TITLE: Question of Hartshorne proposition II6.6
QUESTION [6 upvotes]: Let $X$ be a scheme which is a noetherian integral separated.
In hartshorne's book, $X \times_\mathbb{Z}\mathbb{A}_\mathbb{Z}$ is also a noetherian integral separated.
I understand $X \times_\mathbb{Z}\mathbb{A}_\mathbb{Z}$ is a noetherian and separated.
But I don't know that $X \times_\mathbb{Z}\mathbb{A}_\mathbb{Z}$ is integral...

REPLY [5 votes]: We need two simple algebraic facts:

If $R$ is an integral domain, then $R[T]$ is an integral domain.
If $R \to S$ is an injective homomorphism, then also $R[T] \to S[T]$ is injective.

Now let $X$ be an integral scheme, i.e. reduced and irreducible. If $X$ is affine, then 1. shows that $X[T] := X \times \mathbb{A}^1$ is also integral. In general, let $\emptyset \neq U \subseteq X$ be an open affine subset. Then $U$ is integral, hence $U[T]$ is integral. Being reduced is a local property, so we already know that $X[T]$ is reduced. Now we have to prove that the generic points of $U[T]$, where $\emptyset \neq U \subseteq X$ is open affine, are all the same in $X[T]$. It suffices to check that if $\emptyset \neq V \subseteq U$ is another open affine, then $V[T] \to U[T]$ preserves the generic points. But this is precisely 2. Thus there is a unique point in $X[T]$ which is the generic point in $U[T]$ for every open affine $\emptyset \neq U \subseteq X$. Since these $U[T]$ cover $X[T]$, we see that this point is generic. Hence $X[T]$ is irreducible.