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First, $R$ is a commutative ring with unit. I have two $R$-algebras, $A$ and $B$. I have an isomorphism $$ A\otimes_R\mathbb{Z}\longrightarrow B\otimes_R\mathbb{Z}. $$ Is there a theorem that says when I can conclude that $A\cong B$? What if everything is free?

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    So $\mathbf Z$ is an $R$-algebra/module somehow?2012-07-06
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    As Dylan has noted, you first need that $\mathbb Z$ has an $R$-algebra structure for the tensor to make sense. That essentially means a homomorphism $R\rightarrow \mathbb Z$.2012-07-06
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    @DylanMoreland Yes, $R$ is augmented. In my case $R$ is a graded ring with $R_0=\mathbb{Z}$.2012-07-06
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    Well if "everything is free" means $A$ and $B$ are free $R$ modules, then $A\otimes_R \mathbb{Z}\cong \mathbb{Z}^n$ and likewise for $B\otimes_R\mathbb{Z}$, but I'm not sure you want that to happen. (?)2012-07-06
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    The short answer is: No, there is no such theorem. The long answer will be given if you reveal the *real* problem you are working on to us.2012-07-07

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Take $R = \mathbb Z[t]$, $R\to \mathbb Z$ the augmentation $t\mapsto 0$, $A = R$ and $B = \mathbb Z[t, (1-t)^{-1} ]$. The $A\otimes_R \mathbb Z = \mathbb Z = B\otimes_R \mathbb Z$. But $A$ and $B$ are not isomorphic.

By tensoring with $\mathbb Z$, we evaluated at $0$ so we can't see what happens at $1$: we inverted $1-t$.

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    This is a good counterexample. I had to refine my conditions to obtain a proof.2012-07-17