(A ring is an associative ring with identity.)
In a book, the author claims that
- giving the $R$-module structure to a $k$-module $M$ is equivalent to giving a $k$-algebra homomorphism $R \to \mathrm{End}_k (M)$, where $k$ is a commutative ring and $R$ is a $k$-algebra.
But why?
To be precise,
- definition: a (left) $R$-module, where $R$ is a ring, is a ring homomorphism $R \to \mathrm{End}_{\mathbb Z} (M)$ for some abelian group $M$;
- definition: a $k$-algebra, where $k$ is a commutative ring, is a ring homomorphism $k \to Z (R) \hookrightarrow R$ for some ring $R$, where $Z (R) = \{ r \in R \mid x r = r x \ \forall x \in R \}$ is the center of $R$;
- proposition: then a ring homomorphism $\phi : R \to \mathrm{End}_{\mathbb Z} (M)$ induces a $k$-algebra homomorphism $\phi : R \to \mathrm{End}_k (M)$, where $k$ is a commutative ring, $R$ is a $k$-algebra, and $M$ is a $k$-module.
Let $k$ be a commutative ring, $R$ a $k$-algebra, and $\psi : k \to \mathrm{End}_{\mathbb Z} (M)$ a $k$-module.
In the sequel, $a \in R, \lambda \in k$.
To show that a ring homomorphism $\phi : R \to \mathrm{End}_{\mathbb Z} (M)$ determines the certain $k$-algebra homomorphism, we have to show that
$$
\phi (\lambda) = \psi (\lambda) \quad \forall \lambda \in k
$$
because then
- since $\phi (a) \circ \psi (\lambda) = \phi (a) \circ \phi (\lambda) = \phi (\lambda) \circ \phi (a) = \psi (\lambda) \circ \phi (a)$, each $\phi (a)$ is a $k$-endomorphism;
- since $\phi (\lambda a) = \phi (\lambda) \circ \phi (a) = \psi (\lambda) \circ \phi (a)$, $\phi$ is in fact a $k$-homomorphism.
But I have no idea of the proof of the equation $\phi (\lambda) = \psi (\lambda)$. I suspect that there is a case in which $\phi (\lambda) \neq \psi (\lambda)$, but I do not have examples of it, too.
Please tell me the proof (or a counterexample) of the proposition.
Best Answer
I would assume the author meant to say something similar to the following:
Let $k$ be a commutative ring, $R$ a $k$-algebra and $M$ a $k$-module. Then giving an $R$-module structure to $M$ which extends the already given $k$-module structure is equivalent to giving a $k$-algebra homomorphism $R\to \text{End}_k(M)$.
The added 'which extends...' part is exactly the condition $\phi(\lambda) = \psi (\lambda)$ for $\lambda \in k$ which you found.
To see that this addition is actually necessary, let $R = k = M = \mathbb{F}_{p^2}$ where the $k$-algebra structure of $k$ is given by the identity map. We can consider $k$ with the usual $k$-module structure $$ k \to \text{End}_\mathbb{Z}(k), a \mapsto (b \mapsto ab) $$ but we also have another module structure given by $$ k \to \text{End}_\mathbb{Z}(k), a \mapsto (b \mapsto a^p b)$$ and since these are different, they do not extend one another.