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From Algebra by Hungerford:

Let $R$ and $S$ be rings and $\phi: R \rightarrow S$ a ring homomorphism. Then every $S$-module $A$ can be made into an $R$-module by defining $rx$ $(x \in A)$ to be $\phi(r)x$. One says that the $R$-module structure of $A$ is gen. by pullback along $\phi$.

I find this a bit confusing, because $\phi(r) \in S$ and not in $R$. Then how do we get an $R$-module if the scalars are from $S$?

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    $\begingroup$ By defining $rx$ to be $\phi(r)x$... $\endgroup$ – anon Aug 26 '13 at 21:34
  • $\begingroup$ @anon But if we define $rx$ to be $\phi(r)x$, then we are not multiplying elements of $A$ by scalars from $R$...we are multiplying them by scalars from $S$, since $\phi(r) \in S$... $\endgroup$ – user58289 Aug 26 '13 at 21:38
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    $\begingroup$ We define multiplication on $A$ by scalars from $R$ in terms of the already known action of $S$ on $A$. We are defining the multiplication of by scalars from $R$! $\endgroup$ – anon Aug 26 '13 at 22:46
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    $\begingroup$ Yes, that's right. $\endgroup$ – Ted Aug 27 '13 at 2:10
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    $\begingroup$ Perhaps a better way of understanding an $R$-module structure is $R\to{\rm End}_{\Bbb Z}(A)$, i.e. an action of $R$ on an abelian group $A$. If $A$ is an $S$-module and $R\to S$ a homomorphism then we have an $R$-module structure as $R\to S\to{\rm End}(A)$. But really, "$rx$ is defined to be $\phi(r)x$" should by itself be sufficient explanation of what's going on. $\endgroup$ – anon Aug 27 '13 at 2:33
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You already have an $S$-module, so $\phi(r)x$ is already defined. By defining $rx$ to have this same value, you have defined how to multiply elements of $A$ by scalars in $R$. After checking some equations, you find that this new multiplication makes $A$ into an $R$-module.

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The datum of an $S$-module structure on an abelian group $A$ is precisely a ring homomorphism $\rho: S \rightarrow \operatorname{End}_{\text{Ab}}A$ where $\operatorname{End}_{\text{Ab}}A$ denotes the endomorphism ring of $A$. Indeed, defining $\rho(s)(a) = s \cdot a$ for every $s \in S$ and $a \in A$, it is straightforward to check that the module axioms are equivalent to the assertion that $\rho$ is a ring homomorphism.

Therefore given a ring homomorphism $\phi: R \rightarrow S$, the composite ring homomorphism $ \rho \circ\phi: R \rightarrow \operatorname{End}_{\text{Ab}}A$ puts an $R$-module structure on $A$.

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    $\begingroup$ Hey Cihan, your answer makes sense to me. But I just want know more about the equivalence of the module axioms with the ring homomorphism from Ring R to End(A). Thanks $\endgroup$ – Mat He Mat Cian Oct 7 '13 at 14:53
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    $\begingroup$ A ring homomorphism from $R$ to $End(A)$ sends $1$ to the identity map, so $1 \cdot a = a$ for all $a \in A$. It preserves addition, so $(r+s) \cdot a = r \cdot a + r \cdot s$. It preserves multiplication (note that multiplication in $End(A)$ is composition), so $(rs) \cdot a = r \cdot (s \cdot a)$. Finally the action of $r$ is additive (since it is mapped to an element in $End(A)$) hence $r \cdot (a+b) = r \cdot a + r \cdot b$. $\endgroup$ – Cihan Oct 7 '13 at 21:59
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    $\begingroup$ To be clear, I start with a ring homomorphism $\rho: R \rightarrow End(A)$ and given $r \in R$ and $a \in A$ define $r \cdot a = \rho(r)(a)$. For the other direction, if we are given a map $R \times A \rightarrow A$ which satisfies the module axioms, we can construct a ring homomorphism $\rho:R \rightarrow End(A)$. This can be checked in the same way. $\endgroup$ – Cihan Oct 7 '13 at 22:02
  • $\begingroup$ @Cihan how would one define an $R$-linear arrow of $R$-modules when the latter are defined as ring actions on abelian groups? $\endgroup$ – Arrow Oct 25 '18 at 14:37
  • $\begingroup$ @Arrow If $\rho_i : R \rightarrow End(A_i)$ is a ring homomorphism for $i=1,2$, an $R$-linear map from $A_1$ to $A_2$ is a group homomorphism $f:A_1 \rightarrow A_2$ such that $f \circ \rho_{1}(r) = \rho_2(r) \circ f $ for all $r \in R$. $\endgroup$ – Cihan Oct 25 '18 at 17:50

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