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Let $R$ and $S$ be rings with $R \subseteq S$, and let $_S M$ be a left $S$-module. If we restrict our scalars to $R$, we naturally have a left $R$-module structure on $M$, $_R M$, given by restricting the $S$-module structure. Now we may extend by scalars back to $S$, and define $_S\hat M = S \otimes_R (_R M)$.

Are there any conditions under which $_S\hat M \cong {_S M}$ in the category of $S$-modules?

Discussion:

Let $\mathcal R: {_S\text{Mod}} \to {_R\text{Mod}}$ be the functor given by restricting scalars, and $\mathcal E = S \otimes_R \square : {_R\text{Mod}} \to {_S\text{Mod}}$ be the extension by scalars. The way that $\mathcal E$ is defined, it seems that $\mathcal{RE} = 1_{{_R\text{Mod}}}$ (object-wise this is fine, though I haven't checked what happens to Hom-sets, so maybe this is wrong?). However, it seems very unlikely that the other direction ($\mathcal{ER} = 1_{{_S\text{Mod}}}$) should hold, since otherwise this would define an isomorphism of categories.

Also, if anyone has an example of the above situation succeeding or failing (especially failing) that would be great; that is, I would really love to see an example of an $S$-module $M$ wherein restricting to a subring $R$ and then extending by scalars gives a different $S$-module.

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    $\begingroup$ Succeed:R=Z ≤ S=Q. Fail: R=R ≤ S=C. $\endgroup$ – Jack Schmidt Mar 18 '14 at 19:21
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Adding to Jack Schmidt's answer, note that $$S\otimes_R M= S \otimes_R (S\otimes_S M)=(S\otimes_R S)\otimes_S M.$$ It follows that if $S\otimes_R S\cong S$, then the functors are invertible to each other. For example, if $R$ is an integral domain and $S$ is a localization of $R$ in some multiplicative set (without zero), then this holds (in particular this is true for the field of fractions of $R$). If this is not true, then taking $M=S$ you get a counter example. For example, if $L/F$ is any nontrivial field extension, then $L\otimes_F L \not \cong L$ (if the extension is finite this follows since the modules have different dimensions over $F$). For $\mathbb{C}/\mathbb{R}$ we have that $\mathbb{C}\otimes_\mathbb{R} \mathbb{C} \cong \mathbb{C}\oplus \mathbb{C} \not \cong \mathbb{C}$.

Also, note that $\mathcal{RE}$ is not the identity. For example taking $R=\mathbb{Z}$, $S=\mathbb{Q}$ and $M=\mathbb{Z}$, you have that $\mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Z}\cong \mathbb{Q}$ which is not isomorphic to $\mathbb{Z}$ (the module) over $\mathbb{Z}$ (the ring).

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  • $\begingroup$ @JackSchmidt : Thanks, added the counter example. $\endgroup$ – Ofir Mar 19 '14 at 19:40
  • $\begingroup$ Is the emphasized isomorphism an isomorphism of $S$-modules? (In this case, is trivial that the associativity holds?) It is obviously an isomorphism of $R$-modules so it follows that if $S\otimes_R S\not\cong S$ as $R$-modules, they are also not isomorphic as $S$-modules, but the other implication of the "iff" statement is not clear to me. $\endgroup$ – user8463524 Apr 29 '15 at 7:54

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