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I'm looking for an example of modules $A,B$ such that $A\otimes_{R}B\neq A\otimes_{\mathbb{Z}}B$.

I've thought about some tensor products I know but I haven't been able to produce an example. What I've considered are as follows: $\mathbb{Z}_{n}\otimes\mathbb{Z}_{m}\cong\mathbb{Z}_{\gcd(n,m)}$, $\mathbb{Q}\otimes\mathbb{Q}\cong\mathbb{Q}$, $A\otimes\mathbb{Q}=0$ for any torsion abelian group $A$, $A\otimes_{R}R\cong A$ if $R$ is a ring with 1.

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2 Answers 2

up vote 5 down vote accepted

Generally $M\otimes_R N$ is "bigger" than $M\otimes_SN$ when $R$ is a subring of $S$. This is because the scalar ring being tensored over is what creates relations on the symbols $m\otimes n$, and the bigger the ring the more relations result. There are two obstacles to this always being true: divisibility prevents growth and torsion collapses things altogether.

If $S$ contains any "fractions" of $R$ elements (as seen in a localization or fraction field), then for those nonunits of $R$ that are invertible in $S$, we can say that the modules $M,N$ are "divisible" with respect to them since we can apply the reciprocals indefinitely. The reason no further relations are introduced by fractions of $R$ elements is that sliding a scalar reciprocal $1/r$ from left to right side of the $\otimes$ symbol, i.e. $nr^{-1}\otimes m=n\otimes r^{-1}m$, is the same as sliding $r$ from the right side to left side of the tensor symbol, explicitly $nr^{-1}\otimes m=nr^{-1}\otimes r(r^{-1}m)=(nr^{-1})r\otimes r^{-1}m=n\otimes r^{-1}m$, and the same idea holds for moving reciprocals from right to left.

As you have seen, any incompatible torsion between $N$ and $M$ over the scalar ring will result in the tensor product simply being the zero space, since we can introduce the zero scalar in any symbol willy-nilly with some manipulation.

If you want a counterexample to $M\otimes_{\bf Z}N\cong M\otimes_SN$, you'll want to avoid torsion and go from $R$ to $S$ via more than just introducing fractions. Fields are very nice, well-behaved objects to work with, comparatively anyway. In particular, let $K$ be a number field (which is a finite extension of $\bf Q$) having degree $n>1$ (in particular then $\dim_{\,\bf Q}K=n$ as a vector space). Then

$$K\otimes_{\bf Z}K\cong K\otimes_{\bf Q}K\cong{\bf Q}^{n^2}\not\cong{\bf Q}^n\cong K\cong K\otimes_KK.$$

So $R=\bf Z$, $S=N=M=K$ gives a counterexample when $K$ is a number field.

Some background: Tensoring a $K$-vector space by a field extension $L/K$ is called "ascent," finding a $K$-subspace $W$ of an $L$-vector space $V$ such that every $K$-basis of $W$ is an $L$-basis of $V$ is called "descent." Tensoring a real vector space against $\bf C$ is called complexification. Tensoring in this way is a nice way to change the underlying scalars of a space.

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Hint: look at the last example you mention. It may help to pick a special $A$....

Incidentally, this turns out to be one of those things that, in the grand scheme of things, getting equality of the two tensor products is a very rare thing -- it's just that you get equality for a number of "nice" constructions, and so if you only consider the nice ones you won't find an example.

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Perhaps the polynomial ring $\mathbb{Z}[x]$ might work? – user44532 Feb 23 '13 at 15:30
@CYC: There is a simple example with $R = \mathbb{Z}[x]$. I would go with $A = \mathbb{Z}$ in comparison, although you'll have to find an $R$-module structure on $A$. (Hint: $A$ is a quotient of $R$ in many ways) – Hurkyl Feb 23 '13 at 15:35
I can make $\mathbb{Z}$ a $\mathbb{Z}[x]$-module via the ring homomorphism $\mathbb{Z}[x]\rightarrow\mathbb{Z},f(x)\mapsto f(0)$, i.e. $f(x)n:=f(0)n$. – user44532 Feb 23 '13 at 15:58

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