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I'm trying to prove this statement

Let $ f,g:X\rightarrow Y$ be two $ S $-scheme morphisms that agree on $ U $, a dense open subset of $ X $. If $ X $ is reduced and $ Y$ separated, then $ f = g $.

I've gone so far as showing that the locus of agreement of $ f $ and $ g $ must be all of $ X $. The locus of agreement is a closed subscheme of $ X $ defined via this universal property

If $h:Z\rightarrow X$ is another $ S $-morphism such that $ f\circ h=g\circ h $, then $ h $ factors uniquely through the locus of agreement.

How can I show that since the locus of agreement is all of $ X$, then $ f=g$? It sounds intuitive, but I'm unable to show it via the universal property above.

Thank you

PeterM
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  • Hint: Think of this locally. I.e., let $f,g:A\to B$ be two ring homomorphisms such that $f^{-1}({\frak{p}})=g^{-1}({\frak{p}})$ for every prime ${\frak{p}}\subseteq B$. Show that if $B$ is reduced, then $f=g$. – rfauffar Jan 02 '15 at 05:05
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    The hint above is not true. Take for example $A=B=\mathbb{C}$, $f(z)=z$ and $g(z)=\bar{z}$. – nowhere dense Dec 17 '17 at 21:37

1 Answers1

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I was able to answer this soon after posting. The fact that $ X $ is the locus of agreement gives a commutative diagram. By composing each path in the diagram by an appropriate projection I see that the morphisms are indeed equal.

PeterM
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