# Some Homological Algebra Computations

In this post, I’m going to write down the detailed proofs of some of the exercises in Rotman’s Homological Algebra. They were asked in ML and then answered by me.

1. Let $A$ be a torsion abelian group. Then $\text{Ext}^1_\mathbb{Z}(A, \mathbb{Z}) \cong \text{Hom}_\mathbb{Z}(A,S^1)$, where $S^1$ is the unit circle.

One point is that the structure of the circle group is $\mathbb{Q}/\mathbb{Z} \oplus \mathbb{R}.$ Now, since $A$ is torsion, then $\text{Hom}(A,\mathbb{R})=0.$ Therefore, it is enough to show that $\text{Ext}^1_\mathbb{Z}(A, \mathbb{Z}) \cong \text{Hom}_\mathbb{Z}(A,\mathbb{Q}/\mathbb{Z}).$

We know that $\text{Ext}(A,-)$ functor is the right derived functor of the left exact (covariant) functor $\text{Hom}(A,-).$ Consider the following exact sequence of abelian groups (injective resolution)

$0 \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Q} \longrightarrow \mathbb{Q}/\mathbb{Z} \longrightarrow 0$

where obviously, the second arrow is inclusion and the third one is the projection.

Applying our left exact (covariant) functor $\text{Hom}(A, -)$ to the above exact sequence, we will obtain the following long exact sequence

$0 \longrightarrow \text{Hom}(A,\mathbb{Z}) \longrightarrow \text{Hom}(A,\mathbb{Q}) \longrightarrow \text{Hom}(A,\mathbb{Q}/\mathbb{Z}) \longrightarrow \text{Ext}^1(A, \mathbb{Z}) \longrightarrow 0$

Notice that, $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$ are divisible groups, hence are injective $\mathbb{Z}$-modules, thus, $\text{Ext}^i(A, \mathbb{Q})=\text{Ext}^i(A, \mathbb{Q}/\mathbb{Z})=0$ for $i>0.$

Similarly, since $A$ is torsion then $\text{Hom}(A,\mathbb{Z})=\text{Hom}(A,\mathbb{Q})=0.$ Hence,

$\text{Ext}^1(A, \mathbb{Z}) \cong \text{Hom}(A,\mathbb{Q}/\mathbb{Z})$

as desired.

2. Let $A, B$ be finite abelian groups. Prove that $\text{Ext}^1_\mathbb{Z} (A,B) \cong A \otimes_\mathbb{Z} B.$

By the classification of finitely generated abelian groups, we can assume that $A \cong \mathbb{Z}/q_1 \oplus \mathbb{Z}/q_2 \oplus \cdots \oplus \mathbb{Z}/q_k$ and $B\cong \mathbb{Z}/p_1 \oplus \mathbb{Z}/p_2 \oplus \cdots \oplus \mathbb{Z}/p_l$ where $q_i, p_j$ are (not necessarily distinct) prime numbers.

Lemma: Let $A$ be a finite abelian group, then

$\text{Ext}^1(A, \mathbb{Z}/m) \cong \frac{\text{Hom}(A, \mathbb{Q}/\mathbb{Z})}{m \text{Hom}(A, \mathbb{Q}/\mathbb{Z})}.$

Proof of the lemma: Consider the following injective resolution for $\mathbb{Z}/m$

$0 \longrightarrow \mathbb{Z}/m \longrightarrow \mathbb{Q}/\mathbb{Z} \longrightarrow \mathbb{Q}/\mathbb{Z} \longrightarrow 0$

where the second arrow is multiplication by $1/m$ and the third one is multiplication by $m.$

Applying $\text{Hom}(A, -)$ to the above short exact sequence, we will get

$0 \longrightarrow \text{Hom}(A, \mathbb{Z}/m) \longrightarrow \text{Hom}(A,\mathbb{Q}/\mathbb{Z}) \longrightarrow \text{Hom}(A,\mathbb{Q}/\mathbb{Z}) \longrightarrow \text{Ext}^1(A, \mathbb{Z}/m) \to 0$

Note that, injectivity of $\mathbb{Q}/\mathbb{Z},$ implies that $\text{Ext}^i(A,\mathbb{Q}/\mathbb{Z})=0$ for $i>0.$ Hence, the required isomorphism. (Notice that the third arrow is the induced multiplication by $m$ from our resolution.)

In particular, setting $A=\mathbb{Z}/n$ in the lemma and using the result of the question 1), we get

$\text{Ext}^1(\mathbb{Z}/n, \mathbb{Z}/m) \cong \frac{\text{Hom}(\mathbb{Z}/n, \mathbb{Q}/\mathbb{Z})}{m \text{Hom}(\mathbb{Z}/n, \mathbb{Q}/\mathbb{Z})} \cong \frac{\text{Ext}^1(\mathbb{Z}/n, \mathbb{Z})}{m \text{Ext}^1(\mathbb{Z}/n, \mathbb{Z})} \cong \frac{ \mathbb{Z}/n}{m \mathbb{Z}/n} \cong \mathbb{Z}/d$

Note that, $\text{Ext}^1(\mathbb{Z}/n, \mathbb{Z}) \cong \mathbb{Z}/n$ and $\mathbb{Z}/d \cong \mathbb{Z}/n \otimes \mathbb{Z}/m$ where $d=\gcd(n,m).$ Therefore,

$\text{Ext}^1(\mathbb{Z}/n, \mathbb{Z}/m) \cong \mathbb{Z}/n \otimes \mathbb{Z}/m \;\;\;\; (\star)$

Utilizing the commutativity of $\text{Ext}$ and tensor product with (finite) direct sum, and $( \star),$ we obtain the required isomorphism,

$\text{Ext}^1(A,B)=\text{Ext}^1(\mathbb{Z}/q_1 \oplus \mathbb{Z}/q_2 \oplus \cdots \oplus \mathbb{Z}/q_k, \mathbb{Z}/p_1 \oplus \mathbb{Z}/p_2 \oplus \cdots \oplus \mathbb{Z}/p_l) \cong \bigoplus_{i,j} \text{Ext}^1(\mathbb{Z}/q_i, \mathbb{Z}/p_j) \cong \bigoplus_{i,j} (\mathbb{Z}/q_i \otimes \mathbb{Z}/p_j) \cong (\mathbb{Z}/q_1 \oplus \mathbb{Z}/q_2 \oplus \cdots \oplus \mathbb{Z}/q_k) \otimes (\mathbb{Z}/p_1 \oplus \mathbb{Z}/p_2 \oplus \cdots \oplus \mathbb{Z}/p_l) \cong A \otimes B.$

3. Let $_RF$ be a flat left $R$-module and $P$ be its projective covering, i.e. there is some $R$-module epimorphism $p: P \rightarrow F$. Prove that if $P$ is flat, then $\text{Ker}(p)$ is flat.

Rotman’s definition of projective cover of a module $F$ is indeed, an ordered pair $(P, p),$ where $P$ is projective and $p:P \to F$ is a surjective morphism with $\text{ker}(p)$ a superfluous submodule of $P.$

I will use the following homological characterization for (left) flat $R$-modules,

$\text{Tor}^R_n(-,M)=0 \;\; \text{for all} \;\; n \geq 1 \iff _RM \;\; \text{is a flat left} \; R\text{-module}$

By assumption, $0 \longrightarrow M \hookrightarrow P \longrightarrow F \longrightarrow 0$ is a short exact sequence of left $R$-modules, where $M:=\text{ker}(p).$

Let $A$ be a arbitrary right $R$-module, then after applying the right exact functor $A \otimes_R -$ to the above exact sequence, we will have

$0 \to \text{Tor}^R_1(A,M) \to \text{Tor}^R_1(P,M) \to \text{Tor}^R_1(F,M) \to A \otimes_R M \to A \otimes_R P \to A \otimes_R F \to 0$

By the characterization, $\text{Tor}^R_n(-,P)=\text{Tor}^R_n(-,F)=0$ for $n\geq 1,$ therefore, $\text{Tor}^R_1(A,M)=0,$ as well as $\text{Tor}^R_n(-,M)$ for $n\geq 1.$ Hence, $_RM$ is a flat module.

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