Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Do you mean $Y = X \times_k K$ ? What is L? – jspecter Aug 13 '11 at 18:42
yes. That's what I meant. I'll change it. – Oen Aug 13 '11 at 18:45
up vote 12 down vote accepted

Yes. Recall this important fact:

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 the 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 $\leq [K : k]$.

share|cite|improve this answer
Very nice, Andrea. – Georges Elencwajg Aug 26 '11 at 14:22
Andrea, would you recommend any resources where one could read more about this theorem? – justin Apr 18 at 20:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.