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If we have a ring homomorphism $f\colon R_{1}\rightarrow R_{2}$, and if $M$ is an $R_{1}$-module, my question is: Can we show that the $R_{1}$-module $R_{2}\otimes_{R_{1}}M$ is somehow also an $R_{2}$-module?

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Yes, in the natural way: $a(b\otimes m) = (ab)\otimes M$, extend bilinearly. – Arturo Magidin Jun 21 '12 at 15:17
Adding one thing to the good comment and answer from Arturo and Dylan. The key here is that the ring $R_2$ is a right $R_1$-module, the structure coming via the homomorphism $f$. But it is also a left $R_2$-module, and what's more, the two module actions commute, so $R_2$ is an $(R_2,R_1)$-bimodule. Forming the tensor product in a way `consumes' the $R_1$-action, but the $R_2$-action survives, and can be used to give the tensor product the structure of an $R_2$-module. This process is very common also in representation theory of finite groups (look up induced representation). – Jyrki Lahtonen Jun 21 '12 at 18:55
up vote 5 down vote accepted

Yes. This is called extending scalars to $R_2$ and the $R_2$-module structure is as Prof Magidin has described. To be completely formal, if $r \in R_2$ then the map $t_r\colon R_2 \to R_2$ that is multiplication by $r$ is $R_1$-linear and you can define for $x \in R_2 \otimes_{R_1} M$ that $rx = (t_r \otimes \operatorname{id}_M)(x)$.

On the scheme side of things, this corresponds to the concept of base change.

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