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Let $(R,\mathfrak{m})$, $(S,\mathfrak{n})$ and $\phi:R\to S$ be a ring homomorphism such that $S$ is a finitely generated $R$-module. If $R$ is Noetherian, is $S$ integral over $R$ and $\mathfrak{n}^{c}=\mathfrak{m}$?

This is my opinion: $s_{1},\dots,s_{n}\in S$ are called the finite generated as $R$-module. Since $S$ is finitely generated as $R$-module and $R$ is Noetherian, $S$ is a Noetherian $R$-module. Moreover, $R[s_{i}]$ is $R$-submodule of $S$, then $R[s_{i}]$ is finitely generated as $R$-module or $s_{i}$ is integral over $R$. That means $S$ is integral over $R$, using Lying Over theorem, there exists an ideal $I\in S$ such that $I^{c}=\mathfrak{m}$. But $\mathfrak{m}\subseteq \mathfrak{m}^{ec} \subseteq \mathfrak{n}^c\subseteq \mathfrak{m}$, hence $\mathfrak{n}^{c}=\mathfrak{m}$.

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  • $\begingroup$ Nothing excepting $\mathfrak m^e\ne S$ is necessary, and this follows e.g. from Lying Over. $\endgroup$ – user26857 May 21 '17 at 9:23
  • $\begingroup$ But is my solution about $S$ is integral over $R$ true? It is the main condition for using Lying Over $\endgroup$ – Soulostar May 21 '17 at 13:00
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If $S$ is a finite $R$-algebra, it is integral over $R$, whether $R$ is noetherian or not, and even if $S$ is non-commutative (Bourbaki, Commutative Algebra, ch. 5, §1, prop.1).

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