for any $x,y \in S$ , either $x\le y$ or $y\le x$ .

A partially ordered set (or just poset) is a set together with an assigned partial ordering. Similarly, we have totally ordered sets.

Given a partially ordered set $(S, \le)$ , for $x,y\in S$ , we write:

$x < y$ if $x\le y$ and $x\ne y$ ;
$x\ge y$ if $y \le x$ ;
$x > y$ if $y < x$ .

Examples

The set of real numbers (or rational numbers, or integers) forms a totally ordered set under the usual arithmetic ordering.
If X is a set, let P(X) be the collection of all subsets of X. This forms a partially ordered set under inclusion. If X has more than one element, P(X) is not totally ordered.
Any subset T of a partially ordered set S gives a partially ordered set. If S is totally ordered, so is T.

Lemma 1 (Duality).

If $(S, \le)$ is a poset, then so is $(S, \ge)$ .

Proof.

Easy exercise. ♦

Duality allows us to cut our work by half in a lot of cases.

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Bounds

Let $(S, \le)$ be a poset.

Definition.

The maximum of S is an $m_1 \in S$ such that for any $x\in S$ we have $x\le m_1$ .

The minimum of S is an $m_0 \in S$ such that for any $x\in S$ we have $x \ge m_0$ .

A maximal element of S is an $m\in S$ such that for any $x\in S$ , if $x \ge m$ then $x=m$ .

A minimal element of S is an $m'\in S$ such that for any $x\in S$ , if $x \le m'$ then $x=m'$ .

warning The naming is a little confusing, but one must differentiate between the maximum of a set and the maximal elements of a set. In the poset below, S has three maximal elements but no maximum.

For example, let X = {a, b, c} and let S be the following subsets of P(X) under inclusion.

sample_poset

The following properties are obvious.

Lemma 2.

The maximum (resp. minimum) of a poset is unique if it exists.

If the maximum (resp. minimum) of a poset exists, it is the unique maximal (resp. minimal) element.

Proof

Easy exercise. ♦

Exercise

Suppose S has a unique maximal element $m_1$ . Must $m_1$ be the maximum of S?

Finally, for a subset T of an ordered set S, we define the following.

Definition.

An upper bound (resp. lower bound) of T in S is an $x\in S$ satisfying: for all $y\in T$ , we have $y\le x$ (resp. $y \ge x$ ).

Clearly, upper and lower bounds are not unique in general. E.g. for $S = \mathbb Z$ under the arithmetic ordering, the subset T of even integers has no upper or lower bound. The subset T’ of squares has lower bounds 0, -1, -2, … but no upper bound.

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Noetherian Sets

This will be used a few times in our study of commutative algebra.

Definition.

A poset S is said to be noetherian if every non-empty subset T of S has a minimal element (in T).

It is easy to see that every finite poset S is noetherian.

Let $T\subseteq S$ be any non-empty subset. Pick $x_0 \in T$ . If $x_0$ is a minimal element of T we are done; otherwise there exists $x_1 \in T$ , $x_1 < x_0$ . Again if $x_1$ is a minimal element of T we are done; otherwise we repeat. The process terminates since T is finite because we cannot have $x_0 > x_1 > \ldots$ in T. Thus T has a minimal element.

Examples

The set $\mathbb N$ of positive integers under ≤ is noetherian.
The set $\mathbb N \times \mathbb N$ is noetherian, where $(m, n) \le (m', n')$ if $m\le m'$ and $n \le n'$ .
The set $\mathbb N$ is noetherian, where we take $m\le n$ to mean $m|n$ .

Note that examples 2 and 3 are not totally ordered. The key property of noetherian sets is the following.

Theorem (Noetherian Induction).

Let T be a subset of a noetherian poset S satisfying the following.

If $x\in S$ is such that $(y \in S, y < x \implies y \in T)$ , then $x\in T$ .

Then T = S.

Note

To paraphrase the condition in words: if T contains all elements of S smaller than x, then it must contain x itself.

Proof

If $T\ne S$ , $U = S - T$ is non-empty so it has a minimal element x. By minimality, any $y\in S$ with $y < x$ cannot lie in U so $y\in T$ . But by the given condition this means $x\in T$ , which is a contradiction. ♦

Here is an equivalent way of expressing the noetherian property.

Proposition 1.

A poset S is noetherian if and only if the following hold.

For any sequence of elements $x_1 \ge x_2 \ge \ldots$ in S, there is an n such that $x_n = x_{n+1} = x_{n+2} = \ldots$ .

Proof

(⇒) Suppose S is noetherian and $x_1 \ge x_2 \ge \ldots$ are elements of S. The set $\{x_n : n=1,2,\ldots\}$ thus has a minimal element $x_n$ . Since we have $x_n \ge x_{n+1} \ge \ldots$ , equality must hold by minimality.

(⇐) Suppose S is not noetherian; let $T\subseteq S$ be a non-empty subset with no minimal element. Pick $x_1 \in T$ ; it is not minimal, so we can find $x_2 \in T$ with $x_2 < x_1$ . Again since $x_2\in T$ is not minimal we can find $x_3 \in T$ with $x_3 < x_2$ . Repeat to obtain an infinitely decreasing sequence. ♦

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Zorn’s Lemma

Finally we have the following critical result.

Theorem (Zorn’s Lemma).

Let S be a poset. A chain in S is a subset which is totally ordered. Suppose S is a poset such that every chain $T\subseteq S$ has an upper bound in S.

Then S has a maximal element.

A typical application of Zorn’s lemma is the following.

Proposition 2.

Every vector space V over a field k has a basis.

Note

This looks like an intuitively obvious result, but try finding a basis for the ℝ-space of all real functions $f:\mathbb R\to \mathbb R$ . Using Zorn’s lemma, one can show that a basis exists but describing it does not seem possible. Generally, results that require Zorn’s lemma are of this nature: they claim existence of certain objects or constructions without exhibiting them explicitly. Some of these are rather unnerving, like the Banach-Tarski paradox.

Proof

Let $\Sigma$ be the set of all linearly independent subsets of V. We define a partial order on $\Sigma$ by inclusion, i.e. for $C, D \in \Sigma$ , $C\le D$ if and only if $C\subseteq D$ . Next, we claim that every chain $\Sigma' \subseteq \Sigma$ has an upper bound.

Let $D =\cup_{C \in \Sigma'} C$ ; we need to show that D is a linearly independent subset of V, so that $D\in \Sigma$ is an upper bound of $\Sigma'$ . Suppose

$\alpha_1 v_1 + \ldots + \alpha_n v_n = 0, \quad v_1, \ldots, v_n \in D, \alpha_1, \ldots, \alpha_n \in k.$

For each $i=1, \ldots, n$ , we have $v_i \in C_i$ for some $C_i \in \Sigma'$ . But $\Sigma'$ , being a chain, is totally ordered so without loss of generality, we assume $C_1 \supseteq C_i$ for all $1 \le i \le n$ . This means $v_1, \ldots, v_n \in C_1$ ; since $C_1$ is linearly independent, we have $\alpha_1 = \ldots = \alpha_n = 0$ .

Now apply Zorn’s lemma and we see that $\Sigma$ has a maximal element C. We claim that a maximal linearly independent subset of V must span V. If not, we can find $v\in V$ outside the span of C. This means $C \cup \{v\}$ is linearly independent, thus contradicting the maximality of C. ♦

warning Zorn’s lemma holds vacuously when S is empty. However, in many applications of the lemma, constructing an upper bound for a chain $T\subseteq S$ requires T to be non-empty. In such instances, we will mentally replace Zorn’s lemma with the following equivalent version.

If S is a non-empty poset such that every non-empty chain in S has an upper bound in S, then S has a maximal element..

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Posted in Advanced Algebra | Tagged axiom of choice, noetherian, posets, totally ordered sets, zorn's lemma | Leave a comment

Commutative Algebra 11

Posted on March 29, 2020 by limsup

Coordinate Rings as k-algebras

Let k be an algebraically closed field. Recall that a closed subset $V \subseteq \mathbb A^n_k$ is identified by its coordinate ring k[V], which is a finitely generated k-algebra since

$k[V] = k[X_1, \ldots, X_n] / I(V).$

Definition.

An affine k-variety is a finitely generated k-algebra A which is a reduced ring. Formally, we write V for the variety and $A = k[V]$ for the algebra instead.

We say V is irreducible if A is an integral domain. A morphism of affine k-varieties $f : V\to W$ is a homomorphism of the corresponding k-algebras $f^* : k[W] \to k[V].$

Thus, each closed set V gives an affine variety; isomorphic closed sets give isomorphic affine varieties. Conversely, we have:

Lemma.

Any affine k-variety A is the coordinate ring of some closed $V\subseteq \mathbb A^n_k$ .

Proof

Indeed since A is finitely generated as a k-algebra, there is a surjective homomorphism (of k-algebras) $k[X_1, \ldots, X_n] \to A$ ; its kernel $\mathfrak a$ is a radical ideal because A is a reduced ring. Hence $A \cong k[V(\mathfrak a)]$ . ♦

Note that taking the set:

$B \mapsto \mathrm{Hom}_{k\text{-alg}}(B, k),$

recovers the set of points for B = k[V]. This gives another application of the “Hom” construction.

Now, the reader might question the utility of this point of view, since we are just back to dealing with closed subsets $V\subseteq \mathbb A^n$ . The advantage here is that now we can expand the class of objects of interest.

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Affine k-Schemes

Recall that to study multiplicity of intersection, we should really be looking at general ideals of k[V] and not just radical ones. Hence we define the following.

Definition.

An affine k-scheme is a finitely generated k-algebra A. Again we write V for the scheme and $A = k[V]$ .

An affine k-scheme V is a k-variety if and only if $k[V]$ is reduced.

From our understanding of closed sets, we may now define the following for any affine k–schemes V and W. Note that now we have the capacity to define more general constructions (see the last few rows).

What we say	What we actually mean
P is a point on V.	$\mathfrak m_P$ is a maximal ideal of k[V]; equivalently we have a k-algebra homomorphism $k[V] \to k$ .
$\phi : V \to W$ is a morphism of affine k-schemes.	$\phi^* : k[W] \to k[V]$ is a homomorphism of k-algebras.
W is a closed subvariety of V.	$k[W] = k[V] /\mathfrak a$ for some radical ideal $\mathfrak a$ of $k[V]$ .
W is an irreducible closed subvariety of V.	$k[W] = k[V]/\mathfrak p$ for some prime ideal $\mathfrak p$ of $k[V]$ .
$W = \cap W_i$ is a set-theoretic intersection in V.	For $k[W_i] = k[V] / \mathfrak a_i$ , we have $k[W] = k[V] / r(\sum_i \mathfrak a_i)$ .
$W = W_1 \cup W_2$ is a union in V.	For $k[W_i] = k[V]/\mathfrak a_i$ , we have $k[W] = k[V]/(\mathfrak a_1 \cap \mathfrak a_2)$ .
W is a closed subscheme of V.	$k[W] = k[V]/\mathfrak a$ for some ideal $\mathfrak a$ of $k[V]$ .
$W = \cap W_i$ is a scheme intersection in V.	For $k[W_i] = k[V]/\mathfrak a_i$ , we have $k[W] = k[V] / (\sum_i \mathfrak a_i)$ .

Exercise A

Let A and B be any rings. Prove the following.

An ideal of $A\times B$ must be of the form $\mathfrak a \times \mathfrak b$ , where $\mathfrak a$ (resp. $\mathfrak b$ ) is an ideal of A (resp. B).
A prime ideal of $A\times B$ must be of the form $\mathfrak p \times B$ or $A \times \mathfrak q$ , where $\mathfrak p$ (resp. $\mathfrak q$ ) is a prime ideal of A (resp. B).
A maximal ideal of $A\times B$ must be of the form $\mathfrak m \times B$ or $A \times \mathfrak n$ , where $\mathfrak m$ (resp. $\mathfrak n$ ) is a maximal ideal of A (resp. B).

Define the corresponding construction for disjoint union $V\amalg W$ in the above table.

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Tangent Spaces

The affine k-variety for a singleton point $*$ is just $k[*] = k$ . A point on an affine k-scheme V is thus a morphism $* \to V$ .

Now we take the affine k-scheme # such that $k[\#] := k[X]/(X^2)$ . We will write this as $k[\#] = k[\epsilon]$ with $\epsilon^2 = 0$ . Geometrically, $\epsilon$ corresponds to a “small perturbation” such that $\epsilon^2 = 0$ . Note that # has exactly one point, so there is a unique map $* \to \#$ ; the corresponding homomorphism $k[\epsilon] \to k$ takes $\epsilon \mapsto 0$ .

dot_and_arrow

We imagine # as a point with an arrow sticking out. Thus $\# \to V$ takes the point to a point P on V and the arrow to a tangent vector at P.

Definition.

A tangent vector on the k-scheme V is a morphism of k-schemes $\phi : \# \to V$ .

The base point of the tangent vector $\phi$ is given by composing $* \to \# \stackrel{\phi}\to V$ . The tangent space of a point $P : * \to V$ is the set of all tangent vectors with base point P, denoted by $T_P V$ .

Note

In terms of k-algebras, a tangent vector is a k-algebra homomorphism

$\phi^* : k[V] \to k[\epsilon] = k[X]/(X^2)$ .

If $\phi^*$ has base point P, then $\mathfrak m_P \mapsto 0$ in the composition $k[V] \stackrel {\phi^*}\to k[\epsilon] \to k$ . This means $\phi^*(\mathfrak m_P) \subseteq k\cdot \epsilon$ and so $\phi^*(\mathfrak m_P^2) = 0$ . Thus $\phi^*$ factors through the following:

$\phi^* : k[V] \longrightarrow k[V]/\mathfrak m_P^2 \stackrel f\longrightarrow k[\epsilon].$

So it suffices to consider all k-algebra homomorphisms $f : k[V]/\mathfrak m_P^2 \to k[\epsilon]$ which satisfies $f(\mathfrak m_P/\mathfrak m_P^2) \subseteq k\cdot \epsilon$ . This gives a k-linear map $\mathfrak m_P / \mathfrak m_P^2 \to k$ , i.e. the dual space of $\mathfrak m_P / \mathfrak m_P^2$ as a k-vector space.

Conversely, a k-linear map $g:\mathfrak m_P / \mathfrak m_P^2 \to k$ also gives a k-algebra homomorphism $k[V]/\mathfrak m_P^2 \to k[\epsilon]$ by mapping k to k. Hence we have shown:

Proposition.

$T_P V$ is parametrized by the set of all k-linear maps $\mathfrak m_P / \mathfrak m_P^2 \to k$ . In particular, it forms a finite-dimensional vector space over k.

Exercise B

Explain why $\mathfrak m_P / \mathfrak m_P^2$ is finite-dimensional over k.

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Example 1

Consider a simple example: $V = \mathbb A^2$ with $k[V] = k[X, Y]$ . A point $P = (\alpha, \beta)\in V$ corresponds to $\mathfrak m_P = (X - \alpha, Y - \beta) \subset k[X, Y]$ . A tangent vector at P then corresponds to a k-linear map

$g : \mathfrak m_P / \mathfrak m_P^2 \longrightarrow k.$

But $\mathfrak m_P^2$ is generated by $(X-\alpha)^2, (X-\alpha)(Y-\beta), (Y-\beta)^2$ so g is uniquely determined by $c := g(X-\alpha) \in k$ and $d := g(Y-\beta) \in k$ . Clearly $\dim_k \mathfrak m_P / \mathfrak m_P^2 = 2$ .

To compute the resulting $\phi^* : k[X, Y] \to k[\epsilon]$ we take, for each $f(X, Y) \in k[X, Y]$ ,

$f(X, Y) = f(\alpha, \beta) + \overbrace{\left.\frac{\partial f}{\partial X}\right|_P}^{\in k}\cdot (X - \alpha) + \overbrace{ \left.\frac{\partial f}{\partial Y}\right|_P} ^{\in k}\cdot (Y - \beta) + \ldots$

ignoring the terms of degree 2 or more in $(X-\alpha), (Y-\beta)$ . Hence

$\phi^*(f) = f(\alpha, \beta) + \left.\frac{\partial f}{\partial X}\right|_P\cdot c + \left.\frac{\partial f}{\partial Y}\right|_P \cdot d.$

Example 2

Suppose $V \subset \mathbb A^2$ is cut out by $Y^2 = X^3 - X$ , so $k[V] = k[X, Y]/(Y^2 - X^3 + X)$ . Take the point P = (0, 0), which corresponds to $\mathfrak m_P = (X, Y)$ . Let us compute the space of tangent vectors at P.

graph_1_plot

[Edited from GeoGebra plot.]

As before we have $\mathfrak m_P^2 = (X^2, XY, Y^2) \subset k[V]$ . Instead of looking at $\mathfrak m_P$ and $\mathfrak m_P^2$ , we look at their preimages $\mathfrak n_P = (X, Y), \mathfrak n_P' = (X^2, XY, Y^2, Y^2 - X^3 + X)$ as ideals of $k[X, Y]$ . (Remember the correspondence of ideals between a ring and its quotient!) Thus

$\mathfrak n_P' = (X^2, XY, Y^2, Y^2 - X^3 + X) = (X, Y^2) \implies \dim_k \mathfrak n_P / \mathfrak n_P' = 1$

since Y is a basis for this space. Hence $\dim_k T_P V = 1$ .

Example 3

Now take $V\subset \mathbb A^2$ cut out by $Y^2 = X^3 + X^2$ so $k[V] = k[X, Y]/(Y^2 - X^2 - X^3)$ . Take P = (0, 0) again.

graph_2_plot

[Edited from GeoGebra plot.]

As before $\mathfrak m_P = (X, Y)$ and $\mathfrak m_P^2 = (X^2, XY, Y^2)$ . Again, we take their preimages $\mathfrak n_P = (X, Y)$ and $\mathfrak n_P' = (X^2, XY, Y^2, Y^2 - X^2 - X^3)$ as ideals of $k[X, Y]$ . This time we get $\mathfrak n_P' = (X^2, XY, Y^2)$ and so

$\dim T_P V = \dim_k \mathfrak m_P /\mathfrak m_P^2 = \dim_k \mathfrak n_P / \mathfrak n_P' = 2$ .

Summary.

Geometrically, the tangent space at P is larger than the dimension when there is a singularity at P, as the reader can see from the graphs above. Thus this gives an algebraic definition of singularity at various points. We will of course need an algebraic definition of dimension for this.

Exercise C

In each of the following varieties, compute the dimension of the tangent space at the origin. Intuitively, which varieties are singular at the origin?

$V = \{ (x, y) \in \mathbb A^2 : y^2 = x^3\}$ .
$V = \{ (x, y, z) \in \mathbb A^3 : y + z = z^2 + xz + xyz\}$ .
$V = \{ (w, x, y, z) \in \mathbb A^4 : w^2z + x^2 = z + x + y^2, z + y^3(w+1) = x^3 - w^2\}$ .

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Posted in Advanced Algebra | Tagged algebraic geometry, algebras, cotangent spaces, maximal ideals, tangent spaces, varieties | Leave a comment

Commutative Algebra 10

Posted on March 28, 2020 by limsup

Algebras Over a Ring

Let A be any ring; we would like to look at A-modules with a compatible ring structure.

Definition.

An $A$ –algebra is an $A$ -module $B$ , together with a multiplication operator $\times : B \times B \to B$ such that

$(B, +, \times)$ becomes a commutative ring (with 1);

multiplication is compatible with scalar multiplication:

$a\in A, b, b'\in B \implies a\cdot(b\times b') = (a\cdot b)\times b'.$

In the second condition, we have used · for module multiplication and × for ring multiplication. That condition gives us

$a \in A, b, b'\in B \implies b\times (a\cdot b') = (a\cdot b') \times b = a\cdot (b'\times b) = a\cdot (b\times b')$ .

Hence, for any $a_1, \ldots, a_k \in A$ and $b_1, \ldots, b_k \in B$ we have

$(a_1 \cdot b_1) \times \ldots \times (a_k \cdot b_k) = (\overbrace{a_1 \ldots a_k}^{\times \text{ in } A})\cdot (\overbrace{b_1 \times\ldots \times b_k} ^{\times \text{ in } B}).$

An alternative way of describing A-algebras is as follows.

Lemma 1.

If B is an A-algebra, the map $\phi : A\to B$ , $\phi(a) = a\cdot 1_B$ is a ring homomorphism.

Conversely, for any ring homomorphism $\phi: A\to B$ , B takes the structure of an A-algebra, where the A-module map is:

$A\times B\to B, \quad a\cdot b := \phi(a) \times b.$

Proof

First claim: clearly $\phi$ is additive and takes 1 to 1. Also

$\phi(a a') = aa'\cdot (1_B \times 1_B) = (a\cdot 1_B) \times (a'\cdot 1_B) = \phi(a) \times \phi(a').$

Second claim: verify the axioms as follows.

$\begin{aligned}\phi(1_A) \times b &= 1_B\times b = b\\ \phi(a+a')\times b &= (\phi(a) + \phi(a'))\times b = (\phi(a) \times b) + (\phi(a') \times b),\\ \phi(a)\times (b+b') &= (\phi(a) \times b) + (\phi(a) \times b'), \\ \phi(aa')\times b = \phi(a) \times \phi(a') \times b &= \phi(a) \times (\phi(a') \times b), \\ \phi(a)\times (b\times b') &= (\phi(a) \times b)\times b'.\end{aligned}$

This completes the proof. ♦

Exercise A

Prove that the constructions in the lemma are mutually inverse.

Definition.

If B is an A-algebra, an A-subalgebra is a subset of B which is an A-submodule as well as a subring.

Exercise B

If the A-algebra B corresponds to $\phi : A\to B$ , prove that an A-subalgebra of B is precisely a subring of B containing $\phi(A)$ .

P. S. Please do the above exercises. Don’t be lazy. 🙂

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Homomorphisms

Definition.

Let B and B’ be A-algebras. A homomorphism of A-algebras from B to B’ is an A-linear map $f:B \to B'$ which is also a ring homomorphism.

If we consider the alternative way of defining A-algebras, we get:

Proposition 1.

Suppose B and B’ are A-algebras corresponding to ring homomorphisms $\phi : A\to B$ and $\phi' : A\to B'$ . A homomorphism of A-algebras is precisely a ring homomorphism $f:B\to B'$ such that $f\circ \phi = \phi'$ .

Proof

(⇒) Suppose $f:B\to B'$ is a homomorphism of A-algebras. For $a\in A$ ,

$f(\phi(a)) = f(a\cdot 1_B) = a\cdot f(1_B) = a\cdot 1_{B'} = \phi'(a).$

(⇐) Suppose f is a ring homomorphism such that $f\circ \phi = \phi'$ , so for all $a\in A$ , $f(a\cdot 1_B) = a\cdot 1_{B'}$ . Hence for $a\in A$ , $b\in B$ ,

$f(a\cdot b) = f((a\cdot 1_B) \times b) = f(a\cdot 1_B) \times f(b) = (a\cdot 1_{B'}) \times f(b) = a\cdot f(b).$ ♦

The following result follows from the corresponding results for A-modules and rings.

First Isomorphism Theorem.

If $f:B\to B'$ is a homomorphism of A-algebras, the image of f is an A-subalgebra of B’ and we get an isomorphism of A-algebras.

$\overline f : A/\mathrm{ker} f \longrightarrow \mathrm{im} f, \quad a + \mathrm{ker} f \mapsto f(a).$

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Representability

For A-algebras B and C, let $\mathrm{Hom}_{A-\text{alg}}(B, C)$ be the set of A-algebra homomorphisms $f : B\to C$ . Unlike the case of modules, this set has no canonical additive structure, since if $f_1, f_2 : B\to C$ are homomorphisms of A-algebras, $f_1 + f_2$ usually is not (e.g. it does not take 1 to 1).

However, $\mathrm{Hom}_{A-\text{alg}}(B, C)$ is useful for classifying certain constructions from A-algebras.

As an example fix $B = A[X, Y]/(XY - 1)$ . Any A-algebra homomorphism $f:B\to C$ corresponds to a pair of elements $(\alpha, \beta) \in C \times C$ such that $\alpha\beta = 1$ . Thus we have a bijection:

$\begin{aligned}\mathrm{Hom}_{A-\text{alg}}(B, C) &\stackrel \cong\longrightarrow U(C),\\ f &\mapsto f(X), \end{aligned}$

where U(C) is the set of all units of C. Furthermore, any A-algebra homomorphism $g : C\to C'$ induces a map

$g_* : \mathrm{Hom}_{A-\text{alg}}(B, C) \longrightarrow \mathrm{Hom}_{A-\text{alg}}(B, C'), \quad f \mapsto g\circ f,$

which translates to the map g restricted to $U(C) \to U(C')$ .

Thus, we obtain a natural isomorphism $U(-) \cong \mathrm{Hom}_{A-\text{alg}}(B, -)$ . To express this rigorously, one needs the language of category theory.

Exercise C

For any A-algebra C, let

$\begin{aligned} SL_2 C &= \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a, b, c, d \in C, ad - bc = 1\right\}, \\ GL_2 C &= \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a, b, c, d \in C, ad - bc \text{ unit}\right\}. \end{aligned}$

Find A-algebras B and B’ such that

$SL_2(-) \cong \mathrm{Hom}_{A-\text{alg}}(B, -), \quad GL_2(-) \cong \mathrm{Hom}_{A-\text{alg}}(B', -)$ .

Note

Finally observe that $U(C), SL_2 C, GL_2 C$ are not just sets, but have group structures as well. To specify these group structures rigorously, one requires B to have a Hopf algebra structure.

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Generating a Subalgebra

If B is an A-algebra and $C_i \subseteq B$ is a collection of subalgebras, their intersection $C = \cap_i C_i$ is clearly also a subalgebra of B. This follows from the fact that intersection of submodules (resp. subring) is a submodule (resp. subring).

Definition.

For subset $S \subseteq B$ , let $\Sigma$ be the collection of all A-subalgebras of B containing S. Their intersection $\cap \{C : C\in S\}$ is the A-subalgebra of B generated by S; it is denoted by $A[S].$

This generated subalgebra can be concretely described as follows. For a finite sequence $\alpha_1, \ldots, \alpha_k \in S$ , and polynomial $f(X_1, \ldots, X_k) \in A[X_1, \ldots, X_k]$ in k variables, we take the element $f(\alpha_1, \ldots, \alpha_k) \in B$ . Taking the set of all these elements over all k, all $\alpha_1, \ldots, \alpha_k\in S$ and all $f(X_1, \ldots, X_k)$ then gives us $A[S]$ .

Definition.

An A-algebra B is finitely generated (or of finite type) if $B = A[S]$ for some finite subset $S\subseteq B$ .

If $S = \{\alpha_1, \ldots, \alpha_n\}$ , it suffices to take only polynomials f in exactly n variables and consider all $f(\alpha_1, \ldots, \alpha_n)$ , in which case we have a surjective homomorphism of A-algebras

$A[X_1, \ldots, X_n] \longrightarrow B, \qquad f(X_1, \ldots, X_n) \mapsto f(\alpha_1, \ldots, \alpha_n).$

warning In future articles, if B is an A-algebra, we say B is a finite A-algebra if it is finitely generated as an A-module; we say it is of finite type if it is finitely generated as an A-algebra.

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Direct Product of Algebras

Let $(B_i)_{i\in I}$ be a collection of A-algebras. Let $B := \prod_{i\in I} B_i$ be the set-theoretic product; we saw earlier that B has a natural structure of an A-module. It also has a natural structure of a ring via component-wise multiplication: $(b_i)_i \times (b'_i)_i := (b_i b'_i)_i$ , which is compatible with its A-module structure. Thus:

Theorem (Universal Property of Products).

For each $j\in I$ define the projection map

$\pi_j : \prod_{i\in I} B_i \longrightarrow B_j, \quad (b_i)_i \mapsto b_j,$

clearly a homomorphism of A-algebras. This collection $(B, (\pi_i : B \to B_i))$ satisfies the following.

For any A-algebra C and collection $(C, (\phi_i : C \to B_i))$ where each $\phi_i$ is an A-algebra homomorphism, there is a unique A-algebra homomorphism $f : C \to B$ such that $\pi_i\circ f = \phi_i$ for each $i\in I$ .

Proof

Easy exercise. ♦

Thus $\prod_i B_i$ is the A-algebra analogue of the direct product. Is there an A-algebra analogue of the direct sum?

The natural inclination is to take the direct sum $B = \oplus_{i\in I} B_i$ as an A-module then define multiplication component-wise, but it would not work since $\oplus_i B_i$ does not contain $(1, 1, \ldots)$ . The correct answer is obtained via tensor products, which will be covered in a later article.

In the next article, we will revisit algebraic geometry with a more general framework.

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Commutative Algebra 9

Posted on March 27, 2020 by limsup

Direct Sums and Direct Products

Recall that for a ring A, a sequence of A-modules $M_1, \ldots, M_n$ gives the A-module $M = M_1 \times \ldots \times M_n$ where the operations are defined component-wise. In this article, we will generalize the construction to an infinite collection of modules. Throughout this article, let $(M_i)$ denote a collection of A-modules, indexed by $i\in I$ .

Definition.

The direct product $\prod_{i\in I} M_i$ is the set-theoretic product of the $(M_i)_{i\in I}$ , with the structure of an A-module given by:

$(m_i) + (m_i') := (m_i + m_i'), \quad a\cdot (m_i) := (am_i)$ .

The direct sum $\oplus_{i\in I} M_i$ is the submodule of $\prod_{i\in I} M_i$ comprising of all $(m_i)_{i\in I}$ such that $m_i \ne 0$ only for a finite number of $i\in I$ .

Since there are only finitely many non-zero terms for an element $(m_i) \in \oplus_i M_i$ , one often writes the element additively, i.e. $\sum_i m_i$ .

For a simple example, note that if $M_i = M$ for a fixed module M, then $\prod_{i\in I} M$ is the set of all functions $f : I\to M$ . On the other hand, $\oplus_{i\in I} M$ is the set of all f such that $f(i) \ne 0$ for only finitely many $i\in I$ . [We say f has finite support.]

Exercise A

For an infinite collection $(M_i)$ of A-modules and an ideal $\mathfrak a\subseteq A$ , which of these claims is true?

$\mathfrak a \left( \bigoplus_i M_i\right) = \left( \bigoplus_i \mathfrak a M_i\right), \quad \mathfrak a \left( \prod_i M_i\right) = \left( \prod_i \mathfrak a M_i\right).$

[Note: if it seems too hard to find a counter-example, argue qualitatively why equality does not hold.]

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Universal Property

At first glance, it seems like the direct product is the right definition since it generalizes directly from direct product of groups, rings, topological spaces etc. However, generally the direct sum is better behaved for our needs. The above exercise should provide some evidence of that.

For now, we will show that the direct sum and direct product are actually dual of each other, by showing the corresponding universal properties.

Theorem (Universal Property of Direct Product).

Let $M = \prod_{i\in I} M_i$ . For each $j\in I$ , take the projection map

$\pi_j : \prod_{i\in I} M_i \to M_j, \quad (m_i)_{i\in I} \mapsto m_j.$

Now the collection of data $(M, (\pi_i : M \to M_i)_{i\in I})$ satisfies the following.

For any A-module N and collection of data $(N, (\phi_i : N \to M_i)_{i\in I})$ where each $\phi_i$ is A-linear, there is a unique A-linear $f : N\to M$ such that $\pi_i\circ f = \phi_i$ for each $i \in I$ .

Note

The idea is that homomorphisms $N \to \prod_i M_i$ “classify” the collection of all I-indexed tuples $(\phi_i : N\to M_i)_{i\in I}$ .

universal_property_direct_product

Proof.

For existence, define $f : N\to \prod_i M_i$ to be $n \mapsto (\phi_i(n))_{i\in I} \in \prod_i M_i$ . Now for any $j\in I$ we have

$\pi_j (f(n)) = \pi_j ((\phi_i(n)_i) = \phi_j(n)$ .

For uniqueness, if $j\in I$ , the j-th component of $f(n)$ is $\pi_j(f(n)) = \phi_j(n)$ so the tuple $f(n)$ must be $(\phi_i(n))_{i \in I}$ . ♦

The following result shows why universal properties are important.

Proposition 1.

Let M’ be an A-module and $(M', (\pi'_i : M' \to M_i))$ be a collection of data satisfying the above universal property. Then there is a unique isomorphism $f : M\to M'$ such that $\pi_i' \circ f = \pi_i$ for all $i\in I$ .

Proof

1. We apply the universal property to $N = \prod M_i = M$ and $\phi_i = \pi_i$ for each i. There exists a unique $f : M \to M$ such that $\pi_i \circ f = \pi_i$ . We know one such f, namely $1_M$ . Hence this is the only possibility.

2. Next apply the universal property to N = M’ and $\phi_i = \pi_i'$ . There exists a unique $g : M'\to M$ such that $\pi_i\circ g = \pi_i'$ for each i.

3. But we know $(M', (\pi_i')_i)$ also satisfies this same universal property. Swapping M and M’ there exists a unique $f : M\to M'$ such that $\pi_i' \circ f = \pi_i$ for each i.

4. Now $\pi_i\circ (g\circ f) = \pi_i'\circ f = \pi_i$ for each i. By step 1, we have $g\circ f = 1_M$ . By symmetry we get $f\circ g = 1_{M'}$ too. ♦

Note

In fact, we have proven something stronger: that there is a unique homomorphism $f : M\to M'$ such that $\pi_i' \circ f = \pi_i$ for all i and this f must be an isomorphism.

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Direct Sum

Similarly, the direct sum satisfies the following.

Theorem (Universal Property of Direct Sum).

Let $M = \oplus_{i\in I} M_i$ . For each $j\in I$ , take the embedding

$\epsilon_j : M_j \longrightarrow M, \quad m_j \mapsto (\ldots, 0, 0, m_j, 0, 0, \ldots).$

Thus $\epsilon_j(m_j)_i$ has component $m_j$ at $i = j$ and 0 if $i\ne j$ . The collection of data $(M, (\epsilon_i : M_i \to M)_{i\in I})$ satisfies the following.

For any A-module N and collection of data $(N, (\alpha_i : M_i \to N)_{i\in I})$ where each $\alpha_i$ is A-linear, there is a unique A-linear $f : M\to N$ such that $f\circ \epsilon_i = \alpha_i$ for each $i\in I$ .

Proof.

Left as an exercise. ♦

We also have uniqueness for the above universal property.

Proposition 2.

Let M’ be an A-module and $(M', (\epsilon'_i : M_i \to M))$ be a collection of data satisfying the above universal property. Then there is a unique isomorphism $f: M\to M'$ such that $f\circ \epsilon_i = \epsilon'_i$ for each $i\in I$ .

Proof.

Left as an exercise. ♦

The following exercise tests your conceptual understanding of universal properties.

Important Exercise

Suppose we messed up and take the following maps instead:

embeddings $\epsilon^0_j : M_j \to \prod_{i \in I} M_i$ which take $m_j$ to $(\ldots, 0, 0, m_j, 0, 0, \ldots)$ ;
projections $\pi^0_j : \oplus_{i\in I} M_i \to M_j$ which take $(m_i)_{i\in I}$ to $m_j$ .

Explain why $(\prod_i M_i, (\epsilon^0_i)_{i\in I})$ fails the universal property of direct sum, and $(\oplus_i M_i, (\pi^0_i)_{i\in I})$ fails the universal property of direct product.

[Hint (highlight to read): one of them fails the uniqueness property, the other fails the existence property.]

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Optional Example

Let I be the set of all primes. Consider the direct sum of $\mathbb Z$ -modules $A := \bigoplus_{p\in I} \mathbb Z / p\mathbb Z$ . Consider the $\mathbb Z$ -linear homomorphism (i.e. additive homomorphism)

$\bigoplus_{p\in I} (\mathbb Z / p\mathbb Z) \longrightarrow \mathbb Q/\mathbb Z, \quad (m_p)_{p\in I} \mapsto \sum_{p\in I} \frac{m_p}{p}.$

Note that addition is well-defined because only finitely many $m_p$ are non-zero. Also, $\frac{m_p}p \in \mathbb Q/\mathbb Z$ is independent of our choice of integer $\equiv m_p \pmod p$ .

The map is injective: suppose $\frac{m_1}{p_1} + \ldots + \frac{m_k}{p_k} = 0$ where $p_i$ are distinct primes and each $m_i \not\equiv 0 \pmod {p_i}$ . Multiplying throughout by $p_1 p_2 \ldots p_k$ gives

$(p_1\ldots p_{k-1})m_k + (\text{some terms}) = 0$

where each term in brackets is divisible by $p_k$ . Since $p_1\ldots p_{k-1}$ is coprime to $p_k$ we have $m_k \equiv 0 \pmod {p_k}$ which is a contradiction.

Claim: the image of the map is the set G of all rational $\frac m n$ where n is square free. Indeed, suppose n in $\frac m n$ is square free; write $n = p_1 \ldots p_k$ as a product of distinct primes.

Now the subgroup $H_n := \{\frac m n : 0 \le m < n\}$ of G has order n.
On the other hand, the submodule $B_n := (\mathbb Z/p_1 \mathbb Z) \times \ldots \times (\mathbb Z/p_k \mathbb Z) \subset A$ maps into G injectively with image with $H_n$ .
Since $|B_n| = |H_n|$ , we see that $H_n$ lies in the image.

Thus $\bigoplus_p (\mathbb Z / p\mathbb Z) \cong G$ .

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Posted in Advanced Algebra | Tagged direct products, direct sums, modules, rings, universal properties | 5 Comments

Commutative Algebra 8

Posted on March 26, 2020 by limsup

Generated Submodule

Since the intersection of an arbitrary family of submodules of M is a submodule, we have the concept of a submodule generated by a subset.

Definition.

Given any subset $S\subseteq M$ , let $\Sigma$ denote the set of all submodules of M containing S. Note that $\Sigma$ is non-empty since $M\in \Sigma$ . Now, the submodule of M generated by S is:

$(S) := \cap \{N : N\in \Sigma\}.$

This is the “smallest submodule of M containing S“; to be precise, (a) it is a submodule of M containing S, and (b) any submodule N of M containing S must also contain (S). With this, we see that given a family of submodules $N_i \subseteq M$ , the sum $\sum_i N_i$ is simply the submodule of M generated by their union $\cup_i N_i$ .

Now (S) has a very simple description.

Lemma 1.

For any subset S of M, the submodule $(S)\subseteq M$ consists of the set of all finite linear combinations

$a_1 m_1 + \ldots + a_k m_k, \quad a_i \in A, m_i \in S.$

Thus we may sometimes call (S) the span of S (over the base ring A).

Definition.

A module M is said to be finitely generated (or just finite) if $M = (S)$ for some finite subset S of M.

A finitely generated ideal is an ideal which is finitely generated as an A-module.

warning It is not true that every submodule of a finitely generated module must be finitely generated. In fact, there exist rings with ideals which are not finitely generated, so this gives $\mathfrak a \subseteq A$ where A = (1) is finitely generated but $\mathfrak a$ is not.

E.g. let $A = \mathbb R[x^{1/n} : n = 1, 2, \ldots]$ ; more explicitly

$A = \{ c_0 x^{e_0} + c_1 x^{e_1} + \ldots + c_k x^{e_k} : c_0, \ldots, c_k \in \mathbb R, e_i \in \mathbb Q_{\ge 0} \}.$

Let $\mathfrak m\subset A$ be the set of all sums with only positive $e_i$ . This ideal is not finitely generated, since any finite subset $S \subset \mathfrak m$ is contained in $x^{1/n} \mathbb R[x^{1/n}]$ for some n > 0 so $x^{1/(n+1)} \in \mathfrak m$ lies outside (S).

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Linear Dependence

Note that if $S = \{m_1, \ldots, m_k\} \subseteq M$ is a finite subset, there is a unique A-linear

$f : A^n \to M, \qquad (a_1, \ldots, a_n) \mapsto a_1 m_1 + \ldots + a_k m_k.$

In fact, such a map exists for any finite sequence $m_1, \ldots, m_k$ , i.e. some of the elements may repeat.

This map is surjective if and only if S generates M. Thus M is finitely generated if and only if there is a surjective A-linear map $f : A^n \to M$ for some n > 0. When is f injective?

Proposition 1.

Let $S = \{m_1, \ldots, m_k\} \subseteq M$ and f be as above; f is injective if and only if:

for any $a_1, \ldots, a_k \in A$ , if $a_1 m_1 + \ldots + a_k m_k = 0$ , then all $a_i = 0$ .

When that happens, we say that $m_1, \ldots, m_k$ are linearly independent.

Proof. Easy exercise. ♦

We make two important definitions here, inherited from linear algebra.

Definition.

If S is linearly independent and generates M, we say that S is a basis.

We say M is finite free if it has a finite subset which is a basis.

The rank of a finite free M is the size of S for any basis $S\subseteq M$ .

Note: the zero module is a free module of rank zero since $\emptyset$ generates M.

Subtle question: is the rank well-defined? In other words, if finite subsets $S, S'\subseteq M$ are both bases, must we have $|S| = |S'|$ ? This question will be answered a few articles later.

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Two More Isomorphism Theorems

Just like the case of group theory, we have the following.

Theorem.

Let M be any A-module.

(Second Isomorphism Theorem) For any submodules $N, N'\subseteq M$ we have

$N / (N\cap N') \cong (N + N') / N', \quad n + (N\cap N') \mapsto n + N'.$

(Third Isomorphism Theorem) If we have submodules $P\subseteq N \subseteq M$ , then treating $N/P$ as a submodule of $M/P$ we have

$(M/P)\ /\ (N/P) \cong\ M/N, \quad \overline m + (N/P) \mapsto m + N.$

Note

We have written $m +P \in M/P$ as $\overline m$ to avoid cluttering up the notation.

Proof

For the first claim, map $N \to (N + N')/N'$ by $n\mapsto n + N'$ .

The kernel of this map is $\{n \in N : n + N' = 0 + N'\} = N\cap N'$ .
The map is surjective since any $(n + n') + N'$ for $n\in N, n'\in N'$ is just $n+N'$ .

Hence by the first isomorphism theorem, we get our result.

The second claim is similar: map $M/P \to M/N$ via $m+P \mapsto m+N$ , which is a well-defined homomorphism. Clearly the map is surjective; the kernel is just N/P; now apply the first isomorphism theorem. ♦

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Correspondence of Submodules

Finally, for any submodule P of M, there is a bijective correspondence between

submodules of M containing P, and
submodules of M/P.

Specifically, if N satisfies $P \subseteq N \subseteq M$ in the first case, it corresponds to $N/P \subseteq M/P$ in the second.

submodules_correspondence

[Diagram: correspondence between submodules – arrows indicate inclusion.]

The correspondence respects:

containment: $N/P \subseteq N'/P$ if and only if $N\subseteq N'$ ;
intersection: $\cap_i (N_i/P) = (\cap_i N_i)/P$ ;
sum: $\sum_i (N_i/P) = (\sum_i N_i)/P$ .

Tip: you can prove all these directly, or you can deduce the second and third properties from the first by noting: the intersection of submodules $N_i$ is the largest submodule contained in every $N_i$ . A similar property exists for the sum.

Finally the case for product is tricky: even if N contains P, $\mathfrak a N$ may not contain P. However, $\mathfrak a N + P$ does and we have:

$\mathfrak a (N/P) = (\mathfrak a N + P)/P.$

Exercise. Prove this result.

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Hom Module

For A-modules M and N, let $\mathrm{Hom}_A(M, N)$ be the set of all A-linear maps $f:M\to N$ .

Proposition 2.

$\mathrm{Hom}_A(M, N)$ is an A-module, under the following operations. For A-linear maps $f, g : M\to N$ and $a\in A$ , we have

$f+g : M\to N, \ (f+g)(m) := f(m) + g(m)$ ,

$a\cdot f : M\to N, \ (af)(m) := a\cdot f(m).$

Proof. Easy exercise. ♦

This generalizes the case of vector spaces, where the set of all linear maps $f:V\to W$ of vector spaces, forms a vector space.

The Hom construction thus creates a new A-module out of two existing ones. We have:

Proposition 3.

Let $f:M \to M'$ and $g:N\to N'$ be A-linear maps of A-modules. These induce A-linear maps

$\begin{aligned} f^* : \mathrm{Hom}_A(M', N) \longrightarrow \mathrm{Hom} _A(M, N), \quad &h \mapsto h\circ f\\ g_* : \mathrm{Hom}_A(M, N) \longrightarrow \mathrm{Hom}_A(M, N'), \quad &h\mapsto g\circ h.\end{aligned}$

Proof. Easy exercise. ♦

Example

Suppose $M = A/aA$ for some $a\in A$ , considered as an A-module (not a ring!). Then $\mathrm{Hom}_A(M, N)$ corresponds to the submodule $\{n\in N: a n = 0\}$ of N. Indeed any A-linear map $f : A/aA \to N$ is determined by the image $f(1 + aA) \in N$ . This can be any element $n\in N$ as long as $a n =0$ .

[We usually say n is annihilated by $a\in A$ . This will be covered in greater detail later.]

E.g. if $A = \mathbb Z$ and $M = \mathbb Z/2\mathbb Z$ , then $\mathrm{Hom}_A(M, N)$ picks out the 2-torsion subgroup of N.

Now if $g : N\to N'$ is an A-linear map, the above map $g_*$ can be interpreted as follows:

equation_hom_1

where the composed map is simply g restricted to $\{n : N : an = 0\}$ .

The upshot is: the process of taking the a-torsion elements of a module can be expressed by Hom. We write this process as $N \mapsto \mathrm{Hom}_A(A/aA, N)$ , or simply $\mathrm{Hom}_A(A/aA, -)$ . All these can be formalized in the language of category theory.

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Posted in Advanced Algebra | Tagged free modules, generated submodules, homomorphism, modules, quotient modules, rings, submodules | Leave a comment

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Correspondence of Submodules

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