# Nilpotent Ideal and Surjective Module Homomorphisms

## Problem 431

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.

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## Proof.

Since the homomorphism $\bar{\phi}:M/IM \to N/IN$ is surjective, for any $b\in N$ there exists $a\in M$ such that

\[\bar{\phi}(\bar{a})=\bar{b}, \tag{*}\]
where $\bar{a}=a+IM$ and $\bar{b}=b+IN$.

By definition of $\bar{\phi}:M/IM \to N/IN$, we have

\[\bar{\phi}(\bar{a})=\overline{\phi(a)}=\phi(a)+IN.\]
Thus, it follows from (*) that

\[\phi(a)+IN=b+IN,\]
or equivalently

\[b-\phi(a)\in IN.\]
Thus we have

\[b\in \phi(M)+IN.\]

Now we claim that for any $b\in N$ and any positive integer $k$, we have

\[b\in \phi(M)+I^kN.\] We prove this claim by induction on $k$.

The base case $k=1$ is proved above.

Suppose that $b\in \phi(M)+I^nN$. Then we prove that $b\in \phi(M)+I^{n+1}N$.

Since $b\in \phi(M)+I^nN$, we have

\[b=\phi(a)+\sum_{i}\alpha_i c_i,\] where the sum is finite and $\alpha_i\in I^n$ and $c_i\in N$.

Since each $c_i\in N$, we have $c_i \in \phi(M)+IN$ by the base case.

Hence we have

\[c_i=\phi(a_i)+\sum_{j_i}\beta_{j_i}d_{j_i}\] for some finite pairs $(\beta_{j_i}, d_{j_i})\in (I, N)$.

It follows that we have

\begin{align*}

b&=\phi(a)+\sum_{i}\alpha_i c_i\\

&=\phi(a)+\sum_{i}\alpha_i \left(\, \phi(a_i)+\sum_{j_i}\beta_{j_i}d_{j_i} \,\right)\\

&=\phi(a)+\sum_i\alpha_i\phi(a_i)+\sum_{i}\sum_{j_i}\alpha_i\beta_{j_i}d_{j_i}\\

&=\phi(a)+\sum_i\phi(\alpha_ia_i)+\sum_{i, j_i}(\alpha_i\beta_{j_i})d_{j_i}\\

&=\phi\left(\, a+\sum_i\alpha_ia_i \,\right)+\sum_{i, j_i}(\alpha_i\beta_{j_i})d_{j_i},

\end{align*}

where the last two equalities follows since $\phi$ is an $R$-module homomorphism.

Since $\alpha_i\in I^n$ and $\beta_{j_i}\in I$, the product $\alpha_i\beta_{j_i}\in I^{n+1}$.

Hence the above expression of $b$ yields that

\[b\in \phi(M)+I^{n+1}N,\] and this completes the induction step and the claim is proved.

Now, since $I$ is a nilpotent ideal by assumption, there is a positive integer $n$ such that $I^n$ is the zero ideal of $R$. Thus, it follows from the claim that for any $b\in N$ we have

\begin{align*}

b\in \phi(M)+I^nN=\phi(M).

\end{align*}

This implies that $\phi:M\to N$ is surjective as required.

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