Tensor product of two finitely generated modules

Question

How can I show that if $M$ and $N$ are finitely generated $A$-modules, then so is $M\otimes_AN$? I understand that I have assumption that there are integers $n,m$ such that there are surjections $A^n\to M$ and $A^m\to N$ but how do these implies that there is an integer $l$ such that there is a surjection $A^l\to M\otimes_AN$?

Answer 1

Let $f:A^m\to M$ and $g:A^n\to N$ be surjective. Then $f\otimes g:A^{mn}\cong A^m\otimes A^n\to M\otimes_A N$ is surjective. More specifically, if $x_1,\ldots,x_m$ generate $M$ and $y_1,\ldots,y_n$ generate $N$, then the set of all $x_i\otimes y_j$ generates the tensor product. This can be seen by the bilinearity relations used in the construction of the tensor product.

Answer 2

Suppose $$M = Aa_1+ \ldots + A a_k$$ $$N=Ab_1+ \ldots + A b_h$$then if $v \otimes_A w \in M \otimes_A N$ we have $$v = r_1a_1 + \ldots + r_ka_k \ \ r_i \in A \ \ \forall\ i$$ $$w = s_1b_1 + \ldots + s_hb_h \ \ s_i \in A \ \ \forall\ i$$ Thus $$v \otimes_A w = \sum_{i,j} r_is_j (a_i \otimes_A b_j)$$This means that all decomposable tensors are generated by $\lbrace a_i \otimes_A b_j \rbrace_{i,j}$ and thus all tensors are generated by $\lbrace a_i \otimes_A b_j \rbrace_{i,j}$