This is true when $R$ is reasonable. The properties that I use are:
- $R$ is Noetherian;
- $\tilde R$ is Noetherian;
- $\tilde R$ 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:
- There is a unique irreducible divisor $V(\mathfrak q) \subseteq \operatorname{Spec} \tilde R$ dominating the irreducible divisor $V(\mathfrak p) \subseteq \operatorname{Spec} R$;
- The locus $V(p) \subseteq \operatorname{Spec} \tilde R$ does not split off a new component of higher codimension;
- 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.
