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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.

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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

1 Answer 1

up vote 10 down vote accepted

Yes. Recall this important fact:

Theorem. Let $A \to B$ be a ring homomophism. 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 $\leq [K : k]$.

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3  
Very nice, Andrea. –  Georges Elencwajg Aug 26 '11 at 14:22

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