This is true when $R$ and $\tilde R$ are both Noetherian, e.g. when $R$ is Noetherian and Japanese. It might be possible to weaken some of these hypotheses.

 >>**Proposition.** Let $R$ be a domain such that $R$ and $\tilde R$ are Noetherian. 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, this implies that $\mathfrak p$ has height $1$, hence $R_\mathfrak p$ 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 $\mathfrak q$. We clearly have $(p) = \mathfrak p\tilde R \subseteq \mathfrak q$; we want to show that this is an equality.

Consider the map $R/\mathfrak p \to \tilde R/\mathfrak p\tilde R$, and note that $\mathfrak q/\mathfrak p\tilde R \subseteq \tilde R/\mathfrak p\tilde R$ is a minimal prime. If $\mathfrak q' \subseteq \tilde R$ is a different prime that is minimal over $\mathfrak p\tilde R = (p)$, then by Krull's hauptidealsatz $\mathfrak q'$ has height $1$ in $\tilde R$. If $\mathfrak p' = \mathfrak q' \cap R$, then going up for $R_{\mathfrak p'} \to (\tilde R)_{\mathfrak p'}$ shows that $\mathfrak p'$ has height $1$ in $R$. Since $\mathfrak p \subseteq \mathfrak p'$, we conclude that $\mathfrak p = \mathfrak p'$ for height reasons, hence $\mathfrak q = \mathfrak q'$.

Thus, $\mathfrak q/\mathfrak p\tilde R$ is the unique minimal prime in $\tilde R/\mathfrak p\tilde R$, so $\sqrt{\mathfrak p\tilde R} = \mathfrak q$. 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 $\mathfrak q = (p)$. $\square$

**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).

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**References.**

[AM] <cite authors="Atiyah, M.F.; Macdonald, I.G.">Atiyah, M.F.; Macdonald, I.G., *Introduction to commutative algebra*. Addison-Wesley Publishing Company (1969). [ZBL0175.03601](https://zbmath.org/?q=an:0175.03601).</cite>

[Eis] <cite authors="Eisenbud, D.">Eisenbud, D., *Commutative algebra with a view toward algebraic geometry*. Graduate Texts in Mathematics **150**, Springer-Verlag (1995). [ZBL0819.13001](https://zbmath.org/?q=an:0819.13001).</cite>