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

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?

• I've noticed you frequently include pictures of text in your questions, and I would just like to suggest including the text itself in the body of your questions. This makes it easier to search for your question in the future, and makes it easier for potential answerers to pick apart the relevant portions of your question. – Sriotchilism O'Zaic Oct 12 '17 at 0:35
• Also for those of us living in the relative stone age, text renders better than pics – mdave16 Oct 12 '17 at 0:49
• Ok , I promise I will respect this as long as I have time @EpsilonNeighborhoodWatch. – Intuition Oct 12 '17 at 1:13
• @mdave16 hahaha , I promise I will respect this as long as I have time. – Intuition Oct 12 '17 at 1:13

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 $p,q\in R[X]$.
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 $f(nx)=f(-mx)=f(m(-x))=mf(-x)=m(-f(x))=-mf(x)=nf(x)$.