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This question is from Abstract Algebra book by Pierre Grillet (2nd edition).

Let $\phi:R\to S$ be a homomorphism of rings with identity and let $A$ be a unital left $S$-module. Make $A$ a unital left $R$-module.

Aim is to find a map $g:R \times A \to A$ that satisfies module conditions.

We have $\phi(1_R)=1_S$ and since $A$ is a unital left $S$-module, we have a function $f: S \times A \to A$ with $s \in S, a\in A, f(s,a)=sa$, and since it is unital, we also have $f(1_S,a)=a $ for all $a\in A$.

Can I define the function $g$ as $g:f\circ\phi$, the composition?

If it is possible, we have $(1_R,a)\to (1_S,a)\to a$, which gives unitarity.

Also commutativity is done, since for $r_1,r_2\in R, a\in A$, $R$ action is like, $(r_1+r_2)a=\phi(r_1+r_2)a=[\phi(r_1)+\phi(r_2)]a=\phi(r_1)a+\phi(r_2)a$ which is equal to $r_1a+r_2a$

But $r_1(r_2a)=(r_1r_2)a$ doesn't seem to hold because $\phi$ is a ring homomorphism and we can't have $\phi(r1r2)=r_1\phi(r_2)$.

I might be totally wrong on my work, if so, can someone help me find the function?

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  • $\begingroup$ We have $(r_1r_2)a=\phi(r_1r_2)a=(\phi(r_1)\phi(r_2))a=\phi(r_1)(\phi(r_2)a)=r_1(r_2a)$. $\endgroup$ – Aweygan Aug 31 '17 at 12:57
  • $\begingroup$ You're making it too complicated. First think about the case where $R$ is a subring of $S$ (so $\phi$ is the inclusion map). Then obviously every $S$-module is an $R$-module. All you do is restrict the scalar multiplication to $R$. The general case is straightforward from there. $\endgroup$ – D_S Aug 31 '17 at 13:58
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It may be simpler. Given $r \in R$ and $a \in A$, define your scaling as $$ r \cdot a= \phi(r) \cdot a. $$ Now check that everything is kosher because $\phi$ is a 1-preserving ring map.

My point being you don't need to necessarily write the structure as a composition (which in your case of $f \circ \phi$ isn't defined).

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Giving an $S$-module action on a abelian group $A$ is equivalent to giving a ring homomorphism $\alpha: S \to \mbox{End}(A)$.

Therefore, an $R$-module structure on $A$ is given by the composite ring homomorphism $\alpha \circ \phi : R \to S \to \mbox{End}(A)$.

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