Number of generators in a tensor product

Question

'Commutative Algebra' by Atiyah and Mcdonald, mentions if $ \lbrace x_{i} \rbrace_{i ∈ I}$ and $\lbrace y_{j} \rbrace_{j ∈ J}$ generate $M$ and $N$ as $A-$modules, respectively, then $x_{i} \otimes y_{j} $ over all $i,j$ 's generate $M ⊗ N$. this is not obvious to me. To get the tensor product $T$, we take the free $A-$module $C$ generated by set $M×N$ and go modulo some submodule $D$. since $C$ is generated by elements of type $(m,n)$ where $m∈ M,n∈ N. C/D$ is generated by its image in $C/D$ which is written as $m⊗n$. To get the stated result from this method,we would need to show that $C$ is generated by $(x_{i},y_{j})$,$x_{i} \in M, y_{j} ∈ N$, which is clearly not true ( as $x_{1} + x_{2} ∈ M , y_{1} + y_{2} ∈ N , (x_{1} + x_{2},y_{1} + y_{2}) \in C$ but it cannot be written as some $A-$linear combination of $(x_{i},y_{j}) $ until such an element was explicitly present in the generating set).

Answer 1

This is true if by generating you mean "generate as $A$-modules". If by generating you mean "generate as groups," this is clearly not true (just take $\mathbb{R}\otimes_\mathbb{R}\mathbb{R}$, which has as a group only uncountable sets of generaotrs). I also assume that the tensor product is meant over the ring $A$, i.e. as $\otimes_A$. I also suspect you assume $A$ to be a commutative ring, so I assume that as well.

Take an arbitrary elementary tensor $m\otimes n \in M\otimes N$. Then there are $a_i, b_j \in A$ such that $m=\sum_{i=1}^ka_ix_i, n=\sum_{j=1}^lb_jy_j$. It follows that $$m\otimes n=\left(\sum_{i=1}^ka_ix_i\right)\otimes \left(\sum_{j=1}^lb_jy_j\right)=\sum_{i=1}^k\sum_{j=1}^l(a_ix_i)\otimes(b_jy_j)=\sum_{i=1}^k\sum_{j=1}^l a_ib_j\cdot x_i\otimes y_j.$$

Since the elementary tensors are generators of $M\otimes N$, it follows that $x_i\otimes y_j$ generate the tensor product.

(The reason this may not seem intuitively clear is the fact that the standard construction of the tensor product is done as a quotient of the free group, where every element $(m,n)$ is a free generator. However, the relations of the quotient degenerate this group a lot - in fact, precisely these relations were used in the computation above).