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I have trouble with the exercise mentioned:

Let $S$ be a scheme, let $X$ be a reduced scheme over $S$, and let $Y$ be a separated scheme over $S$. Let $f$ and $g$ be two $S$-morphisms of $X$ to $Y$ which agree on an open dense subset of $X$. Show that $f = g$. Give examples to show that this result fails if either (a) $X$ is non-reduced, or (b) $Y$ is non-separated. [Hint: Consider the map $h : X \rightarrow Y \times_S Y$ obtained from $f$ and $g$.]

I can solve the problem in case ``agree on an open dense subset of $X$'', say $U$, means that $f|_U = g|_U$ as morphisms i.e. not only do the continuous maps agree but the sheaf parts also agree and I don't seem to need the assumption that $X$ is reduced. I guess the problem only means the continuous maps agree. In that case, how does one make use of the assumption that $X$ is reduced to solve the problem?

I have tried many examples and it seems that when $X$ is reduced, the only open dense set is the whole space $X$ in which case there is nothing to do. But I don't think this is true in general.

An Hoa
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  • There are many dense open sets. For example, on an irreducible scheme all nonempty open sets are dense! – Mariano Suárez-Álvarez Feb 21 '16 at 23:07
  • I just realized that I make a computational mistake. @MarianoSuárez-Alvarez So an integral scheme (reduced + irreducible) always have that property. It is a bit unfortunate in terminology that "reduced" and "irreducible" have little to do with each other. – An Hoa Feb 21 '16 at 23:16
  • The thing with reduced and irreducible is a battle that was lost ages ago ;-) – Mariano Suárez-Álvarez Feb 21 '16 at 23:50
  • @TakumiMurayama I don't think that $\text{Spec } k[x]/(x^2)$ is a reduced scheme. – An Hoa Feb 22 '16 at 00:29
  • You will need that $X$ is reduced, when you want to show that the sheaf maps are equal. $f=g$ on an dense open tells you that $\mathcal V((f^{#}-g^{#})(a))$ is all of $X$. To deduce $(f^{#}-g^{#})(a)=0$, we need $X$ to be reduced, since any nilpotent element has the property, that its zero locus is everything. – MooS Feb 22 '16 at 07:54
  • Consider the map $h:X \to Y \times_S Y$,note that X is reduced,the scheme-theoretic image of h is exactaly the closure of $h(X)$ with reduced induced structure, the result of Ex3.11d is useful. – Jiabin Du Aug 14 '17 at 01:15

1 Answers1

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Here is an example showing we need the condition on sheaves.

Let $X = \operatorname{Spec} k[t]_{(t)}$ and let $Y = \mathbf{A}^1 = \operatorname{Spec} k[x]$. Note $X$ has two points: a closed point, and the generic point $\eta$, which is open. Consider the maps \begin{align*} k[x] &\to k[t]_{(t)}\\ x &\mapsto t - a\\ x &\mapsto t - b \end{align*} for $a \ne b$. On schemes, we have $X \to Y$ matching set-theoretically on the generic point $\eta \in X$, while the closed point maps to two different points, given by the coordinates $x = -a$ and $x = -b$.

Takumi Murayama
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