Nilpotent Ideal and Surjective Module Homomorphisms

Module Theory problems and solutions

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.

 
LoadingAdd to solve later
Sponsored Links

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.


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Module Theory
Module Theory problems and solutions
Difference Between Ring Homomorphisms and Module Homomorphisms

Let $R$ be a ring with $1$ and consider $R$ as a module over itself. (a) Determine whether every module...

Close