# Difference Between Ring Homomorphisms and Module Homomorphisms

## Problem 422

Let $R$ be a ring with $1$ and consider $R$ as a module over itself.

(a) Determine whether every module homomorphism $\phi:R\to R$ is a ring homomorphism.

(b) Determine whether every ring homomorphism $\phi: R\to R$ is a module homomorphism.

(c) If $\phi:R\to R$ is both a module homomorphism and a ring homomorphism, what can we say about $\phi$?

## Solution.

### (a) Determine whether every module homomorphism $\phi:R\to R$ is a ring homomorphism.

Consider the ring of integers $R=\Z$. Then the map
$\phi: \Z \to \Z$ defined by
$\phi(x)=2x$ is a $\Z$-module homomorphism.

In fact, we have for $x, y, r\in R$
\begin{align*}
\phi(x+y)=2(x+y)=2x+2y=\phi(x)+\phi(y)
\end{align*}
and
\begin{align*}
\phi(rx)=2(rx)=r(2x)=r\phi(x).
\end{align*}

However, the map $\phi$ is not a ring homomorphism since $\phi(1)=2\neq 1$.
(Every ring homomorphism sends $1$ to itself.)

Thus, we conclude that not every module homomorphism $\phi:R\to R$ is a ring homomorphism.

### (b) Determine whether every ring homomorphism $\phi: R\to R$ is a module homomorphism.

Let us consider the polynomial ring $R=\Z[x]$.
Consider the map
$\phi:\Z[x] \to \Z[x]$ defined by
$\phi\left(\, f(x) \,\right)=f(x^2)$ for $f(x)\in \Z[x]$.

Then $\phi$ is a ring homomorphism because we have
\begin{align*}
&\phi\left(\, f(x)+g(x) \,\right)=f(x^2)+g(x^2)=\phi\left(\, f(x) \,\right)+\phi\left(\, g(x) \,\right), \text{ and }\\
&\phi\left(\, f(x)g(x) \,\right)=f(x^2)g(x^2)=\phi\left(\, f(x) \,\right)\phi\left(\, g(x)\right).
\end{align*}

However, the map $\phi$ is not a $\Z[x]$-module homomorphism. If it were a $\Z[x]$-module, then we would have
\begin{align*}
&x^2=\phi(x)\\
&=x\phi(1) && \text{since $\phi$ is a $\Z[x]$-module homomorphism}\\
&=x\cdot 1 && \text{since $\phi$ is a ring homomorphism}\\
&=x,
\end{align*}
and thus we have $x=x^2$, which is a contradiction.

Thus, the conclusion is that not every ring homomorphism $\phi:R\to R$ is a module homomorphism.

### (c) If $\phi:R\to R$ is both a module homomorphism and a ring homomorphism, what can we say about $\phi$?

Suppose that $\phi:R\to R$ is both a ring homomorphism and an $R$-module homomorphism.

Then for any $x\in R$, we have
\begin{align*}
&\phi(x)=x\phi(1) && \text{since $\phi$ is an $R$-module homomorphism}\\
&=x\cdot 1 && \text{since $\phi$ is a ring homomorphism}\\
&=x,
\end{align*}
and it follows that $\phi$ must be the identity map.

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