I've made some progress on the following problem:

Let $R$ be a Noetherian ring, $R \subseteq S$ an extension of rings, and $x \in S$. Show that $x$ is integral over $R$ if for every minimal prime $\mathfrak q$ of $S$, $\overline{x} \in S/\mathfrak q$ is integral over $R/ \mathfrak q \cap R$.

I believe I've solved the problem under the hypothesis that $S$ is Noetherian. In this case, there are only finitely many minimal prime ideals $\mathfrak q_1, ... , \mathfrak q_s$ of $S$. By hypothesis, for each $i$, there exists a monic polynomial $f_i(X) \in R[X]$ such that $f_i(x) \in \mathfrak q_i$. So then $$f_1(x) \cdots f_t(x) \in \mathfrak q_1 \cap \cdots \cap \mathfrak q_t = \sqrt{0}$$ so $f_1(x)^m \cdots f_t(x)^m = 0$ for some $m \geq 1$. But then $x$ is a root of the polynomial $f_1(X)^m \cdots f_t(X)^m$, which is monic, so $x$ is integral over $R$.

If $S$ is not Noetherian, I can't expect $S$ to have finitely many minimal prime ideals. I haven't figured out how to use the hypothesis that $R$ is Noetherian yet. Is there some way I can adapt the above argument?

  • $\begingroup$ I think the question for $q \cap R$. For instance, $x \in S$ is integral over $R$ if there exists a monic (non zero) polynomial $f(X) \in R[X]$ such that $f(x) = 0 \in R$. The question you need to answer is that for every minimal prime $p$ of $R$, there exists a minimal prime $q$ of $S$ lying over $p$. $\endgroup$
    – Youngsu
    Commented Nov 14, 2015 at 21:44
  • $\begingroup$ Thanks for the suggestion. But if $S \supseteq R$ is not an integral extension, are you sure we can expect something like that? Not every prime in $R$ has a prime in $S$ lying over it, in general. $\endgroup$
    – D_S
    Commented Nov 14, 2015 at 21:46
  • $\begingroup$ I did not say the lying over property holds for every prime ideal of $R$. Let $W = R \setminus p$ where $p$ is a minimal prime of $R$. Then since localization is flat we have $W^{-1}R = R_p \subseteq W^{-1}S$. Since $W^{-1}S$ is a ring with $1$, it has at least one prime ideal say $q$. Then $q \cap R_p$ is a prime ideal in $R_p$, but there is only one prime ideal in $R_p$, $pR_p$. Pulling back $q$ back to $S$, call it $Q$. Then $Q$ lies over $p$. $\endgroup$
    – Youngsu
    Commented Nov 14, 2015 at 21:55
  • $\begingroup$ Okay, but I still don't see why it suffices to look at just those minimal primes of $S$ which lie over minimal primes of $R$. $\endgroup$
    – D_S
    Commented Nov 15, 2015 at 4:57
  • $\begingroup$ The product of functions $f_i(X)$ evaluated at $x$ is in $\sqrt{0_R}$. $\endgroup$
    – Youngsu
    Commented Nov 16, 2015 at 4:51

1 Answer 1


We have the tower of rings $R \subseteq R\left[x\right] \subseteq S$. For every minimal prime ideal $\mathfrak{p}$ of $R\left[x\right]$, there exists a minimal prime ideal $\mathfrak{q}$ of $S$ lying over $\mathfrak{p}$ (this is true for any ring extension, as Youngsu proves in the comments).

Being a finitely generated $R$-algebra, $R\left[x\right]$ is Noetherian. Hence, $R\left[x\right]$ has only finitely many minimal prime ideals, say $\mathfrak{p}_{1}, \ldots, \mathfrak{p}_{t}$. Let $\mathfrak{q}_{1}, \ldots, \mathfrak{q}_{t}$ be minimal prime ideals of $S$ with $\mathfrak{q}_{i} \cap R\left[x\right] = \mathfrak{p}_{i}$ for each $i$.

As you state, for each $i$, there exists a monic polynomial $f_{i}\left(X\right) \in R\left[X\right]$ with $f_{i}\left(x\right) \in \mathfrak{q}_{i}$. Since $x \in R\left[x\right]$ and all the coefficients of $f_{i}\left(X\right)$ are in $R \subseteq R\left[x\right]$, it follows that $f_{i}\left(x\right) \in R\left[x\right]$. Thus, $f_{i}\left(x\right) \in \mathfrak{q}_{i} \cap R\left[x\right] = \mathfrak{p}_{i}$.

From here, you can proceed exactly as you suggest in your question, using that $f_{1}\left(x\right)\cdots f_{t}\left(x\right) \in \sqrt{0_{R\left[x\right]}}$.


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