# Image of a Weil divisor (height 1 prime) under normalization map

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!

• By the way: In the paper arxiv.org/abs/math/0406384 by Roth-Vakil in the proof of Lemma 2.3 in the second part it seems to me they are using that something like this holds... – Jorge Jul 4 '16 at 18:19
• Maybe $R$ is normal? – Hoot Jul 4 '16 at 19:58
• Can I ask a dumb question, what's the easiest example where height of primes is not preserved (or can you give me a reference). I thought that height of primes is preserved in geometric settings. – Karl Schwede Jul 9 '16 at 14:26
• The answer to the question is actually no as shown here: mathoverflow.net/questions/206723/…. @Karl: Height is preserved for example if the "dimension formula" holds in R and R' is finite over R. This holds for example if R is universally catenary and japanese, e.g., an algebra of finite type over a field. – Uname Jul 11 '16 at 9:59
• Ok, that's what I thought. Thanks. – Karl Schwede Jul 11 '16 at 12:00