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Exactness and tensor/Hom

We have seen that each of $-\otimes_R M$, $\operatorname{Hom}_R(M,-)$, and $\operatorname{Hom}_R(-,M)$ is an additive functor.

Left exactness of   $\operatorname{Hom}_R(M,-)$

This is almost tautological from the universal property of kernels. For a morphism $\phi : M \to N$, we know that \(0 \to \operatorname{Hom}_R(T,\ker \phi) \to \operatorname{Hom}_R(T,M) \to \operatorname{Hom}_R(T,N)\) is exact. In other words, we preserve kernels: \(\operatorname{Hom}_R(T,\ker \phi) \cong \ker \left( \operatorname{Hom}_R(T,M) \to \operatorname{Hom}_R(T,N) \right)\)

The same argument in a general abelian category shows that $\operatorname{Hom}_{\mathcal A}(A,-)$ is left exact.

Left exactness (in the opposite category) of   $\operatorname{Hom}_R(-,M)$

In the other entry, we get a functor $\left( \operatorname{Mod} R \right)^{op} \to \operatorname{Ab}$. The opposite category $\left( \operatorname{Mod} R \right)^{op}$ has the structure of abelian category where we exchange kernels and cokernels.

From the universal property of cokernels we know that \(\operatorname{Hom}_R(\operatorname{coker} \phi, T) \cong \ker \left( \operatorname{Hom}_R(N,T) \to \operatorname{Hom}_R(M,T) \right)\) Reversing the arrows, we see that $\operatorname{Hom}_R(-,M)$ takes the kernels in $\left( \operatorname{Mod} R \right)^{op}$ to kernels.

Again, the same argument works in a general abelian category.

Right exactness of   $- \otimes_R M$

Assume we have an exact sequence \(M \overset{\mu}{\to} N \overset{\nu}{\to} C \to 0\) in $\operatorname{Mod} R$. Then $C \cong M/N$. If we apply $- \otimes_R T$ we get \(M \otimes_R T \overset{\mu \otimes 1}{\to} N \otimes_R T \overset{\nu \otimes 1}{\to} C \otimes_R T \to 0\) and we want to check exactness.

We need to first check that each element of $C \otimes_R T$ is in the image of $N \otimes_R T$. Pick some element $\sum c_i \otimes t_i$. Since $N \to C$ is surjective, we can find $n_i$ with $n_i \mapsto c_i$ for each $i$. Then \(\sum n_i \otimes t_i \mapsto \sum c_i \otimes t_i.\)

Assume we can construct a map $\lambda: C \otimes_R T \to (M \otimes_R T)/(N \otimes_R T)$ making the diagram

where the arrows labelled with $=$ are the identity maps. Assume that we have some element $ x \in N \otimes_R T$ with $(\nu \otimes 1)(x) = 0$. Then, we also know that $\pi(x) = 0$. So there exists a $y \in M \otimes_R T$ with $(\mu \otimes 1)(y) = x$ giving exactness at $N \otimes_R T$.

Now we turn to making $\lambda$ using the universal property of tensor products. Consider \(\begin{aligned} C \times T & \to (N \otimes_R T)/(M \otimes_R T) \\ (c,t) & \mapsto \overline{\tilde{c} \otimes t} \end{aligned}\) where $\tilde{c}$ is a choice of element of $N$ such that $\nu(\tilde{c}) = c$. We need to check this is well-defined and $R$-bilinear. Any other lift of $c$ is of the form $\tilde{c} + n$ for some $n \in N$. We have \(\overline{(\tilde{c} + n) \otimes t} = \overline{\tilde{c} \otimes t + n \otimes t} = \overline{\tilde{c} \otimes t}.\) Thus the function is well-defined. Since $\tilde{c}r$ is a lift of $cr$, we see that we have an $R$-bilinear function.

Thus, we get a unique $\lambda : C \otimes_R T \to (N \otimes_R T)/(M \otimes_R T)$ with \(\lambda(c \otimes t) = \overline{\tilde{c} \otimes t}\) This gives us the desired diagram and finishes the argument.


Consider $k[x]$ for $k$ a field. And write $k = k[x]/(x)$. We have an exact sequence \(0 \to k[x] \overset{x}{\to} k[x] \to k \to 0\) If we apply $\operatorname{Hom}_{k[x]}(k,-)$ this turns into \(0 \to 0 \to 0 \to k \to 0\) which is not exact at $k$. Hence $\operatorname{Hom}_R(M,-)$ is not right exact in general.

If we apply $- \otimes_{k[x]} k$ to our exact sequence instead we get \(0 \to k \overset{0}{\to} k \to k \to 0\) which show that $- \otimes_R M$ is not left exact in general.

Similarly, if we apply $\operatorname{Hom}_{k[x]}(-,k)$, we get \(0 \leftarrow k \overset{0}{\leftarrow} k \leftarrow k \leftarrow 0\) which shows that $\operatorname{Hom}_R(-,M)$ is not right exact in general.

Adapted objects

For what modules $M$ is $\operatorname{Hom}_R(M,-)$ exact? $- \otimes_R M$? $\operatorname{Hom}_R(-,M)$


We already saw that if $F$ is free $R$-module then $\operatorname{Hom}_R(F,-)$ is exact. We can take one step away from this.

Recall that $X$ is a retract of an object $Y$ in a category $\mathcal C$ if there exists maps \(\begin{aligned} i : X & \to Y \\ \pi : Y & \to X \end{aligned}\) with $\pi \circ i = 1_X$.

If $P$ is a retract of $F$, then $\operatorname{Hom}_R(P,-)$ is exact also. Indeed, we have a retraction of functors \(\begin{aligned} \operatorname{Hom}_R(\pi,-) : \operatorname{Hom}_R(P,-) & \to \operatorname{Hom}_R(F,-) \\ \operatorname{Hom}_R(i,-) : \operatorname{Hom}_R(F,-) & \to \operatorname{Hom}_R(P,-) \end{aligned}\) so we only need to establish that retracts of exact complexes are again exact. We should promote chain complexes to a category by providing morphisms between chain complexes.

A morphism $\phi_\bullet : (A_\bullet, \delta^A_\bullet) \to (B_\bullet, \delta^B_\bullet)$ is a collection of morphisms $\phi_i : A_i \to B_i$ for which all squares


Given an abelian category $\mathcal A$, we denote the category of chain complexes by $Ch(\mathcal A)$.

Assume we have a retraction in $Ch(\mathcal A)$. This translates into a commutative diagram of which we prune at bit

Given $a \in A_{i+1}$ with $\delta_{i+1} (a) = 0$, then its image $b$ in $B_{i+1}$ also satisfies $\delta_{i+1}(b) = 0$. From exactness of $B_\bullet$, there exists a $b^\prime$ with $\delta_i(b^\prime) = b$. Applying the map to $A_\bullet$ gives an element of $A_i$, $a^\prime$, with $\delta_i(a^\prime) = a$.

We can represent this “diagram chase” via the following picture:

Definition. A object $P$ of an abelian category is projective if $\operatorname{Hom}_{\mathcal A}(P,-)$ is exact.

Proposition. In $\operatorname{Mod} R$, the projective objects are exactly the retracts of free modules.

Proof. The above shows that in $\operatorname{Mod} R$ retracts of free modules are projective.

Assume that $P$ is a projective object. There exists a surjection $F \overset{\pi}{\to} P \to 0$. We can lift the identity

which exhibits $P$ as a retract of $F$.

This is satisfying statement: we have an explicit construction of all projective objects.

As a consequence, we have enough projectives in $\operatorname{Mod} R$. In an abelian category $\mathcal A$, we say have enough projectives if for any objects $X$ there exists $P \twoheadrightarrow X$ for some projective $P$.

Let us also note that if $P_i$ are projective for $i \in I$, then $\bigoplus_{i \in I} P_i$ is also projective in a general $\mathcal A$. Since \(\operatorname{Hom}_{\mathcal A}\left( \bigoplus P_i, M \right) \cong \prod \operatorname{Hom}_{\mathcal A}(P_i, M)\) and products of exact sequences remain exact.


Definition. In an abelian category, we call an object $I$ injective if $\operatorname{Hom}_{\mathcal A}(-,I)$ is exact.

This is the extension problem

Let’s look at injective objects of $\operatorname{Mod} R$. Assume that $I$ is an injective module.

Then we have a more easily-verified extension problem. For any ideal $J \leq R$, we know that we can extend over inclusions of ideals

This is enough to characterize injectives.

Lemma. (Baer’s Criteria) If $I$ extends over any $J \leq R$, then $I$ is injective.

Proof. Assume we have the diagram

and form a partially order set \(\lbrace (M^\prime, \psi^\prime) \mid N \subseteq M^\prime \subseteq M \text{ and } \psi^\prime|_{N} = \psi \rbrace\) We say $(M_1^\prime, \psi_1^\prime) < (M_2^\prime, \psi_2^\prime)$ if $M_1^\prime \subseteq M_2^\prime$ and $\psi_2^\prime|_{M_1^\prime} = \psi_1^\prime$. Then we have a partially order set and we can form upper bounds for any chain by taking unions. By Zorn’s Lemma, we just need to check the maximal element is all of $M$.

Assume not and let $m \in M \setminus M^\prime$ for maximal $(M^\prime, \psi^\prime)$. Let \(J := Rm \cap M^\prime \leq R/\operatorname{Ann}(m)\) If $J = 0$, we can extend the map by setting $\psi^\prime(m) = 0$. Otherwise, let $\tilde{J} = \pi^{-1}(J)$ for the projection $\pi: R \to R/\operatorname{Ann}(m)$. It suffices to solve the extension problem

By assumption we can do exactly this.

Using this we can get some injectives in $\operatorname{Ab} = \operatorname{Mod} \mathbb{Z}$. Any ideal in $\mathbb{Z}$ is of the form $(m)$ for some $m \in \mathbb{Z}$. Thus, to extend

we need to find a $x \in I$ with $mx = \psi(m)$. In other words, we need to divide $x$ by $m$.

We can always do this for $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$ - giving two injective modules. Using $\mathbb{Q}/\mathbb{Z}$ and some tricks, we demonstrate that $\operatorname{Mod} R$ has enough injectives.

Definition. An abelian category $\mathcal A$ has enough injectives if for any object $X$ there exists $X \hookrightarrow I$ with $I$ injective.

Proposition. $\operatorname{Mod} R$ has enough injectives.

Proof. Let \(I := \operatorname{Hom}_\mathbb{Z}(R,\mathbb{Q}/\mathbb{Z})\)

Note that, by tensor-Hom adjunction, we have \(\operatorname{Hom}_R(M,I) \cong \operatorname{Hom}_\mathbb{Z}(M,\mathbb{Q}/\mathbb{Z})\) Since \(M \mapsto \operatorname{Hom}_\mathbb{Z}(M,\mathbb{Q}/\mathbb{Z})\) is exact, we see that $I$ is injective.

One can check that for any abelian group $A$ \(\operatorname{Hom}_\mathbb{Z}(A,\mathbb{Q}/\mathbb{Z}) = 0\) Pick some $0 \neq a \in A$ and then $\mathbb{Z}a = \mathbb{Z}$ or $\mathbb{Z}/m\mathbb{Z}$.
\(\begin{aligned} \operatorname{Hom}_\mathbb{Z}(\mathbb{Z},\mathbb{Q}/\mathbb{Z}) & = \mathbb{Q}/\mathbb{Z} \neq 0 \\\ \operatorname{Hom}_\mathbb{Z}(\mathbb{Z}/m\mathbb{Z},\mathbb{Q}/\mathbb{Z}) & = (\mathbb{Q}/\mathbb{Z})_m \neq 0 \end{aligned}\) where $(-)_m$ is the $m$-torsion. (Note here is where $\mathbb{Q}$ doesn’t work.) We have a map \(\operatorname{Hom}_\mathbb{Z}(A,\mathbb{Q}/\mathbb{Z}) \twoheadrightarrow \operatorname{Hom}_\mathbb{Z}(\mathbb{Z}a,\mathbb{Q}/\mathbb{Z})\) since $\mathbb{Q}/\mathbb{Z}$ is injective. Thus, $\operatorname{Hom}_\mathbb{Z}(A,\mathbb{Q}/\mathbb{Z}) \neq 0$.

Set $M^\vee := \operatorname{Hom}_R(M,I)$. Since $I$ is naturally a $R$-bimodule, $M^\vee$ remains a $R$-module. We claim the map \(M \to M^{\vee\vee}\) is injective. If $M \neq 0$, we know that $M^\vee \neq 0$. If we take the kernel $K$ of $M \to M^{\vee\vee}$, then \(K^{\vee\vee} \hookrightarrow M^{\vee\vee}\) as $I$ is injective. But $K \to M^{\vee\vee}$ is the zero map. Thus $K \to K^{\vee\vee}$ must also be $0$. This implies that $K^\vee = 0$. Since otherwise, there would be an element $k \in K$ and nonzero function $\phi:K \to I$ with $\phi(k) \neq 0$. Thus \(M \hookrightarrow M^{\vee\vee}\)

To finish, we choose $F \twoheadrightarrow M^\vee$ for a free module $F \cong R^{\oplus X}$. Then, we get \(M \hookrightarrow M^{\vee\vee} = \operatorname{Hom}_R(M^\vee, I) \hookrightarrow \operatorname{Hom}_R(F,I) \cong \prod_X I\) and note that if $I_j$ for $j \in J$ are injective then so is $\prod_J I_J$.

Flat modules

Definition. An $R$-module $M$ is called flat if $- \otimes_R M$ is exact.

We have seen that free modules are flat. Similar to the argument above, retracts of free modules are also flat. Thus, projective modules are flat. But in general flat modules are projective.

Consider the localization $k[x]_x$ as a $k[x]$-module. This is flat as localization is an exact functor. However it is not projective since any map $k[x]_x \to k[x]$ must send $1$ to $p \in k[x]$ which satisfies $x^iq_i = p$ for some $q_i$ any $i \geq 0$. The only choice for $p$ is $0$.

There exists a criteria analogous to Baer’s for flatness.

Proposition. $M$ is flat if and only if \(J \otimes_R M \hookrightarrow R \otimes_R M \cong M\) for any ideal $J \leq R$.

We omit the proof.