Hom ring of any ideal of any integral domain is commutative?

Question

Let $R$ be an integral domain (a commutative ring with unity without non-zero zero divisor) ; $I$ be an ideal of $R$ , so that $I$ is an $R$-module . Let $f,g : I \to I$ be any two $R$-module homomorphisms . Then is it true that $f \circ g = g\circ f $ ? I have no idea where to start ; and I have no intuition as to the validity of the statement . Please help . Thanks in advance

Answer 1

Lemma: in the context above, $fg(y)=gf(y)$ for all $y\in I^2$.

Proof: Let $i,j\in I$. Then $(fg-gf)(ij)=fg(ij)-gf(ij)=f(g(i)j)-g(if(j))=f(j)g(i)-f(j)g(i)=0$. Since this holds for all $i,j$, it holds for all elements of $I^2$.

Notice that the above holds for any commutative ring $R$.

Proof of your problem:

Suppose $(fg-gf)(x)=z\neq 0$ for some $x\in I$. Then $(fg-gf)(x^2)=(fg-gf)(x)x=zx\neq 0$ also. But by the lemma we know this cannot be the case since $x^2\in I^2$.

Now actually I remember the version of this question I solved long ago:

Proposition: Let $I$ be an ideal of a commutative ring $R$ such that $I$ contains a regular element $r$ (that is, $r$ isn't a zero divisor.) Then $End(I_R)$ is commutative.

The proof is the same, except that when you arrive at "Suppose $(fg-gf)(x)=z\neq 0$ for some $x\in I$", you take your special regular element $r\in I$ and say that "$(fg-gf)(xr)=(fg-gf)(x)r=zr\neq 0$, a contradiction."