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Let $R$ be a commutative ring with unity, and $A$ be a commutative algebra over $R$ with unity. Let $M$ be an $A$-module. Then can we say that $M$ is also an $R$-module?

The definition of a commutative algebra over $R$ with unity that I have studied is the following:

A commutative $R$-algebra $A$ is a commutative ring that is also an $R$-module where the scalar multiplication satisfies $$ r.(xy) = (r.x)y = x(r.y) $$ for all $r \in R$ and $x,y \in A$

Saikat
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  • Yes. If $f \colon R \to A$ is the ring homomorphism defining the $R$-algebra structure on $A$, then for any $m \in M$, you can define the action of $r \in R$ on $m$ by $rm := f(r)m$. If you prefer to think of the $A$-module structure on $M$ as a ring homomorphism $A \to \mathrm{End}_{\mathbb{Z}}(M)$, then the induced $R$-module structure on $M$ is just given by the composition of this morphism with $f$. – Alex Wertheim Oct 03 '22 at 04:54
  • @AlexWertheim I have added the definition of a commutative R-algebra I have studied in the post. I read Wikipedia, where it states that the definition via Homomorphism and the one above are equivalent. But I cannot understand how. – Saikat Oct 03 '22 at 05:06
  • As noted in the wikipedia article, given your scalar multiplication, the corresponding homomorphism $R \to A$ is given by $r \mapsto r \cdot 1_{A}$. Where are you stuck in understanding this? – Alex Wertheim Oct 03 '22 at 19:27

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