# When does prime elements remain prime in certain integral extension

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

• 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. – Raymond Cheng Jan 28 '18 at 23:04
• @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 $k$. – R. van Dobben de Bruyn Jan 28 '18 at 23:27
• @R.vanDobbendeBruyn: Great, yes: $R/p$ still has the class represented by $t(t^2 - 1)$. Thanks (: – 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).

References.

[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.

• OK. What does it mean for a ring to be Japanese? Gerhard "Enquiring Mind Seeking Knowledge, Arigato" Paseman, 2018.01.28. – Gerhard Paseman Jan 28 '18 at 23:22
• @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 $K = L$. – R. van Dobben de Bruyn Jan 28 '18 at 23:32
• 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 $R_{\mathfrak p'} \to (\tilde R)_{\mathfrak p'}$ – Georges Elencwajg Feb 2 '18 at 19:35
• 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 ? – user111492 Feb 2 '18 at 19:42
• @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. – R. van Dobben de Bruyn Feb 2 '18 at 19:45