Are the two scalar multiplication on an $R$ module equivalent?

Question

Let $R$ be a commutative ring with unity. Let $M$ be an $R$-module. Then, consider $Hom_{\mathbb{Z}}(R, M)$ (note that the homomorphisms are as $\mathbb{Z}$ modules). This is an $R$ module in two different ways:

$(r*f)(x) = r*(f(x))$

$(r*f)(x) = f(rx)$

Are these the same? My intuition telling me no because if so, then $f(1)$ determines all of $f$, as follows: $f(r) = f(r1) = (r*f)(1) = r*(f(1))$

However, I can't seem to prove this. What's an example of when they are different?

Answer 1

Here’s an easy example showing that the actions are indeed different. Take $R=\mathbb Z[\epsilon]$, where $\epsilon^2=0$, and $M=\mathbb Z^2$. Then $\mathrm{Hom}_{\mathbb Z}(R,M)=\mathbb M_2(\mathbb Z)$.

The first action is induced from the $R$-action on $M$, so $\epsilon$ acts as zero on all matrices.

The second action comes from multiplication on $R$, so if we write $$ \begin{pmatrix}a&b\\c&d\end{pmatrix}\colon x+y\epsilon\mapsto (ax+by,cx+dy) $$ then $$ \epsilon\begin{pmatrix}a&b\\c&d\end{pmatrix} = \begin{pmatrix}b&0\\d&0\end{pmatrix}. $$

Answer 2

Try 2

This is unfortunately slightly less elementary, but not much. (This construction is inspired by ideas around the algebraic version of the cotangent bundle and jet bundles.)

Let $R$ be a commutative ring with 1 and let $M$ be an $R$-module. Recall that a map $D : R \rightarrow M$ is called a derivation if it satisfies $D(r + r') = D(r) + D(r')$ and $D(rs) = rD(s) + sD(r)$. Fix a derivation $D$.

Take $N$ to be $R \oplus M$ as an abelian group, with a $R$-module structure defined by

$r \cdot (s, m) = (rs, rm + sD(r))$

(Aside: This may look slightly mysterious, but it comes from trying to put a ring structure on $R \oplus M$ and is motivated by what the ring structure has to be on the elements of the form $(s, D(s))$.)

It is somewhat lengthy to verify the $R$-module properties, but I don't think much is gained in me writing them out here...

Then, the map $f : R \rightarrow R \oplus M$ sending $s \mapsto (s,0)$ is a morphism of abelian groups, but not of $R$-modules, since

\begin{align*} rf(s) &= r(s,0) = (rs,sD(r)), \mbox{but} \\ f(rs) &= (rs,0) \end{align*}

Try 1 (Incorrect)

(This is incorrect because $M \times M$ isn't actually an $R$-module with the above action.)

I think the following works:

Take $M = R \times R$, with the $R$-module structure given by $r(x,y) = (x,ry)$ (note this isn't the diagonal action).

The map $f : R \rightarrow R \times R$ sending $x \mapsto (x,0)$ is a morphism of abelian groups, but not a morphism of $R$-modules, since

\begin{align*} rf(x) &= r(x,0) = (x,0), \mbox{but} \\ f(rx) &= (rx,0) \end{align*}