Commutative Algebra/Modules, submodules and homomorphisms

Basics
Analogously, one can define right $$R$$-modules with an operation $$R \times M \to M, (r, m) \mapsto mr$$; the difference is only formal, but it will later help us define bimodules in a user-friendly way.

For the sake of brevity, we will often write module instead of left $$R$$-module.


 * Exercise 5.1.1: Prove that every Abelian monoid $$(M, +)$$ together with an operation as specified in 1.) - 4.) of definition 5.1 is already a module.

Submodules
The following lemma gives a criterion for a subset of a module being a submodule.

Lemma 5.3:

A subset $$N \subseteq M$$ is a submodule iff
 * $$\forall r \in R, n, q \in N: rn - q \in N$$.

Proof:

Let $$N$$ be a submodule. Then since $$-q \in N$$ since we have an Abelian group and further $$rn \in N$$ due to closedness under the module operation, also $$rn + (-q) =: rn - q \in N$$.

If $$N$$ is such that $$\forall r \in R, n, q \in N: rn - q \in N$$, then for any $$n, m \in N$$ also $$n + m = n + (-1_R)(-m) \in N$$.

Proof:

Well-definedness: If $$m + N = p + N$$, then $$m - p \in N$$, hence $$r(m - p) = rm - rp \in N$$ and thus $$rm + N = rp + N$$.


 * 1) $$1_R (m + N) = (1_r m) + N = m + N$$
 * 2) $$r (n + N + m + N) = r ((m + n) + N) = r(m + n) + N = rm + rn + N = rm + N + rn + N$$
 * 3) $$(r+s)(m+N) = (r + s)m + N = rm + sm + N = rm +N + rn + N$$
 * 4) analogous to 3. (replace $$+$$ by $$\cdot$$)

Sum and intersection of submodules
We shall now ask the question: Given a module $$M$$ and certain submodules $$\{N_\alpha\}_{\alpha \in A}$$, which module is the smallest module containing all the $$N_\alpha$$? And which module is the largest module that is itself contained within all $$N_\alpha$$? The following definitions and theorems answer those questions.

Proof:

1. $$\sum_{\alpha \in A} N_\alpha$$ is a submodule:


 * It is an Abelian subgroup since if $$\sum_{l=1}^k r_l n_{\alpha_l}, \sum_{j=1}^m s_l n_{\beta_j} \in \sum_{\alpha \in A} N_\alpha$$, then
 * $$\sum_{l=1}^k r_l n_{\alpha_l} - \sum_{j=1}^m s_l n_{\beta_j} = \sum_{l=1}^k r_l n_{\alpha_l} + \sum_{j=1}^m (-s_l) n_{\beta_j} \in \sum_{\alpha \in A} N_\alpha$$.


 * It is closed under the module operation, since
 * $$s \left( \sum_{l=1}^k r_l n_{\alpha_l} \right) = \sum_{l=1}^k (s r_l) n_{\alpha_l} \in \sum_{\alpha \in A} N_\alpha$$.

2. Each $$N_\alpha$$ is contained in $$\sum_{\alpha \in A} N_\alpha$$:

This follows since $$1_r n_\alpha \in \sum_{\alpha \in A} N_\alpha$$ for each $$\alpha \in A$$ and each $$n_\alpha \in N_\alpha$$.

3. $$\sum_{\alpha \in A} N_\alpha$$ is the smallest submodule containing all the $$N_\alpha$$: If $$K \le M$$ is another such submodule, then $$K$$ must contain all the elements
 * $$\sum_{l=1}^k r_l n_{\alpha_l}, k \in \mathbb N, r_l \in R, n_{\alpha_l} \in N_{\alpha_l}$$

due to closedness under addition and submodule operation.

Proof:

1. It's a submodule: Indeed, if $$r \in R, n, p \in \bigcap_{\alpha \in A} N_\alpha$$, then $$n, p \in N_\alpha$$ for each $$\alpha$$ and thus $$n - rp \in N_\alpha$$ for each $$\alpha$$, hence $$n-rp \in \bigcap_{\alpha \in A} N_\alpha$$.

2. It is contained in all $$N_\alpha$$ by definition of the intersection.

3. Any set that contains all elements from each of the $$N_\alpha$$ is contained within the intersection.

We have the following rule for computing with intersections and sums:

Proof:

$$\subseteq$$: Let $$l + n \in (L + N) \cap K$$. Since $$L \subseteq K$$, $$l \in K$$ and hence $$n \in K$$. Since also $$n \in N$$ by assumption, $$l + n \in L + K \cap N$$.

$$\supseteq$$: Let $$l + m \in L + (K \cap N)$$. Since $$L \subseteq K$$, $$l \in K$$ and since further $$m \in K$$, $$l + m \in K$$. Hence, $$l + m \in K \cap (L + N)$$.

More abstractly, the properties of the sum and intersection of submodules may be theoretically captured in the following way:

Lattices
There are some special types of lattices:

In fact, it suffices to require conditions 1. and 2. only for sets $$S$$ with two elements. But as we have shown, in the case that $$L$$ is the set of all submodules of a given module, we have the "original" conditions satisfied.

Proof:

First, we note that least upper bound and greatest lower bound are unique, since if for example $$u, u'$$ are least upper bounds of $$S$$, then $$u \le u'$$ and $$u' \le u$$ and hence $$u = u'$$. Thus, the joint and meet operation are well-defined.

The commutative laws follow from $$\{a, b\} = \{b, a\}$$.

The idempotency laws from clearly $$a$$ being the least upper bound, as well as the greatest lower bound, of the set $$\{a, a\}$$.

The first absorption law follows as follows: Let $$u$$ be the least upper bound of $$\{a, b\}$$. Then in particular, $$u \ge a$$. Hence, $$a$$ is a lower bound of $$\{a, u\}$$, and any lower bound $$l$$ satisfies $$l \le a$$, which is why $$a$$ is the greatest lower bound of $$\{a, u\}$$. The second absorption law is proven analogously.

The first associative law follows since if $$u$$ is the least upper bound of $$\{a, b, c\}$$ and $$v$$ is the upper bound of $$\{a, b\}$$, then $$u \ge v$$ (as $$u$$ is an upper bound for $$\{a, b\}$$) and if $$w$$ is the least upper bound of $$\{v, c\}$$, then $$w = u$$ since $$u$$ is an upper bound and further, $$w \ge v \ge a$$ and $$w \ge b$$. The same argument (with $$a$$ and $$c$$ swapped) proves that $$u$$ is also the least upper bound of the l.u.b. of $$\{b, c\}$$ and $$a$$. Again, the second associative law is proven similarly.

From theorems 5.5-5.7 and 5.10 we note that the submodules of a module form a modular lattice, where the order is given by set inclusion.

Exercises

 * Exercise 5.2.1: Let $$R$$ be a ring. Find a suitable module operation such that $$R$$ together with its own addition and this module operation is an $$R$$-module. Make sure you define this operation in the simplest possible way. Prove further, that with respect to this module operation, the submodules of $$R$$ are exactly the ideals of $$R$$.

Homomorphisms
We shall now get to know the morphisms within the category of modules over a fixed ring $$R$$.

Since we are cool, we will often simply write morphisms instead of homomorphisms where it's clear from the context in order to indicate that we have a clue about category theory.

We have the following useful lemma:

Lemma 5.12:

$$f: M \to N$$ is $$R$$-linear iff
 * $$\forall r \in R, m, p \in M : f(rm + p) = rf(m) + f(p)$$.

Proof:

Assume first $$R$$-linearity. Then we have
 * $$f(rm + p) = f(rm) + f(p) = rf(m) + f(p)$$.

Assume now the other condition. Then we have for $$m, p \in M$$
 * $$f(m + p) = f(1_R m + p) = 1_R f(m) + f(p) = f(m) + f(p)$$

and
 * $$f(rm) = f(rm + 0) = rf(m) + f(0) = rf(m)$$

since $$f(0) = 0$$ due to $$f(0) = f(0 + 0) = f(0) + f(0)$$; since $$M$$ is an abelian group, we may add the inverse of $$f(0)$$ on both sides.

Lemma 5.13:

If $$f: M \to N$$ is $$R$$-linear, then $$\forall m \in M: f(-m) = -f(m)$$.

Proof:

This follows from the respective theorem for group homomorphisms, since each morphism of modules is also a morphism of Abelian groups.

Lemma 5.14:

Let $$f:M \to N$$ be a morphism. The following are equivalent:
 * 1) $$f$$ is an isomorphism
 * 2) $$\ker f = \{0\}$$
 * 3) $$f$$ has an inverse which is an isomorphism

Proof:

Lemma 5.15:

The kernel and image of morphisms are submodules.

Proof:

1. The kernel:


 * $$f(rn - q) = rf(n) + f(-q) = rf(n) - f(q) = 0$$

2. The image:


 * $$rf(m) \overbrace{- f(p)}^{= + f(-p)} = f(rm - p)$$

The following four theorems are in complete analogy to group theory.

Proof:

We define $$\overline \varphi(m + N) := \varphi(m)$$. This is well-defined since $$\ker \varphi \subseteq N$$. Furthermore, this definition is already enforced by $$\overline \varphi \circ \pi = \varphi$$. Further, $$\overline \varphi(m + N) = 0 \Leftrightarrow m \in \ker \varphi$$.

Proof:

We set $$N = \ker f$$ and obtain a homomorphism $$\overline f: M /\ker f \to K$$ with kernel $$N/N$$ by theorem 5.11. From lemma 5.16 follows the claim.

Proof:

Since $$L \le N$$ and $$N \le M$$ also $$L \le M$$ by definition. We define the function
 * $$\varphi: M/L \to M/N, m + L \mapsto m + N$$.

This is well-defined since
 * $$m + L = p + L \Leftrightarrow m - p \in L \Rightarrow m - p \in N \Leftrightarrow m + N = p + N$$.

Furthermore,
 * $$m + L \in \ker \varphi \Leftrightarrow m + N = 0 + N \Leftrightarrow m \in N$$

and hence $$\ker \varphi = N/L$$. Hence, by theorem 5.17 our claim is proven.

Proof:

Consider the isomorphism
 * $$\varphi: L \to (L + N)/N, \varphi(l) := l + N$$.

Then $$\varphi(l) = 0 \Leftrightarrow l \in N$$, which is why the kernel of that homomorphism is given by $$L \cap N$$. Hence, the theorem follows by the first isomorphism theorem.

And now for something completely different:

Proof:

Let $$a, b \in \varphi^{-1}(L)$$. Then $$\varphi(a+b) = \varphi(a) + \varphi(b) \in L$$ and hence $$a + b \in \varphi^{-1}(L)$$. Let further $$r \in R$$. Then $$\varphi(ra) = r \varphi(a) \in L$$.

Similarly:

Proof: Let $$a, b \in \varphi(K)$$. Then $$a = \varphi(i), b = \varphi(j)$$ and $$a + b = \varphi(i + j) \in \varphi(K)$$. Let further $$r \in R$$. Then $$ra = \varphi(ri) \in \varphi(K)$$.

Exercises

 * Exercise 5.3.1: Let $$R, S$$ be rings regarded as modules over themselves as in exercise 5.2.1. Prove that the ring homomorphisms $$\varphi: R \to S$$ are exactly the module homomorphisms $$R \to S$$; that is, every ring hom. is a module hom. and vice versa.

The projection morphism
The following two fundamental equations for $$\pi_N (\pi_N^{-1}(S))$$ and $$\pi_N^{-1}(\pi_N(K))$$ shall gain supreme importance in later chapters, $$S \subseteq M/N$$, $$K \le M$$.

Proof:

Let first $$m + N \in S$$. Then $$m \in \pi^{-1}(S)$$, since $$\pi_N(m) = m + N$$. Hence, $$m + N \in \pi_N(\pi^{-1}(S))$$. Let then $$m + N \in \pi_N (\pi_N^{-1}(S))$$. Then there exists $$m' \in \pi_N^{-1}(S)$$ such that $$\pi_N(m') = m + N$$, that is $$m' + N = m + N$$. Now $$m' \in \pi_N^{-1}(S)$$ means that $$\pi(m') = m' + N \in S$$. Hence, $$m + N = m' + N \in S$$.

Let first $$m \in K + N$$, that is, $$m = k + n$$ for suitable $$k \in K$$, $$n \in N$$. Then $$\pi_N(m) = k + n + N = k + N = \pi_N(k) \in \pi_N(K)$$, which is why by definition $$m \in \pi_N^{-1}(\pi_N(K))$$. Let then $$m \in \pi_N^{-1}(\pi_N(K))$$. Then $$\pi_N(m) = m + N \in \pi_N(K)$$, that is $$m + N = k + N$$ with $$k \in K$$, that is $$m = k + n$$ for a suitable $$n \in N$$, that is $$m \in K + N$$.

The following lemma from elementary set theory have relevance for the projection morphism and we will need it several times:

Lemma 5.24:

Let $$f: S \to T$$ be a function, where $$S, T$$ are completely arbitrary sets. Then $$f$$ induces a function $$2^S \to 2^T$$ via $$A \mapsto f(A)$$, the image of $$A$$, where $$A \subseteq S$$. This function preserves inclusion. Further, the function $$2^T \to 2^S, B \mapsto f^{-1}(B)$$, also preserves inclusion.

Proof:

If $$A' \subseteq A$$, let $$y' \in f(A')$$. Then $$y' = f(x')$$ for an $$x' \in A' \subseteq A$$. Similarly for $$f^{-1}$$.