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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...

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    $\begingroup$ Usually to check integral, you just check irreducible and reduced. $X \times_\mathbb{Z}\mathbb{A}_\mathbb{Z}$ can be seen to be reduced since if $R$ is a reduced ring, then $R\otimes_{\Bbb Z}\Bbb Z[x] \cong R[x]$ which is reduced and reduced is a local property. I think $X\times Y$ irreducible if $X$ and $Y$ are irreducible is a general fact but I have been unable to find a proof for it. $\endgroup$ – PVAL-inactive Aug 11 '13 at 21:27
  • $\begingroup$ Thanks PVAL. I got it. I try to find a proof... $\endgroup$ – Sang Cheol Lee Aug 12 '13 at 0:34
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    $\begingroup$ @PVAL: It does not hold in general. But it is true over alg. closed fields. See math.stackexchange.com/a/356169/1650 for instance. $\endgroup$ – Martin Brandenburg Aug 12 '13 at 8:28
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We need two simple algebraic facts:

  1. If $R$ is an integral domain, then $R[T]$ is an integral domain.

  2. 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.

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  • $\begingroup$ Nice proof: +1. $\endgroup$ – Georges Elencwajg Aug 12 '13 at 11:44
  • $\begingroup$ Is dominant morphism stable under flat base change? $\endgroup$ – Yuchen Liu Aug 12 '13 at 15:00
  • $\begingroup$ Dear @Yuchen, you should ask this as a separate question. $\endgroup$ – Georges Elencwajg Aug 12 '13 at 15:19
  • $\begingroup$ Dear @Georges, I will try to solve it by myself, if I can't solve it I will ask on this site, thanks for your suggestion. $\endgroup$ – Yuchen Liu Aug 12 '13 at 15:28

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