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Let $R$ be a commutative domain and let $M$ and $N$ be torsion-free $R$ modules. I would like to know whether or not $M\otimes_{R}{N}$ is always non-zero. Now, I know this is true in the finitely generated case; in fact we have something stronger, since for finitely generated modules $M$ and $N$, $M\otimes_{R}{N}=\{0\}$ if and only if $\operatorname{Ann}{M}+\operatorname{Ann}{N}=R$.

Back to the general non finitely generated case. Take $m\in{M}$ and $n\in{N}$, both non-zero. Then in particular (using the result stated above or just by the fact that we are tensoring free modules), $m\otimes{n}$ is non-zero as an element of $R{m}\otimes_{R}{R{n}}$. Consider now the short exact sequence

$0\rightarrow{Rn}\rightarrow{N}\rightarrow{N/Rn}\rightarrow{0}$

Let's now tensor this sequence by the module $mR$. Since $M$ is torsion free, this means we are tensoring by a free (hence flat) module, and so we get a short exact sequence

$0\rightarrow{mR\otimes_{R}Rn}\rightarrow{mR\otimes_{R}{N}}\rightarrow{mR\otimes_{R}{N/Rn}}\rightarrow{0}$

and so we deduce that since $m\otimes{n}$ is non-zero in $mR\otimes_{R}Rn$ and this module embeds in $mR\otimes_{R}{N}$, that $m\otimes{n}$ is non-zero as an element of $mR\otimes_{R}{N}$. This is as far as I've managed to get.

A couple of points to note: If I was working over a semiheriditary domain, then modules are flat if and only if they are torsion-free, and so I believe the result should always hold in this case. Thus if a counter example to my question exists, we must work over domains in which every there are non-projective finitely generated ideals. Since I can't find a statement, never mind a proof, of this result anywhere, I feel it's probably false. However, intuitively (to me at least), it should be true, for it feels like whenever I've seen an example about how to show some element $m\otimes{n}$ of a tensor product is zero, it usually involves some sort of argument with the bilinearity properties of the tensor and the torsion properties of the module. I'd very much appreciate if someone could help me work towards finding a proof or counterexample to this claim.

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1 Answer 1

up vote 5 down vote accepted

You have to suppose $M\ne 0$ and $N\ne 0$ of course.

Let $K$ be the field of fractions of $R$. Then $M\otimes_R K$ and $N\otimes_R K$ are non-zero because they contain respectively $M$ and $N$. Now $$ (M\otimes_R K)\otimes_K (N\otimes_R K)\ne 0$$ because we tensor two non-zero vector spaces over $K$. As $$ (M\otimes_R N)\otimes_R K= (M\otimes_R K)\otimes_K (N\otimes_R K)\ne 0$$ we get $M\otimes_R N\ne 0$.

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Thanks, but I'm unsure of exactly why $M$ is contained in $M\otimes_{R}{K}$. This isn't true in general so you must be using the torsion freeness of the modules, could you elaborate slightly please? –  Paul Gilmartin Nov 2 '12 at 15:22
    
Never mind, I think I know why this is true. I believe it follows from the fact that $\operatorname{Tor}_{1}^{R}(Q,M)=0$ for torsion free modules $M$, where $Q=K/M$. If there's a more elementary way to see this, please let me know. –  Paul Gilmartin Nov 2 '12 at 15:44
5  
Dear Paul, a more elementary way is to write $M\otimes_R K=S^{-1}M$ with $S=R\setminus \lbrace 0\rbrace $ and realize that $M\to S^{-1}M: m\mapsto \frac {m}{1}$ is injective because $m/1=0 \iff m=0$ (because $M$ has no torsion). –  Georges Elencwajg Nov 2 '12 at 16:12

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