Definition of variety in Silverman's AEC

Question

In Silverman's AEC, p3, he defines a variety as the following:

An affine algebraic set $V$ is called a variety if $I(V)$ is a prime ideal in $K’[X]$, where $K’$ denotes an algebraic closure of $K$.

But Wikipedia says $V$ is a variety if and only if coordinate ring is an integral domain. Here, the coordinate ring is defined as $K[X]/I(V)$, not $K’[X]/I(V)$.

I think this is weird. Why does Silverman's AEC defines variety like above? I think we should like to change $K'$ to $K$, because $K[X]/I(V)$ being an integral domain does not imply that $I(V)$ is prime in $K'[X]$.

Answer 1

You may be interested to read What is an algebraic variety? to see how there's a wide range of things that people can mean when they say "variety". What's going on here is that Silverman wants to take "geometrically integral", a condition which is stronger than just integral, to be a part of his definition of variety over a non-algebraically closed field. Wikipedia only works over algebraically closed fields, where geometrically integral is equivalent to integral.

One reason that Silverman wants to do this is in AEC is evident from the title: "arithmetic"! Usually when you see "arithmetic" hanging around algebraic geometry, it means that number theory is involved in some critical way. This usually means studying varieties or schemes over a base that has interesting number-theoretical properties, like number fields (finite extensions of $\Bbb Q$) and their rings of integers, or finite fields. Since these fields aren't algebraically closed, we'll frequently have cause to consider base changing to extension fields in the course of proving things, and if we'd like to keep the niceties that integrality gives us, we would like these base changes to still be integral.

This is exactly where the condition of geometrically integral comes in: a variety $X$ over a field $k$ is geometrically integral iff for any field extension $k\subset k'$ we have $X_{k'}$ integral. One of the first things one proves about this is that we actually only have to check the condition when $k'=\overline{k}$, an algebraic closure of $k$. For affine varieties, this is exactly the condition that Silverman writes down.