# Weibel 2.5.1 Equivalent statements of injective $R$-module.

Show that the following are equivalent:

1. $$B$$ is an injective $$R$$-module.

2. $$\operatorname{Hom}_{R}(-, B)$$ is an exact functor.

3. $$\operatorname{Ext}_{R}^{i}(A, B)$$ vanishes for all $$i \ne 0$$ and all $$A$$ ($$B$$ is $$\operatorname{Hom}_{R}(A,-)$$-acyclic for all $$A$$ ).

4. $$\operatorname{Ext}_{R}^{1}(A, B)$$ vanishes for all $$A$$.

1 $$\implies$$ 2. Given an exact sequence $$0\to X\xrightarrow{f} Y \xrightarrow{g} Z\to 0$$ of $$R$$-modules, we need to show that $$0\xleftarrow{} \operatorname{Hom}_{R}(X, B)\xleftarrow{f_* = -\circ f} \operatorname{Hom}_{R}(Y, B) \xleftarrow{g_* = -\circ g} \operatorname{Hom}_{R}(Z, B)\xleftarrow{} 0$$ is exact. The exactness at $$\operatorname{Hom}_{R}(X, B)$$(surjectivity of $$f_*$$) can be implied by injectivity of $$B$$, the exactness at $$\operatorname{Hom}_{R}(Z, B)$$(injectivity of $$g_*$$) can be implied by exactness at $$Z$$(surectivity of $$g$$). How to get the exactness at $$\text{Hom}_R (Y,B)$$?

2 $$\implies$$ 1. Since induced map $$f_∗$$ is surjective whenever $$f$$ is injective. For every $$h \in \text{Hom}_R(X, B)$$, there exists $$t \in \text{Hom}_R(Y, B)$$ such that $$h = t\circ f$$, hence $$B$$ is injective.

3$$\implies$$ 4 is clear.

How to prove other equivalences? Thanks in advance!

• That's at least three more questions. But at least $3\implies 4$ is easy.... Commented Jun 15, 2020 at 16:01
• @AnginaSeng Yes.
– Ryze
Commented Jun 15, 2020 at 16:04
• You know that Ext can be computed from an injective resolution of the second argument? Commented Jun 15, 2020 at 16:18

At a minimum, there are only two things left to prove to complete the equivalences.

I'll prove (1) $$\implies$$ (3) and (4) $$\implies$$ (2), since you know (3) $$\implies$$ (4) and (2) $$\implies$$ (1).

I'm going to state two facts about $$\newcommand\Ext{\operatorname{Ext}}\newcommand\Hom{\operatorname{Hom}}\Ext$$, and if you're not familiar with them, then I'd suggest looking into these, because they're a bit beyond the scope of an answer to reprove here.

Fact 1 If $$B\to I^0\to I^1\to \cdots \to I^n\to\cdots$$ is any injective resolution of $$B$$, then for any $$A$$, $$\Ext^n(A,B)\cong H^n(\Hom(A,I^\bullet))$$

Fact 1 gives (1) $$\implies$$ (3), since $$B\to 0 \to 0 \to 0 \to \cdots$$ is already an injective resolution of $$B$$ when $$B$$ is injective, so $$\Ext^n(A,B) \cong H^n(\Hom(A,B)\to 0 \to 0 \to 0 \to 0),$$ so $$\Ext^i(A,B)=0$$ for $$i>0$$ (and any $$A$$).

Fact 2 If $$0\to A' \to A\to A''\to 0$$ is any short exact sequence of $$R$$-modules, then there is a long exact sequence for $$\Ext$$ for any $$B$$: $$0\to \Hom(A'',B) \to \Hom(A,B)\to \Hom(A',B)\to \Ext^1(A'',B)\to \Ext^1(A,B)\to \cdots$$ $$\Ext^n(A',B)\to\Ext^{n+1}(A'',B)\to \Ext^{n+1}(A,B)\to \Ext^{n+1}(A',B)\to \cdots$$

Fact 2 gives (4) $$\implies$$ (2), since if $$\Ext^1(A,B)=0$$ for all $$A$$, then for any short exact sequence $$0\to A'\to A\to A''\to 0$$, we have the long exact sequence $$0\to \Hom(A'',B)\to \Hom(A,B)\to \Hom(A',B)\to \Ext^1(A'',B)=0,$$ since we assumed that $$\Ext^1(A'',B)=0$$ for any $$A''$$, so $$\Hom(-,B)$$ is an exact functor.

Edit:

Also I missed this when reading your question, but I realized I didn't directly address your first question about proving exactness in the middle when proving (1) $$\implies$$ (2).

This also follows from Fact 2 above, but that's overkill, there is actually an elementary proof that for any short exact sequence $$\newcommand\toby\xrightarrow 0\to A'\toby{f} A\toby{g} A''\to 0$$, and any $$B$$, the sequence $$0\to \Hom(A'',B)\toby{g^*} \Hom(A,B)\toby{f^*} \Hom(A',B)$$ is exact. Then $$B$$ being injective is equivalent to the last map being surjective for all short exact sequences.

Proof

Since $$gf=0$$, we have $$f^*g^*=(gf)^*=0$$, which means $$\newcommand\im{\operatorname{im}}\im g^*\subseteq \ker f^*$$ so we have two things to prove: (a) injectivity of $$g^*$$, and (b) that $$\ker f^*\subseteq \im g^*$$.

(a) If $$\phi : A''\to B$$ is some map, and $$g^*\phi = \phi\circ g =0$$, then, if $$x\in A''$$ is any element, since $$g$$ is surjective, $$x=g(a)$$ for some $$a\in A$$, so $$\phi(x) = \phi(g(a))=0$$. Therefore $$\phi=0$$, so $$g^*$$ is injective.

(b) Suppose $$\phi : A\to B$$ is in the kernel of $$f^*$$, so $$\phi\circ f =0$$. Then since $$f$$ is injective and $$g$$ is surjective, we can regard $$A'$$ as a submodule of $$A$$ and we have that $$A''\cong A/A'$$. Then $$\phi : A\to B$$ is a morphism that is zero on $$A'$$, so we know that it induces a morphism $$\phi' : A''\to B$$ defined by $$\phi'(g(a))=\phi(a)$$. But this is exactly what it means to say $$\phi = g^*\phi'$$, so $$\phi$$ is in the image of $$g^*$$. $$\blacksquare$$

• The $\mathrm{Ext}(A,-)$ functors have been defined as right derived functors of $\mathrm{Hom}(A,-)$. Then we get a long exact sequence for $\mathrm{Hom}(A,-)$, but I don't think Fact 2 is clear yet at this point of the book. Commented May 23, 2023 at 5:12
• @subrosar If you know they are the right derived functors of $\operatorname{Hom}(A,-)$ you can take an injective resolution of $B$, $B\to I_0\to I_1\to\cdots$, and apply $\operatorname{Hom}(-, I_n)$ to your short exact sequence and using the fact that those are all exact functors (by (1) $\implies$ (2) and all the $I_n$ being injective) we get a short exact sequence of chain complexes, and this gives us the desired long exact sequence when we take cohomology.