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I've encountered this problem on my Non commutative algebra handouts wich says:

given $R,S$ rings and $f:R\to S\:$ a ring homomorphism, define a canonical functor $$F:\textbf{Mod-S}\to \textbf{Mod-R}.$$ Where $\textbf{Mod-S}$ is the category of right $S-$modules and similarly $\textbf{Mod-R}$ is the category of right $R-$modules.

Once you've done this, prove that the functor $F$ is faithful.

Now, my problem is that the definition of $F$ doesn't sound natural at all to me. It would be better in my opinion to define $$F:\textbf{Mod-R}\to \textbf{Mod-S}.$$ Am i wrong? and in the case, how to define $F$?

Thank you for your replies.

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“doesn't sound natural at all to me.” Mathematics has a peculiar sense of naturality. A concept which is not natural for a novice is natural for a person who learned more. –  beroal Nov 10 '11 at 12:28

2 Answers 2

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Actually there are canonical functors in both directions. To go from $R$-modules to $S$-modules, you need to use the tensor product (and this functor is not in general faithful), but the other direction (which is suggested in the homework) is easier.

One way to see what the construction is to note that an $S$-module is the same thing as a an abelian group $A$ and a ring homomorphism $S \to \mathrm{End}(A)$. These ring homomorphisms can be precomposed...

If you want something to google, look up restriction and extension of scalars.

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As an instructive example, think about the case where $f$ is an inclusion of a subring $R\subset S$. Then an $S$-module is also an $R$-module. Just forget how elements outside of $R$ acted. This gives a functor $\textbf{Mod-S}\to \textbf{Mod-R}$.

There is, however, no way to extend an $R$-action to an $S$-action on a given underlying abelian group. But, as Dan Petersen mentions, there is extension of scalars. This gives a functor $\textbf{Mod-R}\to \textbf{Mod-S}$ as you suggest.

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