0

This is problem 3.34(ii) from Rotman's Homological Algebra text.

Let $B$ and $C$ be a left $R$-modules. Definite $\nu : Hom(B,R) \otimes C \rightarrow Hom(B,C)$ by $f \otimes c \mapsto \hat{f}$ where $\hat{f}(b) = f(b)c$ for $b \in B$.

Claim

Prove that $\nu$ is an isomorphism if $B$ is finitely generated free.

Attempted Proof

$\nu$ is surjective. Let $B = \langle x_1, \ldots x_n \rangle$ and $g \in Hom(B,C)$ such that $g(x_i) = c_i$ for some $c_i \in C$. Consider $f_i(x_j) = \delta_{ij}$. Then $$ \nu(\sum_i f_i \otimes c_i)(x_j) = \sum_i f_i(x_j)c_i = c_j = g(x_j)$$

$\nu$ is injective. Suppose $f \otimes c \in ker(\nu)$. Then $f(x_i)c = 0$ for all $i$. If $f$ or $c$ is zero injectivity is obviously. Otherwise $c$ is a torsion element of $C$.

Question

I only want a minimal hint to help me show that $\nu$ is injective. I haven't used the fact that $B$ is finitely generated yet, and I believe that $Hom(B,R)$ is also finitely generated free because $Hom(B,R) \cong R^n \cong B$.

IsaacR24
  • 635
  • 1
    minimal hint which actually makes things "worse" at first: when proving that the kernel is zero, it is not enough to check that any simple tensor in the kernel is zero, you must look at an element of the form $\sum f_i \otimes c_i$ – hunter Nov 05 '23 at 13:16
  • @hunter I've considered this before actually -- sorry can you provide an additional minimal hint? – IsaacR24 Nov 05 '23 at 15:28
  • If $f_i(x_j) = \delta_{ij}$ then the $f_i$ generate $\mathrm{Hom}_R(B,R)$. If $\sum_j g_j\otimes c_j$ is in the kernel of $\nu$ then rewrite this sum in terms of the $f_i$ and evaluate on the $x_i$ as you did. – Vincent Boelens Nov 06 '23 at 10:27
  • Thanks @VincentBoelens -- that answers it! – IsaacR24 Nov 06 '23 at 15:44

0 Answers0