Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

'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).

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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 $$m\otimes n=\left(\sum_{i=1}^ka_ix_i\right)\otimes \left(\sum_{j=1}^lb_jy_j\right)=\sum_{i=0}^k\sum_{j=0}^l(a_ix_i)\otimes(b_jy_j)=\sum_{i=0}^k\sum_{j=0}^l a_ib_j\cdot x_i\otimes y_j.$$

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.