# Abstract Nonsense

## Homomorphism Groups of Products and Coproducts (Pt. II)

Point of Post: This is a continuation of this post.

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We’d now like to figure out what and when we can take products/coproducts (not necessarily both) out of the second entry of Hom. So, let’s see if (similar to what we did after the last proof) we can logic out what the theorem we are about to state might not be. For example, I claim that it’s obvious that we’re not going to be able to have the exact same theorem with the second entry as we did for the first. To be particular, I claim that

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$\displaystyle \text{Hom}_R\left(M,\bigoplus_{\alpha\in\mathcal{A}}N_\alpha\right)\not\cong\prod_{\alpha\in\mathcal{A}}\text{Hom}_R(M,N_\alpha)$

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Why? We first observe the following:

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Theorem: Let $R$ be a unital ring and $M$ a unital left $R$-module. Then, $\text{Hom}_R(R,M)\cong M$ are isomorphic as abelian groups, and in the case that $R$ is commutative as left $R$-modules.

Proof: Define $\varphi:\text{Hom}_R(R,M)\to M$ by $\varphi(f)=f(1)$. Clearly $\varphi(f+g)=(f+g)(1)=f(1)+g(1)=\varphi(f)+\varphi(g)$ so that $\varphi$ is a $\mathbb{Z}$-map. If $R$ is further commutative we see that $\varphi(rf)=(rf)(1)=rf(1)=r\varphi(f)$ so that $\varphi$ is an $R$-map. To see why $\varphi$ is injective we merely note that if $f(1)=g(1)$ then $f(r)=rf(1)=rg(1)=g(r)$ for all $r\in R$ so that $f=g$. To see that $\varphi$ is surjective define, for any $m\in M$, the map $f:R\to M$ by $f(r)=rm$. By definition of left $R$-module this is an $R$-map and moreover $\varphi(f)=m$. The conclusion follows. $\blacksquare$

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How does this help us? Well, taking in our hopeful “formula” $M=N_\alpha=R=\mathbb{Q}$ and $\mathcal{A}=\mathbb{N}$ we’d see that by the previous theorem our “formula” says that $\mathbb{Q}^{\oplus\mathbb{N}}\cong\mathbb{Q}^{\mathbb{N}}$. But, in our proof concerning when a module is isomorphic to its dual we proved that for countable fields this can never happen (they don’t have the same cardinality–the left hand side is countable whereas the right hand side has the cardinality of the continuum), and so we get a contradiction.

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Of course, we may wonder if taking a coproduct out into a product just makes things “too large” as it did in the previous example, and if we instead took a coproduct out if we’d get a correct formula. In other words, perhaps the following is true:

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$\displaystyle \text{Hom}_R\left(M,\bigoplus_{\alpha\in\mathcal{A}}N_\alpha\right)\cong\bigoplus_{\alpha\in\mathcal{A}}\text{Hom}_R(M,N_\alpha)\quad\mathbf{(1)}$

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Unfortunately, this is not true either, but its disproof requires a little bit trickier counterexample. Indeed, consider $\displaystyle \bigoplus_{n\geqslant2}\mathbb{Z}_n$. Note that any morphism $f:M\to\mathbb{Z}_n$ is torsion (of finite order) since $nf=0$. Thus, it’s pretty easy to see that every element of $\displaystyle \bigoplus_{n\geqslant2}\text{Hom}_\mathbb{Z}\left(\bigoplus_{n\geqslant2}\mathbb{Z}_n,\mathbb{Z}_n\right)$ is torsion since it’s a finite sum of torsion things. That said, note that there is no $k\in\mathbb{Z}$ for which $kx=0$ for all $x$ in $\displaystyle\bigoplus_{n\geqslant2}\mathbb{Z}_n$. Consequently, the identity map on $\displaystyle \bigoplus_{n\geqslant 2}\mathbb{Z}_n$ is a non-torsion element of $\displaystyle \text{Hom}_\mathbb{Z}\left(\bigoplus_{n\geqslant2}\mathbb{Z}_n,\mathbb{Z}_n\right)$. Since torsion elements are carried to torsion elements under isomorphisms it’s clear from this analysis that $\mathbf{(1)}$ is false with $\mathcal{A}=\mathbb{N}^{\geqslant2}$, $R=\mathbb{Z}$, $N_\alpha=\mathbb{Z}_\alpha$, and $\displaystyle M$ equal to $\displaystyle \bigoplus_{n\geqslant2}\mathbb{Z}_n$.

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With this in mind we may try the following:

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Theorem: Let $R$ be a ring and $M,\left\{N_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a set of left $R$-modules. Then,

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$\displaystyle \text{Hom}_R\left(M,\prod_{\alpha\in\mathcal{A}}N_\alpha\right)\cong\prod_{\alpha\in\mathcal{A}}\text{Hom}_R(M,N_\alpha)$

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where the isomorphism is of abelian groups. If further $R$ is commutative, then it is an isomorphism of left $R$-modules.

Proof: Let $A$ either represent the ring $\mathbb{Z}$ or $R$. This way, when we say $A$-map, it’s clear that we mean a $R$-map in the case that $R$ is commutative and a $\mathbb{Z}$-map otherwise.

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Define the maps, for each $\alpha_0\in\mathcal{A}$, $\displaystyle p_{\alpha_0}:\text{Hom}_R\left(M,\prod_{\alpha\in\mathcal{A}}N_\alpha\right)\to\text{Hom}_R(M,N_{\alpha_0})$ by $p_{\alpha_0}(f)=\pi_{\alpha_0}\circ f$. Clearly this is an $A$-map. We claim that that $\displaystyle \left(\text{Hom}_R\left(M,\prod_{\alpha\in\mathcal{A}}N_\alpha\right),\left\{p_\alpha\right\}_{\alpha\in\mathcal{A}}\right)$ is a product of $\left\{\text{Hom}_R(M,N_\alpha)\right\}_{\alpha\in\mathcal{A}}$ from where the conclusion will follow from the isomorphism uniqueness of products.

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To do this, suppose that $L$ is an $A$-module and $\displaystyle g_\alpha:L\to\text{Hom}_R\left(M,N_\alpha\right)$ be a set of $A$-maps. Define a map $\displaystyle g:L\to\text{Hom}_R\left(M,\prod_{\alpha\in\mathcal{A}}N_\alpha\right)$ by $(g(\ell))(m)=((g_\alpha(\ell))(m))$. Evidently $g(\ell)$ really is an $R$-map $\displaystyle \text{Hom}_R\left(M,\prod_{\alpha\in\mathcal{A}}N_\alpha\right)$ for each $\ell\in L$. Moreover, it’s clear that $g$ is an $A$-map. We note then that evidently

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$((p_{\alpha_0}\circ g)(\ell))(m)=(\pi_{\alpha_0}(g(\ell)))(m)=\pi_{\alpha_0}(((g_\alpha(\ell)(m)))=(g_{\alpha_0}(\ell))(m)$

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for all $m\in M$ and $\ell\in L$ so that $p_{\alpha_0}\circ g=g_{\alpha_0}$. Moreover, it’s clear by the definition of products that this is the unique map for which this is true. Thus, $\displaystyle \left(\text{Hom}_R\left(M,\prod_{\alpha\in\mathcal{A}}N_\alpha\right),\{p_\alpha\}_{\alpha\in\mathcal{A}}\right)$ really is a product of $\displaystyle \left\{\text{Hom}_R\left(M,N_\alpha\right)\right\}_{\alpha\in\mathcal{A}}$ and so the conclusion follows by previous remark. $\blacksquare$

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As a corollary to the above we get the following:

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Theorem: Let $R$ be a commutative unital ring. Then,

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$\text{Hom}_R\left(R^n,R^m\right)\cong R^{nm}$

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as left $R$-modules.

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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