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?

  • $\begingroup$ 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

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).


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 $\alpha \circ \phi : R \to S \to \mbox{End}(A)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.