A homomorphism that is an $R$-module homomorphism but is not a ring homomorphism.

Question

I read the following example in a book.

For any modules the zero map $0:A \to B$ given by $a \mapsto 0$ (for $a \in A$) is a module homomorphism. Every homomorphism of abelian groups is a $Z$-modules homomorphism.

If $R$ is a ring, the map $R[x] \to R[x]$ given by $f \mapsto xf$ (for example, $(x^2 + 1) \mapsto x(x^2 + 1)$) is an $R$-module homomorphism, but not a ring homomorphism.

But I did not understand:

1 - Why should every homomorphism of abelian groups be a $Z$-module homomorphism? Could anyone explain this for me please?

2 - Also I do not understand why the given map is an $R$-module homomorphism but is not a ring homomorphism. Could anyone clarify this for me please?

Answer 1

Because if $f$ is an homorphism between commutative groups, $f(nx)=nf(x)$ where $n$ is an integer.

For the second question $f(pq)$ is not always $f(p)f(q)$ as $x(pq)$ is not always $(xp)(xq)$ for polynomials $p,q\in R[X]$.

Answer 2

1) You have to show that for any $n\in\mathbb{X}$, $f(nx)=nf(x)$. First, we prove it for the naturals by induction. So $f(0x)=f(0)=0=0x$. We assume it is valid for $n$, then $f((n+1)x)=f(nx+1x)=f(nx)+f(x)=nf(x)+f(x)=(n+1)f(x)$. Now we prove it for the negatives, let $n$ be a negative integer, then $n=-m$ with $m$ a positive integer, then $f(nx)=f(-mx)=f(m(-x))=mf(-x)=m(-f(x))=-mf(x)=nf(x)$.

2) As @Tesmo Aristide says.