Exact Sequences

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In the [Multilinear Algebra] category we study the properties of modules. Although we examined properties of modules in [Algebraic Structures], that category focused on what can be defined and proved in common for general algebraic structures, whereas this category focuses more on properties unique to modules.

For example, results for modules defined over a field \(\mathbb{K}\)—that is, \(\mathbb{K}\)-vector spaces—were studied in [Linear Algebra]. Choosing bases for \(\mathbb{K}\)-vector spaces or using them to represent linear maps as matrices are things one can hardly expect in general algebraic structures.

From this perspective, the main goal of the [Multilinear Algebra] category may be said to generalize the results of [Linear Algebra] as far as possible by replacing \(\mathbb{K}\) with a general ring \(A\). On the other hand, because the posts in this category were written with non-specialists or lower-year undergraduates in mind, they hardly used the language of [Category Theory]; thus another goal is to recast the contents of [Linear Algebra] in modern language.

In the posts of this category, an \(A\)-module always means a left \(A\)-module, and the same arguments can be carried out for right \(A\)-modules in an appropriate way. When a left \(A\)-module and a right \(A\)-module must appear together in the same post, we do not make this omission in order to avoid confusion; however, even in such cases one can unfold the same argument by interchanging left \(A\)-modules with right \(A\)-modules and vice versa.

Sum of Modules

The goal of this post is to introduce several exact sequences and define the notion of a split exact sequence. We begin with the following simple lemma.

Lemma 1 For a family \((N_i)_{i\in I}\) of submodules of an \(A\)-module \(M\), the intersection \(\bigcap_{i\in I} N_i\) is a submodule of \(M\).

Proof

The proof of [Linear Algebra] §Bases of Vector Spaces, ⁋Lemma 3 did not use the fact that \(\mathbb{K}\) is a field, so we may use the same proof.

Definition 2 Let a subset \(X\) of an \(A\)-module \(M\) be given. Then we write \(\langle X\rangle\) for the smallest submodule of \(M\) containing \(X\), and call it the submodule generated by \(X\). In this case we call \(X\) a generating set of \(\langle X\rangle\).

If for some submodule \(N\) of \(M\) one can find a finite set \(X\) such that \(N=\langle X\rangle\), then we call \(N\) finitely generated.

Then the following holds.

Proposition 3 For any subset \(X\) of an \(A\)-module \(M\), \(\langle X\rangle\) equals the set of all linear combinations of elements of \(X\).

Proof

The proof is the same as that of [Linear Algebra] §Bases of Vector Spaces, ⁋Lemma 4.

In particular, let a family \((N_i)_{i\in I}\) of submodules of an \(A\)-module \(M\) be given. Then we write \(\sum N_i\) for the submodule \(\left\langle \bigcup N_i\right\rangle\) of \(M\) generated by the subset \(\bigcup N_i\). Now for each \(i\), since \(N_i\) is a submodule of \(M\), there are inclusions \(N_i \hookrightarrow M\). From these there arises the following canonical morphism

\[\bigoplus_{i\in I} N_i \rightarrow M\]

On the other hand, \(\bigoplus N_i\) is generated by the union of the canonical inclusions \(N_i\hookrightarrow\bigoplus_{i\in I}N_i\), and the image of this union under the above canonical morphism is \(\sum N_i\), so the following holds.

Proposition 4 The image of the canonical morphism \(\bigoplus N_i \rightarrow M\) defined above is \(\sum N_i\).

That is, there is an exact sequence

\[\bigoplus_{i\in I} N_i \rightarrow M \rightarrow M\bigg/\sum_{i\in I} N_i \rightarrow 0\tag{$\ast$}\]

Now let a family \((N_i)_{i\in I}\) of submodules of an \(A\)-module \(M\) be given, and let canonical surjections \(M \twoheadrightarrow M/N_i\) be given. Then from these there arises the following canonical morphism

\[M \rightarrow \prod_{i\in I} M/N_i\]

The following proposition is also obvious.

Proposition 5 The kernel of the canonical morphism \(M \rightarrow \prod_{i\in I} M/N_i\) defined above is \(\bigcap N_i\).

That is, there is an exact sequence

\[0 \rightarrow \bigcap_{i\in I} N_i \rightarrow M \rightarrow\prod_{i\in I} M/N_i\]

Direct Sum and Sum of Modules

For a given \(A\)-module \(M\) and a family \((N_i)_{i\in I}\) of submodules of \(M\), if the canonical morphism \(\bigoplus N_i \rightarrow M\) is an isomorphism, we call \(M\) the direct sum of the \(N_i\). For this to be the case, we first know from \((\ast)\) that \(M=\sum N_i\) must hold. That is, every element of \(M\) must be expressible as a linear combination of elements of the \(N_i\). On the other hand, one can check that the canonical morphism \(\bigoplus N_i \rightarrow M\) being injective is equivalent to the expression of such a linear combination being unique. More generally, the following holds.

Proposition 6 In the situation above, the following are all equivalent.

1. \(\sum_{i\in I} N_i=\bigoplus_{i\in I} N_i\).
2. If \(x_i\in N_i\) satisfy \(\sum_{i\in I} x_i=0\), then \(x_i=0\) for all \(i\).
3. For any \(j\in I\), the intersection of \(N_j\) and \(\sum_{i\neq j} N_i\) is \(0\).

Proof

The equivalence of the first two conditions is obvious, and writing \(\bigoplus N_i\) coordinate-wise, it is also obvious that the first condition implies the last. Now assume the last condition and prove the second. Let \(x_i\in N_i\) satisfying \(\sum x_i=0\) be given. Then for any \(j\in I\),

\[x_j=\sum_{i\neq j}(-x_i)\]

and assuming the last condition, the above equation forces \(x_j=0\), completing the proof.

Supplementary Submodules

The following proposition helps us examine Proposition 6 a little more intuitively when \(M\) is the direct sum of two submodules \(N_1,N_2\).

Proposition 7 Let two submodules \(N_1,N_2\) of an \(A\)-module \(M\) be given. Then there exist the following two exact sequences

\[0 \longrightarrow N_1\cap N_2 \overset{\Delta}{\longrightarrow} N_1\oplus N_2 \overset{i_1-i_2}{\longrightarrow} N_1+N_2\longrightarrow 0\]

and

\[0 \longrightarrow M/(N_1\cap N_2)\overset{\Delta'}{\longrightarrow}(M/N_1)\oplus(M/N_2)\overset{p_1-p_2}{\longrightarrow}M/(N_1+N_2)\longrightarrow 0\]

Here \(i_k,p_k\) are the canonical morphisms

\[i_k: N_k \rightarrow N_1+N_2,\qquad p_k:M/N_k \rightarrow M/(N_1+N_2)\]

and \(\Delta, \Delta'\) are the morphisms defined respectively by the formulas

\[\Delta(x)=(x,x),\qquad \Delta'(x+(N_1\cap N_2))=(x+N_1,x+N_2)\]

The proof of this is a straightforward computation, so we omit it. Anyway, to explain the first exact sequence,

\[N_1\oplus N_2=\{(x_1,x_2)\mid x_k\in N_k\}\]

so no matter how \(N_1\) and \(N_2\) are situated inside \(M\), the elements of \(N_1\) and \(N_2\) appear inside \(N_1\oplus N_2\) as elements of the form \((x_1,0)\) and \((0,x_2)\) respectively, and hence are treated as distinct. In particular, even if \(x\in N_1\cap N_2\), as long as \(x\neq 0\), \(x\in N_1\) and \(x\in N_2\) are different elements inside \(N_1\oplus N_2\). However,

\[i_1-i_2:N_1\oplus N_2 \rightarrow N_1+N_2;\qquad (x_1,x_2)\mapsto x_1-x_2\]

once we send these to \(N_1+N_2\) via this map, the elements that were treated as distinct in this way must have image \(0\), and therefore the kernel of \(i_1-i_2\) is exactly \(N_1\cap N_2\).

Meanwhile, we make the following definition.

Definition 8 For any \(A\)-module \(M\) and two submodules \(N_1, N_2\) of \(M\), if \(M\) is the direct sum of \(N_1\) and \(N_2\), we call \(N_1,N_2\) supplementary submodules of each other. If a submodule \(N\) of \(M\) has a supplementary submodule, we call \(N\) a direct summand of \(M\).

Thus, in such a situation \(N_1+N_2=M\) and \(N_1\cap N_2=0\). Then using this, one can check that restricting the domain of the canonical morphism \(M \rightarrow M/N_1\) to \(N_2\) is an isomorphism, and similarly that restricting the domain of \(M \rightarrow M/N_2\) to \(N_1\) is an isomorphism.

Split Exact Sequences

Finally, before defining a split exact sequence, we introduce the following lemma. Its proof can be found in [Homological Algebra] §Diagram Chasing, ⁋Proposition 1.

Lemma 9 (Four lemma) Consider a commutative diagram whose rows are exact

and assume that \(\alpha\) is surjective and \(\delta\) is injective. Then

1. If \(\gamma\) is surjective then \(\beta\) is also surjective.
2. If \(\beta\) is injective then \(\gamma\) is also injective.

We now define a split exact sequence via the following proposition. A split exact sequence is an exact sequence of the form

\[0\rightarrow N_1\hookrightarrow N_1\oplus N_2\twoheadrightarrow N_2 \rightarrow 0\]

To spell this out precisely:

Proposition 10 For an exact sequence of \(A\)-modules

\[0\longrightarrow M \overset{u}{\longrightarrow}L \overset{v}{\longrightarrow}N \longrightarrow 0\]

the following conditions are all equivalent.

1. There exists a linear retraction \(r:L \rightarrow M\) of \(u\). ([Set Theory] §Retractions and Sections, ⁋Definition 2)
2. There exists a linear section \(s:N \rightarrow L\) of \(v\). ([Set Theory] §Retractions and Sections, ⁋Definition 2)

3. There exists an isomorphism \(\alpha: L \rightarrow M\oplus N\) making the following diagram commute:

   

Proof

First assume condition 3. Then setting \(r=\pr_M\circ\alpha\) gives condition 1, and similarly, composing the canonical inclusion \(i_N: N \rightarrow M\oplus N\) with \(\alpha^{-1}\) and setting \(s=\alpha^{-1}\circ i_N\) gives condition 2.

For the remaining directions, we assume conditions 1 and 2 respectively and prove condition 3. If condition 1 holds, define \(\beta: M\oplus N \rightarrow L\) by \((x,y)\mapsto u(x)+s(y)\); if condition 2 holds, define \(\alpha:L \rightarrow M\oplus N\) by \(z\mapsto (r(z), v(z))\). Then by Lemma 9 (Four lemma) one sees that \(\alpha,\beta^{-1}\) define the isomorphism required in condition 3.