9
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Do you mean $Y = X \times_k K$ ? What is L? $\endgroup$
    – jspecter
    Aug 13, 2011 at 18:42
  • $\begingroup$ yes. That's what I meant. I'll change it. $\endgroup$
    – Oen
    Aug 13, 2011 at 18:45

1 Answer 1

17
$\begingroup$

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 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 $\leq [K : k]$.

$\endgroup$
2
  • 4
    $\begingroup$ Very nice, Andrea. $\endgroup$ Aug 26, 2011 at 14:22
  • $\begingroup$ Andrea, would you recommend any resources where one could read more about this theorem? $\endgroup$
    – justin
    Apr 18, 2016 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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