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If I have a morphism of schemes $f: X \rightarrow S$ and a point $s$ in $S$, then I can consider the fibre $X_s$ over that point.

If $x$ is a point in $X$ which is also (topologically) in $X_s$, is then the residue field of $x$ as point in $X$ and as point in $X_s$ the same?

Of course, I tried to solve it locally, but was confused by the many localizations and possible isos. Perhaps you can give a hint if the statement is true and a short note how to show it.

Thanks!

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    $\begingroup$ It's better if you try to solve the exercise by yourself. Think about the universal property of fiber product and the following (if you don't already know this beware: this is EXTREMELY important to understand scheme theory). If $X$ is any scheme and $k$ any field, what you need to specify if order to define a morphism $\operatorname{Spec}(k) \to X$? $\endgroup$
    – Ricky
    Sep 7, 2011 at 10:03
  • $\begingroup$ Thank you for your help, Ricky; with your hint it is indeed very easy. One defines a morphism $Spec(k(x)) \rightarrow X_s$ by universal property, as one has naturally a morphism into $X$ and into Spec(k(s)). The commutativity of the received diagram shows that one has two inclusions of the residue fields in question which composed are the identity. So you can conclude. $\endgroup$
    – Descartes
    Sep 7, 2011 at 14:10
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    $\begingroup$ You may want to think through the local argument again, since it shouldn't be that confusing. In local terms, you are asking: if $(A,\mathfrak m) \to (B,\mathfrak n)$ is a local morphism of local rings, then does the natural map from $B \to B/\mathfrak m B$ induce an isomorphism of residue fields? $\endgroup$
    – Emerton
    Sep 7, 2011 at 17:12

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