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I want to prove that surjectivity is stable under base change: if $f:X\to S$ a surjective morphism of scheme and $\varphi:T\to S$ then $f_T:X\times_S T\to T$ is surjective.

Idea 1: I know that for all $t\in T$, $f_T^{-1}(t)\simeq (X\times_S T)\times_T \mathrm{Spec}(k(t))\simeq X\times_S \mathrm{Spec}(k(t))$ and (for the same reason) with $s=\varphi(t)$, $f^{-1}(s)\simeq X\times_S\mathrm{Spec}(k(s))\neq\emptyset$. But how deduce that $X\times_S \mathrm{Spec}(k(t))\neq\emptyset$? Maybe as $k(t)\simeq k(s)$ (topologically) so $\mathrm{Spec}(k(t))\twoheadrightarrow\mathrm{Spec}(k(s))$ and so $X\times_S \mathrm{Spec}(k(t))\twoheadrightarrow X\times_S \mathrm{Spec}(k(s))$?

Idea 2: let $s=\varphi(t)$ and $x\in X$ so that $f(x)=s$ then $f(s)=\varphi(t)$. In the purely set-theoretic construction of $X\times_S T$ this would implicate the existence of $z\in X\times_S T$ with $f_T(z)=t$. But the fibred-product of scheme is not so built...

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What you're trying to prove is that surjectivity is stable under base change, not that it is local on the target. –  Keenan Kidwell Dec 19 '12 at 14:27
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You need the following fact: if $K \to L$ and $K \to L'$ are a pair of field extensions, then there is a field $M$ and a pair of field homomorphisms $L \to M, L' \to M$ making the obvious diagram commute. –  Zhen Lin Dec 19 '12 at 14:34
    
EGA I, Prop 3.6.1. It all comes down to the fact that the tensor product of nontrivial $k$-algebras is nontrivial ($k$ a field). This also shows that $|X \times_S T| \to |X| \times_{|S|} |T|$ is surjective, so that Idea 2 also works. –  Martin Brandenburg Dec 20 '12 at 9:08
    
Ok I understand. I looked for a proof and I have found: $\mathrm{Spec}(k(x)\otimes_k(s) k(t))\to\{z\in X\times_S T|f_T(z)=t,q(z)=x\}$. One get this morphism in reasonning through open affines subsets. One need argument of Zhen Lin for $k(x)\otimes_{k(s)} k(t)\neq\emptyset$ –  Gabriel Soranzo Dec 20 '12 at 11:12

1 Answer 1

up vote 3 down vote accepted

Following your Idea 1 it only remains to show $X\times_S \mathrm{Spec} (k(t))$ is non-empty.

Let $s=\varphi(t)$, then the above fiber product is also $X_s\times_{\mathrm{Spec}(k(s))} \mathrm{Spec}(k(t))$. Take a non-empty affine open subset $U=\mathrm{Spec}(R)$ of $X_s$. It is enough to show $U\times_{\mathrm{Spec}(k(s))} \mathrm{Spec}(k(t))$ is non-empty.

The latter is an affine scheme given by $R\otimes_{k(s)} k(t)$. As $R\ne 0$, $R\otimes_{k(s)} k(t)\ne 0$ because we make a field extension (a vector basis of $R$ over $k(s)$ extends to a vector basis over $k(t)$). By Krull's theorem, $R\otimes_{k(s)} k(t)$ has a maximal (hence prime) ideal, so $U\times_{\mathrm{Spec}(k(s))} \mathrm{Spec}(k(t))\ne\emptyset$.

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