In a faithfully flat ring extension, is $\operatorname{ht}I=\operatorname{ht}IS$ right?

For Noetherian rings $R$ and $S$, let $R\rightarrow S$ be a faithfully flat ring extension and $I$ an ideal of $R$. Does $\operatorname{ht}I=\operatorname{ht}IS$ hold?

Is it a conclusion in some books on commutative algebra?

Solution 1:

It seems that the statement is true when $S$ is Noetherian (so $R$ is consequently Noetherian).

The map $R\to S$ is faithfully flat, we have some basic facts:

1. for any ideal $J\subseteq R$, $JS\cap R=J$.
2. for any prime ideal $\mathfrak{p}$ of $R$, there is a prime $\mathfrak{q}$ of $S$ such that $\mathfrak{q}\cap R=\mathfrak{p}$.
3. the map $R\to S$ has the going-down property.

We prove a statement first.

Lemma. Let $\mathfrak{p}\subseteq R$, $\mathfrak{q}\subseteq S$ be primes such that $\mathfrak{q}\cap R=\mathfrak{p}$ and $\mathfrak{q}$ is minimal over $\mathfrak{p}S$. Then $\operatorname{height}(\mathfrak{p})=\operatorname{height}(\mathfrak{q})$.

Proof. By going-down, $\operatorname{height}(\mathfrak{q})\geq \operatorname{height}(\mathfrak{p})$.

Conversely, let $\operatorname{height}(\mathfrak{p})=r$. Since $R$ is Noetherian, by the dimension theory of Noetherian rings (Generalized Krull's principal ideals theorem, see chapter 10 in GTM 150), we can choose $x_1,\ldots,x_r\in \mathfrak{p} $ such that $\mathfrak{p}$ is minimal over $(x_1,\ldots,x_r)$. Then $\mathfrak{q}$ is also minimal over $(x_1,\ldots,x_r)S$. (Reasons: one can choose integers $N_1,N_2$ such that $\mathfrak{p}^{N_1}R_{\mathfrak{p}}\subseteq(x_1,\ldots,x_r)R_{\mathfrak{p}}$, $\mathfrak{q}^{N_2}S_{\mathfrak{q}}\subseteq \mathfrak{p}S_{\mathfrak{q}}$, thus $(\mathfrak{q}^{N_2})^{N_1}S_{\mathfrak{q}}\subseteq \mathfrak{p}^{N_1}S_{\mathfrak{q}}\subseteq (x_1,\ldots,x_r)S_{\mathfrak{q}}\subseteq \mathfrak{q}S_{\mathfrak{q}}$.) So $\operatorname{height}(\mathfrak{q})\leq r$.

Now we prove $\operatorname{height}(I)=\operatorname{height}(IS)$.

For a prime ideal $\mathfrak{q}\supseteq IS$ such that $\operatorname{height}(IS)=\operatorname{height}(\mathfrak{q})$, let $\mathfrak{p}=\mathfrak{q}\cap R$, so by the Lemma, $\operatorname{height}(\mathfrak{q})=\operatorname{height}(\mathfrak{p})\geq \operatorname{height}(I)$.

Suppose $\mathfrak{p}\supseteq I$ is a prime such that $\operatorname{height}(\mathfrak{p})=\operatorname{height}(I)$, then we can choose a minimal prime $\mathfrak{q}$ over $\mathfrak{p}S$ such that $\mathfrak{q}\cap R=\mathfrak{p}$, then $\operatorname{height}(I)=\operatorname{height}(\mathfrak{p})=\operatorname{height}(\mathfrak{q})\geq \operatorname{height}(IS)$.

We are done.