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Let $k$ be a field (not necessarily algebraically closed). We call $k$-variety a scheme of finite type over $\mathrm{Spec}\left(k\right)$. Let $X$ be a geometrically reduced $k$-variety and $Y$ a $k$-variety.

The rest of the problem statement is as follows: Let $f,g:X\rightarrow Y$ two $k$-morphisms. We suppose that the set-wise applications $X\left(\bar{k}\right)\rightarrow Y\left(\bar{k}\right)$ induced by $f$ and $g$ coincide. We wish to show that $f=g$.

Question: How can it be shown that we can suppose $X$ and $Y$ to be affine?

Jake
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  • @Hoot Yes, but how can it be shown that they are equal locally? Indeed, for any open subset $U$ of $X$ exists a ring $A$ such that $\left(U,\mathcal{O}{X}\left(U\right)\right)$ $\simeq$ $\left(Spec\left(A\right),\mathcal{O}{Spec\left(A\right)}\right)$ and there exists $\phi:\left(U,\mathcal{O}{X}\left(U\right)\right)\rightarrow\left(\mathrm{Spec}\left(k\right),\mathcal{O}{\mathrm{Spec}\left(k\right)}\right)$, but it is not evident that $A=k$, right? – Jake Dec 04 '14 at 15:55
  • Ah, I misunderstood your point of contention. When you get the TeX sorted out maybe I can say something. – Hoot Dec 04 '14 at 15:56
  • @Hoot Okay, the TeX expression is now fixed. – Jake Dec 04 '14 at 15:58
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    I don't think reducing to the affine case is particularly helpful. In Ravi Vakil's notes, this is the reduced to separated theorem. Here's an outline. There is a definition of the locus where any two morphisms agree and it can be shown to be locally closed on the source and actually closed for separated targets (which a variety is assumed to be usually). By your assumption, this locus includes all the closed points of $X$. But in any geometrically reduced $k$-variety, the set of all closed points is dense (this can be seen by base changing to $\overline{k}$ and using the Nullstellensatz.) – Siddharth Venkatesh Dec 05 '14 at 08:43
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    But reducing to the affine case is easy. It suffices to show that the morphisms $f$ and $g$ coincide on an open affine cover of $X$. So, take an affine cover of $Y$ and then take affines covering every preimage of the open affines in the cover for $Y$. The hypotheses still hold since any map from $\mathrm{Spec} (\overline{k})$ into $X$ factors through one of these open affines. So, we reduce to showing that $f = g$ on each of these open affines. – Siddharth Venkatesh Dec 05 '14 at 08:45

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Let $X$ be a geometrically reduced $k$-variety and $Y$ a $k$-variety. Let $(Y_i)$ be an open affine covering of $Y$. Then $f^{-1}(Y_i)$ is an open (by continuity of $f$) covering of $X$, but this covering is not affine in general as $f$ or $g$ might not be quasi-affine morphisms. Nevertheless take an open affine cover $V_{ij, f}$ of $f^{-1}(Y_i)$. The open $V_{ij, f} \cap V_{ik, g}$ are affine, this is implied by the stability of open immersions by base change, see here. Now look at $ f,g : V_{ij, f} \cap V_{ik, g} \rightarrow Y_i$ : these restrictions of $f,g$ verify the hypothesis (exercise, check it !) and are affine so they are equal. From this follows the conclusion.

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Olórin
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  • It is not true that the inverse image of an open affine by a morphism of schemes is affine. Such morphisms are called "affine". Just take a morphism from a non-affine scheme to an afffine scheme to see that this is not necessarily true.

    However, you can cover the inverse images of affines by open affines, and use the same argument.

     – user115940 Jan 01 '15 at 09:21
  • The link you gave shows that that in a morphism between affine schemes, the inverse image of an affine is again affine. This is because affine morphisms are stable under base change. – user115940 Jan 01 '15 at 09:22