Let $R$ be an integral domain and $\bar R$ denote its integral closure in the fraction field (i.e. normalization). If $p\in R$ is a prime element in $R$, then does $p$ remain prime in $\bar R$ also ?

If this is not true in general, then what if we also assume $R$ is Noetherian ?

By a prime element in an integral domain $R$, I mean a non-zero non unit $p\in R$ such that $p |ab $ for some $a,b \in R$ implies $p|a$ or $p|b$ i.e. if $pR$ is a prime ideal in $R$ . I can see that $p$ still remains a non-unit in $\bar R$ , but I'm unable to say anything about the ideal $p\bar R$.

UPDATE : The claim is true for any Noetherian domain. This is Lemma 4.7 in ; On finite generation of $R$-subalgebras of $R[X]$ , Amartya K. Dutta; Nobuharu Onoda; Journal of Algebra 320 (2008) 57- 80. On finite generation of R-subalgebras of R[X] - ScienceDirect https://www.sciencedirect.com › pii

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    $\begingroup$ No, even if $R$ is assumed to be Noetherian. For an example, let $k$ be a field and set $R = k[t^2 - 1, t(t^2 - 1)]$. Then $p = t^2 - 1$ is a prime element since $R/pR = k$ is an integral domain. However, the normalization $\overline R \cong k[t]$ and $p = (t - 1)(t + 1)$ there. This example comes from geometry: $R$ here is the coordinate ring of a nodal cubic, and the prime element $p$ corresponds to the node at the origin. When you normalize, the node splits off into two points, and so the pullback of this prime element factors into two smaller things. $\endgroup$ – Raymond Cheng Jan 28 '18 at 23:04
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    $\begingroup$ @RaymondCheng: your element $p$ is not prime (but it is irreducible). The ring is $k[x,y]/(y^2-x^3-x^2)$, so if $p = x$ then the quotient is $k[y]/(y^2)$, not $k$. $\endgroup$ – R. van Dobben de Bruyn Jan 28 '18 at 23:27
  • $\begingroup$ @R.vanDobbendeBruyn: Great, yes: $R/p$ still has the class represented by $t(t^2 - 1)$. Thanks (: $\endgroup$ – Raymond Cheng Jan 28 '18 at 23:49

This is true when $R$ is reasonable. The properties that I use are:

  1. $R$ is Noetherian;
  2. $\tilde R$ is Noetherian;
  3. $\tilde R$ is catenary and equidimensional (i.e. every maximal chain $0 = \mathfrak q_0 \subsetneq \ldots \subsetneq \mathfrak q_d$ of prime ideals in $\tilde R$ has the same length).

For example, these are all satisfied if $R$ is of finite type over a field or over $\mathbb Z$. It might be possible to weaken some of these hypotheses.

Lemma. Let $f\colon R \to S$ be an integral ring map, let $\mathfrak q \subseteq S$ be a prime, and let $\mathfrak p = f^{-1}(\mathfrak q)$. Then $$\dim R/\mathfrak p = \dim S/\mathfrak q.$$

Proof. This is poset-theoretic, using only the going up theorem for integral maps [AM, Thm. 5.11]. Indeed, the going up theorem implies that a chain $\mathfrak p = \mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \ldots$ of primes of $R$ containing $\mathfrak p$ can be lifted to some chain $\mathfrak q = \mathfrak q_0 \subsetneq \mathfrak q_1 \subsetneq \ldots$ of $S$, whence $\dim S/\mathfrak q \geq \dim R/\mathfrak p$.

Conversely, if $\mathfrak q_1 \subseteq \mathfrak q_2$ are primes of $S$ with $f^{-1}(\mathfrak q_1) = f^{-1}(\mathfrak q_2) = \mathfrak p$, then we must have $\mathfrak q_1 = \mathfrak q_2$. Indeed, they correspond to primes in the integral ring map $\kappa(\mathfrak p) \to S \otimes_R \kappa(\mathfrak p)$, and there are no inclusions between prime ideals of $S \otimes_R \kappa(\mathfrak p)$ [Tag 00GS(3)]. Hence, the inverse image of a chain $\mathfrak q = \mathfrak q_0 \subsetneq \mathfrak q_1 \subsetneq \ldots$ of primes of $S$ containing $\mathfrak q$ is a strict chain $\mathfrak p = \mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \ldots$ of primes of $R$ containing $\mathfrak p$, whence $\dim R/\mathfrak p \geq \dim S/\mathfrak q$. $\square$

Remark. In the proof below, we want to relate the heights of $\mathfrak q$ and $\mathfrak p$ as in the lemma. We can do this under assumption (3), for this forces $\operatorname{ht}(\mathfrak p) = \dim R - \dim R/\mathfrak p$ (and similarly for $\mathfrak q$).

Proposition. Let $R$ be a domain satisfying properties (1)-(3) above. If $p \in R$ is a prime element, then $p$ is a prime element in $\tilde R$.

Proof. By assumption, $\mathfrak p = (p)$ is a prime ideal. By Krull's Hauptidealsatz [AM, Cor. 11.17], this implies that $\mathfrak p$ has height $1$, i.e. $R_\mathfrak p$ is a $1$-dimensional domain. Since its maximal ideal $\mathfrak pR_\mathfrak p$ is principal, we conclude that $R_\mathfrak p$ is a DVR [AM, Prop. 9.2] with uniformiser $p$; in particular $R_\mathfrak p$ is normal.

On the other hand, normalisation commutes with localisation [AM, Prop. 5.12]. Thus, $$(\tilde R)_\mathfrak p = (R_\mathfrak p)^\sim = R_\mathfrak p,$$ since $R_\mathfrak p$ is normal. That is, the natural map $R \to \tilde R$ becomes an isomorphism when tensoring with $R_\mathfrak p$, hence also when tensoring with $\kappa(\mathfrak p) = R_\mathfrak p/\mathfrak pR_\mathfrak p$. The primes of $\tilde R \otimes_R \kappa(\mathfrak p)$ are the primes of $\tilde R$ lying over $\mathfrak p$ [AM, Exc. 3.21(iv)], so we conclude that there is a unique such prime $\mathfrak q$. Note that $\mathfrak q$ is minimal over $\mathfrak p\tilde R$, hence has height $1$ by Krull's Hauptidealsatz.

If $\mathfrak r \subseteq \tilde R$ is another height $1$ prime, then $p \not\in \mathfrak r$. Indeed, if $p \in \mathfrak r$, then $\mathfrak p' = \mathfrak r \cap R$ contains $\mathfrak p$. Applying the lemma and the remark above, we conclude that $\operatorname{ht}(\mathfrak p') = \operatorname{ht}(\mathfrak r) = 1$. Hence $\mathfrak p' = \mathfrak p$ since $\mathfrak p \subseteq \mathfrak p'$ and both have height $1$.

Hence, for a height $1$ prime $\mathfrak r \subseteq \tilde R$, we have $$v_{\mathfrak r}(p) = \left\{\begin{array}{cc} 1, & \mathfrak r = \mathfrak q,\\ 0, & \mathfrak r \neq \mathfrak q, \end{array}\right.$$ since $p$ is a uniformiser of the DVR $\tilde R_\mathfrak q \cong R_\mathfrak p$. If $q \in \mathfrak q$, then $v_\mathfrak r(q) \geq v_\mathfrak r(p)$ for all height $1$ primes $\mathfrak r \subseteq \tilde R$. Hence, $\frac{q}{p} \in \tilde R$ [Eis, Cor. 11.4], which shows that $\mathfrak q \subseteq (p)$. The reverse inclusion follows since $\mathfrak q \cap R = \mathfrak p$, hence $(p) = \mathfrak q$ is prime. $\square$

Remark. In geometric language, we proved:

  1. There is a unique irreducible divisor $V(\mathfrak q) \subseteq \operatorname{Spec} \tilde R$ dominating the irreducible divisor $V(\mathfrak p) \subseteq \operatorname{Spec} R$;
  2. The locus $V(p) \subseteq \operatorname{Spec} \tilde R$ does not split off a new component of higher codimension;
  3. The uniformiser $p$ for the divisor $V(\mathfrak p) \subseteq \operatorname{Spec} R$ remains a uniformiser for $V(\mathfrak q) \subseteq \operatorname{Spec} \tilde R$ (there is no ramification).


[AM] Atiyah, M.F.; Macdonald, I.G., Introduction to commutative algebra. Addison-Wesley Publishing Company (1969). ZBL0175.03601.

[Eis] Eisenbud, D., Commutative algebra with a view toward algebraic geometry. Graduate Texts in Mathematics 150, Springer-Verlag (1995). ZBL0819.13001.

  • $\begingroup$ OK. What does it mean for a ring to be Japanese? Gerhard "Enquiring Mind Seeking Knowledge, Arigato" Paseman, 2018.01.28. $\endgroup$ – Gerhard Paseman Jan 28 '18 at 23:22
  • $\begingroup$ @GerhardPaseman: it means that for every finite extension $K = \operatorname{Frac} R \to L$, the integral closure of $R$ in $L$ is finite over $R$. We only need the case $K = L$. $\endgroup$ – R. van Dobben de Bruyn Jan 28 '18 at 23:32
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    $\begingroup$ Dear Remi, I'm sorry but in line 16 I don't understand to what inclusion of prime ideals in $R_{\mathfrak p'}$ you apply the going-up theorem for the morphism $R_{\mathfrak p'} \to (\tilde R)_{\mathfrak p'}$ $\endgroup$ – Georges Elencwajg Feb 2 '18 at 19:35
  • $\begingroup$ For the kinds of rings you described, do you think for any prime ideal $P$ in $R$, $P\tilde R$ is prime in $\tilde R$ also ? $\endgroup$ – user111492 Feb 2 '18 at 19:42
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    $\begingroup$ @GeorgesElencwajg: I wanted to do the following: if $\operatorname{ht}(\mathfrak p') > 1$, then there is a chain $0 \subsetneq \mathfrak p'' \subsetneq \mathfrak p'$. Applying going up gives a chain of length at least $2$ in $(\tilde R)_{\mathfrak p'}$. But I just realised that this is not necessarily contained in $\mathfrak r$, so the argument indeed seems to have a gap. I'm not sure how to fix it at this moment; I hope it doesn't break the argument. $\endgroup$ – R. van Dobben de Bruyn Feb 2 '18 at 19:45

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