17

Let $k$ be a field and $X,Y$ be two finite-type $k$ schemes such that $X$ is geometrically reduced.

Let $f,g : X \to Y$ be two morphisms of $k$-schemes such that the induced morphisms : $X(\overline{k}) \to Y(\overline{k})$ are equal. How does one show that $f = g$. There is a hint saying that one can reduce this to the case where $X$ is affine and $Y = \mathbb{A}^1_k$ and $k = \overline{k}$.

I have't been able to prove the hint nor the result assuming the hint so I would be very grateful for help or a reference. (this isn't homework)

Also I would like to know what goes wrong with this when $X$ isn't geometrically reduced.

TheloniousM
  • 285

2 Answers2

8

Since $f=g$ iff $f_{\overline{k}}=g_{\overline{k}}$ after base change, we may assume that $k=\overline{k}$. This means that $k$ is algebraiically closed and that $f,g : X \to Y$ agree on all rational i.e. closed points. These are dense. It follows that the underlying maps of $f$ and $g$ agree, if $X$ is separated (I think we need this assumption). We are left to show that the sheaf homomorphisms $f^\#,g^\#$ are equal, i.e. that $f^\#(s)=g^\#(s)$ for all local sections $s \in \mathcal{O}_Y(V)$. By viewing $s$ as a morphism $V \to \mathbb{A}^1$ and making a base change to $\mathbb{A}^1$, we may assume that $Y=\mathbb{A}^1$. Thus, we have to prove:

If $X$ is a reduced finite type $k$-scheme and $s,t$ are global sections of $X$ such that $s(x)=t(x) \in k$ for all $x \in X(k)$, then $s=t$. By considering $s-t$ we may assume $t=0$. By working locally we may assume that $X$ is affine, say $X=\mathrm{Spec}(A)$. Then $s \in A$ is an element which is contained in all maximal ideals of $A$, i.e. in all prime ideals of $A$ (since $A$ is jacobson). Since $A$ is reduced, we get $s=0$.

user2902293
  • 2,659
Martin Brandenburg
  • 163,620
  • That's great thanks a lot. – TheloniousM Jan 08 '14 at 06:42
  • Very interesting: I had never thought of testing equality of morphisms by changing base. Is the result true in general for faithfully flat descent? I can vaguely imagine how this could be proved (although I haven't checked anything) but is there a reference for this? – Georges Elencwajg Jan 08 '14 at 10:21
  • 2
    For field extensions there is an elementary argument. For fpqc morphisms we may use descent theory, see also http://math.stackexchange.com/questions/619791 – Martin Brandenburg Jan 08 '14 at 10:25
  • 1
    Well, yes, it would be descent theory (and probably not very difficult) but although you mention Görtz-Wedhorn in your link I couldn't find the result there: I was wondering if you knew a precise reference in that book or in EGA. – Georges Elencwajg Jan 08 '14 at 11:34
  • 1
    As explained in the link, it suffices to remark that fpqc morphisms are epis (in fact they are universal effective epis). This should be somewhere in EGA IV (sorry I'm too lazy right now), and can be easily found in the stacks project http://stacks.math.columbia.edu/tag/023P – Martin Brandenburg Jan 08 '14 at 11:50
  • @ Martin Brandenburg Could you explain why if $X$ is separated, the $f$ and $g$ agree on the underlying set? Thank you very much! – Mike Oct 11 '19 at 01:32
  • 1
    Consider the difference kernel $K \to X$ of $f$ and $g$. It can be constructed as the fiber product of $(f,g) : X \to Y \times Y$ and the diagonal $\Delta_Y : Y \to Y \times Y$. If $Y$ is separated (apparently, I made a mistake, saying that $X$ has to be separated), $\Delta_Y$ is a closed immersion, so that $K \to X$ is a closed immersion too. Now use density. – Martin Brandenburg Dec 22 '19 at 14:23
4

Here is an example showing why "geometrically reduced" is necessary for the result to hold.

Let $k$ be an algebraically closed field and consider the $k$-algebra of dual numbers $k[\epsilon]=k[T]/(T^2)$. Let $X=Y=\operatorname {Spec}(k[\epsilon ])$ and consider the two $k$-morphisms $f=Id, g:X\to X$ corresponding to the $k$-algebra morphisms $\phi=Id, \gamma:k[\epsilon]\to k[\epsilon]$ where $Id(\epsilon)=(\epsilon)$ and$\gamma(\epsilon)=0$.
The morphisms $f=Id, g$ are different since $\phi=Id, \gamma$ are different.
Nevertheless the induced morphisms $f(k)=Id(k),g(k):X(k)=k\to X(k)=k$ are both equal to the identity of $k$.

Georges Elencwajg
  • 150,790