# Abstract Nonsense

## The Hom Functor is Left Exact

Point of Post: In this post we prove that the Hom functor is left exact.

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Motivation

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We have proven that the tensor functor is ‘right exact’, and moreover we have seen that this partial exactness can be a great boon in our quest to compute things. But, there are other things besides tensor products for which we would like to compute. In particular, we like to compute the homomorphism group/module if not for practical reasons then for purely philosophical ones. Thus, in this post we shall prove that the Hom functor itself has a certain kind of ‘partial exactness’, one which is dual to that of the tensor product: ‘left exactness’. In other words, the Hom functor (in either variance [i.e. the covariant or contravariant]) takes exact sequences of the form $0\to M\to N\to L$ to exact sequences of the same form. There should be, at least some, lack of surprise at this fact considering the somewhat dualistic nature that the Hom and tensor functors share, via their adjointness.

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Left Exactness

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There are two ways in which one could prove this: the conventional and the clever. The conventional is the get-down-and-dirty approach, one just ‘does it’ (as the kids say) and the second where one uses the adjointness of Hom and tensor to do a more hands-off, clean proof. In my opinion, especially considering some of the more categorical clean proofs that have pervaded my posts recently it would help to pursue the latter avenue of proof–let’s get our hands dirty.

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So, the statement:

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Theorem: Let $R$ be some ring and $A,M,N,L$ left $R$-modules. Then, if

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$0\to M\xrightarrow{f}N\xrightarrow{g}L$

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is an exact sequence of left $R$-modules, then

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$0\to\text{Hom}_R(A,M)\xrightarrow{f_\ast}\text{Hom}_R(A,N)\xrightarrow{g_\ast}\text{Hom}_R(A,L)$

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is an exact sequence of abelian groups.

Proof: Let’s first show that $f_\ast$ is an injection. This is easy though since we know every injective $R$-map is a monomorphism, and so if $f\circ h=f_\ast(h)=f_\ast(k)=f\circ k$ then necessarily $h=k$. So we must now show that the induced sequence is exact at $\text{Hom}_R(A,N)$. To do this we first show that $g_\ast\circ f_\ast=0$, but $g\ast\circ f_\ast=(g\circ f)_\ast=0_\ast=0$ so this is easily done. We must now show conversely that if $g_\ast(h)=0$ then $h=f_\ast(k)$ for some $k$. So we have that $g(h(x))=0$ for all $x\in A$. But, since our first sequence is exact we know then that for each $x\in A$ $h(x)=f(y_x)$ for some $y_x\in M$, and moreover this $y_x$ is unique by the injectivity of $f$. Define then $k:A\to M$ by $k(x)=y_x$. Certainly then we have that $k(rm+m')=rk(m)+k(m')$ since $f(rk(m)+k(m'))=rf(k(m))+f(k(m'))=rh(m)+h(m')=h(rm+m')$. Thus, we see that $k\in\text{Hom}_R(A,M)$ and $f_\ast(k)=h$. The conclusion follows. $\blacksquare$

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There is a somewhat similar statement for the contravariant Hom functor, the only issue being that we will have to reverse the direction of our sequences:

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Theorem: Let $R$ be a ring and $A,M,N,L$ be left $R$-modules. Then, if

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$M\xrightarrow{f} N\xrightarrow{g} L\to 0$

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is exact, then so is

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$0\to\text{Hom}_R(L,A)\xrightarrow{g^\ast}\text{Hom}_R(N,A)\xrightarrow{f^\ast}\text{Hom}_R(M,A)$

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Since the proof is in the same spirit as the first proof I’ll omit it.

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As promised, let’s use this to make a previously messy computation easy. Namely, let’s compute $\text{Hom}_\mathbb{Z}(\mathbb{Z}_k,\mathbb{Z}_\ell)$, or really (since we know the answer) prove this is $\mathbb{Z}_{(k,\ell)}$. As we have remarked in the previous computation, it really suffices to compute $\text{Hom}_\mathbb{Z}(\mathbb{Z}_{p^n},\mathbb{Z}_{p^m})$ for $m\leqslant n$ since the rest follows by the multiplicativeness of Hom and the Chinese remainder theorem. To make this computation we merely note that we have the exact sequence

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$\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}_{p^m}\to0$

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where the first map is ‘multiply by $p^m$‘ and the second is the canonical epimorphism. We note then that we may apply the left exactness of $\text{Hom}_\mathbb{Z}(\bullet,\mathbb{Z}_{p^n})$ to see that we have the exact sequence

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$0\to\text{Hom}_\mathbb{Z}(\mathbb{Z}_{p^m},\mathbb{Z}_{p^n})\to\text{Hom}_\mathbb{Z}(\mathbb{Z},\mathbb{Z}_{p^n})\to\text{Hom}_\mathbb{Z}(\mathbb{Z},\mathbb{Z}_{p^n})$

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In particular, we have an injection $j:\text{Hom}_\mathbb{Z}(\mathbb{Z}_{p^m},\mathbb{Z}_{p^n})\to\mathbb{Z}_{p^n}$. But, the image of this map is the kernel of the corresponding map $\mathbb{Z}_{p^n}\to\mathbb{Z}_{p^n}$ which, clearly, is just multiplication by $p^m$. But, it’s clear that the kernel of this map is just the multiples of $p^{n-m}$ less than or equal to $p^n$, of which there are clearly $p^m$. Thus, $j$ is an isomorphism of $\text{Hom}_\mathbb{Z}(\mathbb{Z}_{p^m},\mathbb{Z}_{p^n})$ with a subgroup of $\mathbb{Z}_{p^n}$ of size $p^m$ which (by cyclicity) must be (isomorphic to) $\mathbb{Z}_{p^m}$. Thus, $\text{Hom}_\mathbb{Z}(\mathbb{Z}_{p^m},\mathbb{Z}_{p^n})\cong\mathbb{Z}_{p^m}$ as desired. Much easier, right?

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.

[5] Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.