# degree of morphism of schemes

Let $\phi: Y \to X$ be a finite etale morphism of proper smooth connected schemes over a field $K$ and suppose that the induced morphism $\phi: \overline{Y} \to \overline{X}$ has degree $n$, where $\overline{Y} = Y \times_K \bar{K}$ etc.

Can I conclude that $\phi$ has degree $\leq n$? If not, what other conditions should I impose so that this happens?

Thanks.

## 1 Answer

The answer is yes under much broader assumptions.

Indeed, if $\phi:Y\to X$ is a completely arbitrary finite étale morphism of degree $n$, then $X$ can be covered by affine open subsets $U=\text {Spec}(A)$ such that $\phi^{-1}(U)=\text {Spec} (B)$ is affine with $B$ a free separable $A$-algebra of dimension $n$.
Hence for any base change $X'\to X$ the resulting morphism $X'\times_X Y\to X'$ is again finite étale of degree n: the underlying algebraic reason being that for any $A$-algebra $A'$, the $A'$-algebra $A'\otimes_A B$ is separable and free of dimension $n$ too.

In conclusion, the degree of a finite étale map is invariant under base change, which of course answers your question.

Remark
This answer illustrates the weird fact that often in algebraic geometry (and elsewhere in mathematics too), generalizing a question may lead to an almost trivial answer by forcing one to forget about irrelevant details.

Bibliography
Here is a reference by Lenstra which, although it doesn't explicitly contain the answer to the OP's question, is quite relevant.

• The most shocking example of your remark, for me, is the way algebraic geometers have generalized the meaning of "local" so as to trivialize things. It is a great accomplishment :-) Sep 15, 2014 at 20:49