# Is the number of prime ideals of a zero-dimensional ring stable under base change?

Let $A$ be a zero-dimensional ring of finite type over a field $k$ and let $X= \textrm{Spec} \ A$ be its spectrum. Note that $X$ is a finite set.

Suppose that $k\subset K$ is a finite field extension and let $Y = X \times_k K$. That is, $Y=\textrm{Spec} ( A\otimes_k K)$.

Question. Is $\textrm{card} \ Y \leq [K:k]\textrm{card} \ X$?

Example. Take $k=\mathbf{R}$, $A=\textrm{Spec} (\mathbf{R}[x]/(x^2+1))$ and $K=\mathbf{C}$. Note that $X$ is a singleton in this case and $Y$ consists of the points $i$ and $-i$.

Example. Take $k=A=\mathbf{R}$ and $K=\mathbf{C}$. In this case both $X$ and $Y$ are singletons.

• Do you mean $Y = X \times_k K$ ? What is L? Aug 13, 2011 at 18:42
• yes. That's what I meant. I'll change it.
– Oen
Aug 13, 2011 at 18:45

Theorem. Let $$A \to B$$ be a ring homomorphism. If $$B$$ can be generated by $$n$$ elements as an $$A$$-module, then every fiber of $$\operatorname{Spec} B \to \operatorname{Spec} A$$ has cardinality not greater than $$n$$.
Proof. Let $$P$$ be a prime of $$A$$. The fiber of $$P$$ is the spectrum of $$B \otimes_A k(P)$$. It is clear that $$B \otimes_A k(P)$$ has dimension $$\leq n$$ over $$k(P)$$. So, if we substitute $$A$$ with $$k(P)$$ and $$B$$ with $$B \otimes_A k(P)$$, we can assume that $$A$$ is a field and $$B$$ is a finite $$A$$-algebra.
Let $$Q_1, \dots, Q_r$$ be primes of $$B$$. Since $$B$$ is an artinian ring, $$Q_i$$ is maximal, then by Chinese Remainder Theorem the map $$B \to B / Q_1 \times \cdots \times B / Q_r$$ is surjective. Computing the dimension, we have $$n \geq \dim_A B \geq \sum_{i=1}^r \dim_A B/ Q_i \geq r$$. $$\square$$
Now, in your case, $$A \otimes_k K$$ can be generated as an $$A$$-module by a set of cardinality $$\leq [K : k]$$.