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I'm reading through the stacks project and came across a lemma along these lines: Let X be a scheme over a perfect field k. Then, $X$ is reduced implies $X$ is geometrically reduced.

here is my question: does this lemma extend to schemes over rings of char. 0? for instance, if $X$ is a reduced scheme over $\mathbb{Z}$ and $k$ is a field containing $\mathbb{Z}$, is it true that that $X_{k}=X\times_\mathbb{Z} k$ is a reduced scheme over $k$? A proof (or reference) would be great.

adrido
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    For what it's worth, I don't think that $\mathbb Z$ has characteristic zero, nor of course any other characteristic. Only algebras over a field should have a characteristic, namely that of the field. There is of course a strong temptation to call characteristic of a ring $R$ the smallest integer such that $n\cdot 1_R=0$ (or zero if no such integer exists) but the simplicity of the definition does not save the definition from its drawbacks, which I would sum up under the vague statement of "lack of functoriality" . In a sense $\mathbb Z$ has all characteristics... – Georges Elencwajg Jan 20 '14 at 18:10
  • @ Georges Elencwajg: can you please elaborate on lack of functoriality? You mean the morphisms of schemes over $\mathbb{Z}$ do not extend after base change to fields of characteristic 0? – adrido Jan 21 '14 at 02:01
  • Dear abdrido, it's a long story and I cannot discuss it here. A key point is that morphisms wouldn't preserve that notion of characteristic. – Georges Elencwajg Jan 21 '14 at 07:31
  • Dear @Georges: yes, it indeed appears to be a subtle point. I tried to read about it, with no success. but it's good to have it in mind. – adrido Jan 21 '14 at 07:54

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By the theorem you cite we are immediately reduced to the case $k = \mathbb{Q}$. (Note that if you don't insist that $k$ has characteristic zero the result is false, e.g. consider $\mathbb{Z}[x]/(x^p-1)$. )

Now, let $A$ be an abelian group. Since $\mathbb{Q}$ is flat over $\mathbb{Z}$, an element $f \in A$ maps to $0$ in $A \otimes \mathbb{Q}$ if and only if $f$ is a torsion element.

If $A$ is a $\mathbb{Z}$-algebra (i.e. a ring), $f \in A$, and $f^n = 0 \in A \otimes \mathbb{Q}$, then $f^n$ is torsion in $A$. Suppose $mf^n = 0$. Then $m^nf^n = 0$, so $(mf)^n = 0$. If $A$ is reduced, we conclude that $mf = 0$ in $A$, i.e. $f$ is torsion in $A$, i.e. $f = 0 \in A \otimes \mathbb{Q}$, and thus $A \otimes \mathbb{Q}$ is also reduced.

hunter
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  • I'm a little confused by your mention of $\mathbb{Z}[x]/(x^p-1)$: this ring has characteristic $0$. The reductions (to $k = \mathbb{Q}$, and the ring case) are also not so clear (to me at least): perhaps it might help to explain a bit more there – zcn Jan 20 '14 at 06:08
  • OP asks if, when $k$ is a field containing $\mathbb{Z}$, we can say "reduced over $\mathbb{Z}$ implies reduced over $k$". The example in my remark shows why we really need the ``containing $\mathbb{Z}$'' hypothesis on $k$ since the example is reduced but if we base change this ring to $k = \mathbb{F}_p$ then $x$ becomes nilpotent. – hunter Jan 20 '14 at 06:20
  • The reduction to $k = \mathbb{Q}$ is that any characteristic field contains $\mathbb{Q}$ and the base change of a $\mathbb{Z}$-scheme to $K$ certainly factors through the base change to $\mathbb{Q}$. The stacks project proof cited in the question already shows that reduced over $\mathbb{Q}$ implies reduced over any field extension. – hunter Jan 20 '14 at 06:21
  • Ah, so that's what your example is for, fair enough. It might be useful to have the link to the relevant proof in the stacks project: I have not seen it before (algebraic geometry is not my area of expertise). Finally, could you tell me what you mean by "reduced over $\mathbb{Z}$" and "reduced over $k$"? – zcn Jan 20 '14 at 06:31
  • @user115654: it's lemma 4.3 here http://stacks.math.columbia.edu/download/varieties.pdf. there is also a pointer to the definition of reduced scheme. – adrido Jan 20 '14 at 07:05
  • @abdrido: Thanks for the link, that answers my question – zcn Jan 20 '14 at 07:14
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    er, x-1 becomes nilpotent, i mean, in my comment way above – hunter Jan 20 '14 at 13:22
  • Note that the proof given above works when one replaces $\mathbb{Z}$ with any integral domain with characteristic 0, and these are the only rings where the question makes sense. – RghtHndSd Jan 20 '14 at 15:26