(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.
Basic Exercise Problems in Module Theory
Let $R$ be a ring with $1$ and $M$ be a left $R$-module.
(a) Prove that $0_Rm=0_M$ for all $m \in M$.
Here $0_R$ is the zero element in the ring $R$ and $0_M$ is the zero element in the module $M$, that is, the identity element of the additive group $M$.
To simplify the […]
Annihilator of a Submodule is a 2-Sided Ideal of a Ring
Let $R$ be a ring with $1$ and let $M$ be a left $R$-module.
Let $S$ be a subset of $M$. The annihilator of $S$ in $R$ is the subset of the ring $R$ defined to be
\[\Ann_R(S)=\{ r\in R\mid rx=0 \text{ for all } x\in S\}.\]
(If $rx=0, r\in R, x\in S$, then we say $r$ annihilates […]
Nilpotent Ideal and Surjective Module Homomorphisms
Let $R$ be a commutative ring and let $I$ be a nilpotent ideal of $R$.
Let $M$ and $N$ be $R$-modules and let $\phi:M\to N$ be an $R$-module homomorphism.
Prove that if the induced homomorphism $\bar{\phi}: M/IM \to N/IN$ is surjective, then $\phi$ is surjective.
[…]
Finitely Generated Torsion Module Over an Integral Domain Has a Nonzero Annihilator
(a) Let $R$ be an integral domain and let $M$ be a finitely generated torsion $R$-module.
Prove that the module $M$ has a nonzero annihilator.
In other words, show that there is a nonzero element $r\in R$ such that $rm=0$ for all $m\in M$.
Here $r$ does not depend on […]
Short Exact Sequence and Finitely Generated Modules
Let $R$ be a ring with $1$. Let
\[0\to M\xrightarrow{f} M' \xrightarrow{g} M^{\prime\prime} \to 0 \tag{*}\]
be an exact sequence of left $R$-modules.
Prove that if $M$ and $M^{\prime\prime}$ are finitely generated, then $M'$ is also finitely generated.
[…]
Linearly Dependent Module Elements / Module Homomorphism and Linearly Independency
(a) Let $R$ be a commutative ring. If we regard $R$ as a left $R$-module, then prove that any two distinct elements of the module $R$ are linearly dependent.
(b) Let $f: M\to M'$ be a left $R$-module homomorphism. Let $\{x_1, \dots, x_n\}$ be a subset in $M$. Prove that if the set […]