The following is exercise $1.5$ in the book Commutative Algebra With A View Toward Algebraic Geometry by David Eisenbud.
Exercise 1.5
Let $S$ be a commutative Noetherian ring and $R\subset S$ be a subring which is a summand of $S$ in the sense that there exists an $R$-module homomorphism $\varphi: S\longrightarrow R$ such that $\varphi$ takes every element of $R$ to itself. Then prove that $R$ is also Noetherian.
Proof: Let $\mathfrak{a}$ be an ideal(non-zero) of $R$. Consider the ideal $\mathfrak{a}S$ of $S$. Since $S$ is Noetherian, $\mathfrak{a}S$ is generated by a finite subset $\{s_1,s_2,\ldots,s_m\}$ of $\mathfrak{a}S$. Now consider the ideal $\langle \varphi(s_1),\varphi(s_2),\ldots,\varphi(s_m)\rangle$ of $R$ generated by the elements $\varphi(s_j)\in R$. We claim that $\langle \varphi(s_1),\varphi(s_2),\ldots,\varphi(s_m)\rangle=\mathfrak{a}$. Since $R\subset S$, each element $a$ of $\mathfrak{a}$ is also an element of $\mathfrak{a}S$ and hence can be represented as $$a=s_1a_1+s_2a_2+\cdots+s_ma_m$$ for some elements $a_j$ of $S$. Since $a\in\mathfrak{a}\subset R$, we get, $$a=\varphi(a)=\sum_{j=1}^{m}\varphi(s_j)\varphi(a_j)$$ Since $\varphi(a_j)\in R$, we conclude that any element of $\mathfrak{a}$ can be represented as a linear combination of $\varphi(s_j)$ over $R$. Hence $\mathfrak{a}\subseteq\langle \varphi(s_1),\varphi(s_2),\ldots,\varphi(s_m)\rangle$. The reverse inclusion is trivial. therefore we get $\langle \varphi(s_1),\varphi(s_2),\ldots,\varphi(s_m)\rangle=\mathfrak{a}$. Hence every ideal of $R$ is finitely generated, which means $R$ is Noetherian.
$$\tag*{$\blacksquare$}$$
Can someone tell me if my proof is correct or wrong?
Thanks in anticipation!
Best Answer
Your proof is correct! I want to point out another possible approach: for any ideal $\mathfrak{a}$ of $R$, we have $\mathfrak{a} = \varphi(\mathfrak{a}S)$ (both inclusions are easy). Thus, letting $\mathcal{L}_R$ and $\mathcal{L}_S$ denote the posets of ideals of $R$ and $S$, respectively, the map $$\mathfrak{a} \mapsto \mathfrak{a}S : \mathcal{L}_R \to \mathcal{L}_S$$ is injective. This map is also clearly order-preserving, so we can identify $\mathcal{L}_R$ with a subposet of $\mathcal{L}_S$. Since each ascending chain in $\mathcal{L}_S$ stabilizes, each ascending chain in $\mathcal{L}_R$ must also stabilize.