# 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.

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 $A \to B \stackrel f \to B/\mathfrak q$ 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.

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 $\mathfrak p\in V(\mathfrak a) \cup V(\mathfrak b)$. ♦

# Structure of Spec A

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

Proposition.

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.

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.

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.

# 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.

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

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 $V(\mathfrak a)$.

# Comparison with Algebraic Geometry

Proposition.

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

From the above proposition, prove that we have a bijection

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.

# Some Results on Posets

• 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 $x=y$.

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

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

1. The set of real numbers (or rational numbers, or integers) forms a totally ordered set under the usual arithmetic ordering.
2. 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.
3. Any subset T of a partially ordered set S gives a partially ordered set. If S is totally ordered, so is T.

Lemma (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.

# 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 $x \ge m_0$.
• maximal element of S is an $m\in S$ such that for any $x\in S$, if $x \ge m$ then $x=m$.
• minimal element of S is an $m'\in S$ such that for any $x\in S$, if $x \le m'$ then $x=m'$.

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 = {abc} and let S be the following subsets of P(X) under inclusion.

The following properties are obvious.

Lemma.

• 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.

# 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

1. The set $\mathbb N$ of positive integers under ≤ is noetherian.
2. The set $\mathbb N \times \mathbb N$ is noetherian, where $(m, n) \le (m', n')$ if $m\le m'$ and $n \le n'$.
3. 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 $(y \in S, y < x \implies x \in T)$, 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.

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. ♦

# 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.

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. ♦

# 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:

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

recovers the set of points for Bk[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.

# 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 kschemes 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

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).
• 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).
• 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 $V\coprod W$ in the above table.

# 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 $\epsilon \mapsto 0$.

We imagine # as a “fattened point” with arrows sticking out.

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

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

## 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 $\mathbb V \subset A^2$ 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.

[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.

[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

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

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

# 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.

• 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

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

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. 🙂

# 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.

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  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.

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

# 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. However, it is useful for classifying certain constructions from A-algebras.

As an example fix the A-algebra $B = A[X, Y]/(XY - 1)$. 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 rigourously, one needs the language of category theory.

Exercise

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 rigourously, one requires B to have a Hopf algebra structure.

# 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

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

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.

# 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.

# 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

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.]

# 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 $(\phi_i : N\to M_i)_{i\in I}$.

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.

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 NM’ 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.

# 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.

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 $\pi^0_j : \oplus_{i\in I} M_i$ 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.]

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

# 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.

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.

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

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).

# 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.

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.

# 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. ♦

# 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.

[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.

# 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.

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

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:

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.

# Modules

Having dipped our toes into algebraic geometry, we are back in commutative algebra.

Next we would like to introduce “linear algebra” over a ring A. Most of the proofs should pose no difficulty to the reader so we will skip them. We recommend that the reader prove these properties at least once.

Definition.

An A-module is an abelian group $(M, +)$ together with a map $A\times M \to M$, written as $(a,m) \mapsto am$, such that the following hold for all $a, a' \in A$ and $m, m' \in M$.

• $1_A \cdot m = m$.
• $(a+a')m = am + a'm \in M$.
• $(aa')m = a(a'm) \in M$.
• $a(m + m') = am + am' \in M$.

### Alternative Look

If we fix $a\in A$, we obtain a “multiplication-by-a” map defined by $\mu_a : M\to M, m\mapsto am$. The last property says $\mu_a : M\to M$ is an (additive) homomorphism so we get a map $A \to \mathrm{End}(A)$, where $\mathrm{End}(M)$ is the set of all group homomorphisms $(M, +) \to (M, +)$. Now $\mathrm{End}(M)$ is not just any old set – it has a natural structure of a (non-commutative) ring if we introduce the following definitions:

$\phi, \psi \in \mathrm{End}(M) \implies \begin{cases} (\phi + \psi) : m \mapsto \phi(m) + \psi(m),\\ (\phi\psi) : m \mapsto \phi(\psi(m)). \end{cases}$

In words, we say “addition is defined on the image and product is given by composition”.

Now that $\mathrm{End}(M)$ has a ring structure, the first three axioms of the definition says $A \to \mathrm{End}(A), a\mapsto \mu_a$ is a ring homomorphism.

### Easy Properties

Immediately, we can show the following:

$a\in A, m\in M \implies \begin{cases} 0_A\cdot m = a\cdot 0_M = 0_M,\\ (-a)m = a(-m) = -(am).\end{cases}$

E.g. a(-m) = -(am) because $\mu_a : M\to M$ is an additive homomorphism.

### Examples

1. Since the homomorphism $\mathbb Z \to \mathrm{End}(M)$ is unique, every abelian group is automatically a $\mathbb Z$-module.

2. A module over a field k is the same as a vector space over k.

3. Any ring A is a module over itself. More generally if $A\subseteq B$ is a subring, then B is an A-module.

4. If M and M’ are A-modules, then $M\times M'$ becomes an A-module via

$(m_1, m_1') + (m_2, m_2') = (m_1 + m_2, m_1' + m_2'), \quad a(m, m') = (am, am')$.

5. From examples 3 and 4, we get $M = A^n = \{ (a_1, \ldots, a_n) : a_1, \ldots, a_n \in A\}$ with

$(a_1, \ldots, a_n) + (a_1', \ldots, a_n') = (a_1 + a_1', \ldots, a_n + a_n'), \quad a\cdot (a_1, \ldots, a_n) := (aa_1, \ldots, aa_n)$.

6. Let $A = B \times B'$ be a product of two rings. Then any B-module M gives an A-module $M\times \{0\}$; similarly, any B’-module M’ gives an A-module $\{0\} \times M'$.

7. More generally for $A = B\times B'$, any B-module N and B’-module N’ gives an A-module $M = N\times N'$.

# Basic Constructions

Just as we have subspaces of vector spaces, we have:

Definition.

If M is an A-module, a A-submodule is a subgroup N of $(M, +)$ such that for all $a\in A, m\in N$ we have $am\in N$.

And this gives:

Proposition.

Let N be an A-submodule of M. The quotient of $(M, +)$ by the subgroup N has a natural structure of an A-module, given by:

$A \times (M/N) \to M/N, \quad (a, m + N) \mapsto am + N$.

Thus we obtain the quotient module $M/N$.

Proof. Easy exercise. ♦

Next, we have the linear maps.

Definition.

Let M, N be A-modules. A module homomorphism is a function $f:M\to N$ such that

$a\in A, m, m'\in M \implies \begin{cases} f(m + m') = f(m) + f(m'),\\ f(am) = a\cdot f(m).\end{cases}$

We will also say f is an A-linear map (or just linear map if the base ring is implicit).

A monomorphism (resp. epimorphismisomorphism) of modules is an injective (resp. surjective, bijective) homorphism of modules.

Immediately we have the following.

First Isomorphism Theorem.

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

• The kernel of f, $\mathrm{ker} f := \{x \in M : f(x) = 0\}$, is a submodule of M.
• The image of f, $\mathrm{im} f := \{f(x) \in N : x\in M\}$, is a submodule of N.
• The map f induces an isomorphism of modules

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

Proof. Easy exercise. ♦

Exercise

Let $A = B\times B'$ be a product of two rings. For a B-module N and B’-module N’, take the A-module $M = N \times N'$. Must every submodule of M be of the form $N_1 \times N_1'$ for B-submodule $N_1 \subseteq N$ and B’-submodule $N_1' \subseteq N'$?

# Ideals as Modules

Recall that every non-trivial ring A comes with two modules.

• The zero module is the trivial group (0, +), and $a\cdot 0 = 0$ for all $a\in A$.
• The ring A is a module over itself, where $A\times M\to M$ is given by ring product $A\times A \to A$.

Lemma.

For any ring A, a subset of A is a submodule if and only if it is an ideal of A.

Proof. Easy exercise. ♦

## Some Notes

Although the lemma is a trivial observation, it highlights an important philosophy: modules can be considered as a generalization of ideals. For an ideal $\mathfrak a\subseteq A$, it is geometrically more natural to associate it with $A/\mathfrak a$ as a module, although the reasons are not clear for now.

Note that an ambiguity may arise if we were sloppy with words: for $A/\mathfrak a$, as a module, could mean the ring $A/\mathfrak a$ as a module over itself, or the quotient A-module of $A$ by the submodule $\mathfrak a$. We will try to be extra careful in stating the base ring. Often, the base ring is used as a subscript in notation, e.g. $C_A(M)$ means a certain construction from M by regarding it as an A-module.

For readers with some experience with differentiable manifolds, another geometrical application of modules is as follows. Recall that closed sets V correspond with their coordinate ring k[V]. Modules over k[V] then correspond to “vector bundles” over the set V. Of special interest are those bundles which are “locally free”, i.e. close to each point the vector bundle restricts to a trivial one.

# Operations on Submodules

Finally, the operations on ideals can be generalized.

Proposition.

Let $(N_i)$ be a collection of submodules of M, over some fixed base ring.

• $\cap N_i$ is a submodule of M.
• Let $\sum_i N_i$ be the set of all finite sums $m_{1} + \ldots + m_{k} \in M$, where $m_{1} \in N_{i_1}, \ldots, m_k \in N_{i_k}$. This is a submodule of M.
• If $\mathfrak a$ is an ideal of A, let $\mathfrak a M$ be the set of all finite sums $a_1 m_1 + \ldots + a_k m_k$ for $a_i \in \mathfrak a$ and $m_i \in M$. This is a submodule of M.

Proof. Easy exercise. ♦

Exercise

An A-module is said to be simple if it is non-zero and its only submodules are 0 and itself. Prove that a simple A-module must be isomorphic to $A/\mathfrak m$ for some maximal ideal $\mathfrak m\subset A$.

Note

When MA and each $N_i = \mathfrak a_i$ is an ideal of A, these definitions are consistent with the earlier ones for ideals. Thus we can think of them as generalizations of the corresponding operations on ideals.

Note that product of submodules is not defined since we cannot multiply elements in a module. But as we shall see later, one can forcibly define a “product space” P for modules M and N so that $m\in M$ and $n\in N$ multiply to give an element in P. That leads us to tensor products.

The next result is important, so we will state it separately.

Proposition.

If M is an A-module and $\mathfrak a$ is an ideal of A, then $M/\mathfrak a M$ has a canonical structure of an $(A/\mathfrak a)$-module, via

$(A/\mathfrak a) \times (M/\mathfrak a M) \longrightarrow (M/\mathfrak a M), \quad (a + \mathfrak a, m + \mathfrak a M) \mapsto am + \mathfrak aM$.

Proof. Easy exercise. ♦

The above construction is a “canonical” way of constructing a module over $A/\mathfrak a$ given a module over A. We will state more precisely what canonicity means in later articles, by universal properties.