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Let $C$ be an abelian ring and $E,F$ two finitely generated projective modules. Then $\text{Hom}_C(E,F)$ is a finitely generated projective $C$-module.

First of all, since $C$ is abelian, the abelian group $\text{Hom}_C(E,F)$ is a $C$-module. As $E,F$ are finitely generated projective $C$-modules, there exist free $C$-modules $M,N$, with finite bases, such that $E,F$ are isomorphic, respectively, to direct factors of $M,N$.

So, let $R_1,R_2$ be supplementary submodules of $M$ such that $E\simeq R_1$. On the other hand, let $L_1,L_2$ be supplementary submodules of $N$ such that $F\simeq L_1$. Furthermore, $\text{Hom}_C\left(R_1\oplus R_2,L_1\oplus L_2\right)$ is isomorphic to $$\text{Hom}_C(R_1,L_1)\oplus\text{Hom}_C(R_1,L_2)\oplus\text{Hom}_C(R_2,L_1)\oplus\text{Hom}_C(R_2,L_2).$$

Also, $\text{Hom}_C(E,F)\simeq\text{Hom}_C(R_1,L_1)$.

I am not sure how I can deduce that (i) $\text{Hom}_C(E,F)$ is finitely generated and (ii) $\text{Hom}_C(E,F)$ is projective from the above.

Any hints?

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1 Answer 1

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Hint:

$\operatorname{Hom}_C(R_1, L_1)$ is a direct summand of $\operatorname{Hom}_C(M, N)$ , which is isomorphic to $C^{m\cdot n}$, if $m$ is the rank of the free module $M$ and $n$ the rank of the free module $N$.

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  • $\begingroup$ Thank you for your answer. If $(a_i)_{1\leq i\leq m}$, $(b_i)_{1\leq i\leq n}$ are the bases of $M$ and $N$, respectively, can you please tell me what the isomorphism between $\text{Hom}_C(M,N)$ and $C^{m\cdot n}$ is? I am having difficulty constructing it. $\endgroup$
    – alf262
    Commented May 23, 2020 at 22:05
  • $\begingroup$ As usual with vectors spaces, the correspondence is an $m{\times}n$ matrix, with column vectors the coordinates of the $f(a_i)$s in basis $(b_j)$. $\endgroup$
    – Bernard
    Commented May 23, 2020 at 22:09

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