Let $f:A \to B$ be a surjective homomorphism between rings. If $B$ and ker$f$ has ascending chain condition, then $A$ has it too.

Question

Let $f:A \to B$ be a surjective homomorphism between rings, where $A$ is abelian. Let $B$ and $\text{ker} f$ fulfill the ascending chain condition. I want to show that $A$ fulfills the ascending chain condition, namely that given $$A_{1} \subseteq A_{2}\subseteq ...\subseteq A,$$ there is an integer $N$ such that $A_{M}=A_{N}$ for $M \geq N$. Here's my attempt. Let $$A_{1} \subseteq A_{2}\subseteq ...\subseteq A.$$
Since $f$ preserves ideals, we have that $$f(A_{1}) \subseteq f(A_{2}) \subseteq ... \subseteq f(A).$$ Since $B$ satisfies the ascending chain condition, there is an integer $N$ such that $$f(A_{M})=f(A_{N})$$ for $M \geq N.$ Is it valid to say that $$f ^{-1} (f(A_{M}))=A_{M}=f ^{-1}( f(A_{N}))=A_{N}$$ for $M \geq N,$ and thus $A$ also satisfies the ascending chain condition? I'm not sure where to go from here and how to use the fact that ker $f$ is acc. I also thought about proving the contrapositive or getting a contradiction. I take this statement to logically be
$$f \text{ surjective homomorphism } \land B \text{ acc} \land \text{ker} f \text{ acc} \implies A \text{ acc} . $$ Therefore the contrapositive would be $$A \text{ not acc} \implies f \text{ not surjective homomorphism } \lor B \text{ not acc } \lor \text{ker} f \text{ not acc.} $$ But if we assume that $B$ isn't acc, then given any ideals $$A_{1} \subseteq A_{2}\subseteq ...\subseteq A$$ they never stop increasing. But this contraducts our assumption that ker $f$ is acc. Is this reasoning correct?

Answer 1

Recall that an $R$-module $M$ is defined to be Noetherian if the following equivalent conditions hold:

1. $M$ has the ascending chain condition.
2. Every submodule of $M$ is finitely generated.
3. Every set of submodules of $M$ has a maximal element (we won't use this condition).

The ring $R$ is Noetherian if it is Noetherian as a module over itself, which (by 2.) is equivalent to all of its ideals being finitely generated.

Let $I = \operatorname{ker}f$. Then $I$ is an ideal of $A$ and $B \cong A/I$ by the First Isomorphism Theorem. So we are assuming that $I$ and $A/I$ have the ascending chain condition. That is, $I$ is a noetherian $A$-module and $A/I$ is a Noetherian ring.

Let $J$ be an ideal of $A$. Then by the Second Isomorphism Theorem, $$ \frac{J}{I\cap J} \cong \frac{I+J}{I}. $$ Since $I\cap J$ is a submodule of $I$, which is Noetherian, so $I\cap J$ is finitely generated as an $A$-module (since submodules of Noetherian modules are finitely generated). Suppose that $I \cap J$ is generated by $f_1,\ldots, f_m \in I\cap J$ as an $A$-module.

Now, $\frac{I+J}{I}$ is an ideal of $A/I$, so since $A/I$ is Noetherian it is finitely generated. By our isomorphism above, $J/(I\cap J)$ is also finitely generated, say by $\bar{g}_1,\ldots, \bar{g}_n$. For each $i$, there is an element $g_i\in J$ such that $\bar{g}_i = g_i + I\cap J$.

Now let $x \in J$. Then $x + I\cap J \in J/(I\cap J)$, so there exist $a_1,\ldots, a_n \in A$ such that $$ x + I\cap J = a_1g_1 + \ldots + a_ng_n + I\cap J. $$ Then $x - a_1g_1 - \ldots a_ng_n \in I\cap J$, so there are elements $b_1,\ldots, b_m \in A$ such that $$ x - a_1g_1 - \ldots - a_ng_n = b_1f_1 + \ldots + b_mf_m. $$ It follows that $x \in M$, where $M$ is the $A$-module generated by the $f_i$ and the $g_i$. Since $x \in J$ was arbitrary, we have $J\subseteq M$. But each $f_i \in I\cap J \subseteq J$ and $g_i \in J$, so $M\subseteq J$. Therefore $J = M$ is finitely generated.

So every ideal of $A$ is finitely generated, which means that $A$ is Noetherian, so $A$ has the ascending chain condition.