Let $R$ be a noetherian integral domain and let $R'$ be the normalization of $R$ in a finite field extension of the fraction field of $R$. Let $\varphi:Spec(R') \rightarrow Spec(R)$ be the corresponding morphism. Is it true that if $\mathfrak{p}' \in Spec(R')$ is of height 1 in $R'$, then $\varphi(\mathfrak{p}') = R \cap \mathfrak{p}'$ is of height 1 in $R$?

I know that in general the height of primes is not preserved but would like to know if this holds in this setting.

My idea: Since $\varphi$ is integral, we have $dim V(\mathfrak{p}') = dim V(\mathfrak{p})$, so $ht( \mathfrak{p}) = codim_{Spec(R)} V(\mathfrak{p}) \leq \dim Spec(R) - dim V(\mathfrak{p}) = dim Spec(R') - \dim V(\mathfrak{p}')$.

Is it true for some reason that $dim Spec(R') - dim V(\mathfrak{p}') = 1 = ht(\mathfrak{p'})$? This is about a height 1 prime in a Krull domain, so I was hoping that this holds. I actually also always thought of Weil divisors (in a normal noetherian scheme at least) as really having "naive" codimension 1, i.e., their dimension is one less than the dimension of the base scheme. I never thought about that this might actually be wrong!