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The example is wrong
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Ricky
Ricky
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The following example does not work, it hasn't geometrically connected fibers. Sorry.

In general the answer is no. Take $k \subset K$, a finite extension of field (so the morphism $\operatorname{Spec}(K)\to\operatorname{Spec}(k)$ is proper). Let $X$ be an affine variety over $k$. Let now $x$ be a $K$-point of $X$ that is not defined over $k$. The corresponding morphism $\operatorname{Spec}(K)\to X$ does the job.

In general the answer is no. Take $k \subset K$, a finite extension of field (so the morphism $\operatorname{Spec}(K)\to\operatorname{Spec}(k)$ is proper). Let $X$ be an affine variety over $k$. Let now $x$ be a $K$-point of $X$ that is not defined over $k$. The corresponding morphism $\operatorname{Spec}(K)\to X$ does the job.

The following example does not work, it hasn't geometrically connected fibers. Sorry.

In general the answer is no. Take $k \subset K$, a finite extension of field (so the morphism $\operatorname{Spec}(K)\to\operatorname{Spec}(k)$ is proper). Let $X$ be an affine variety over $k$. Let now $x$ be a $K$-point of $X$ that is not defined over $k$. The corresponding morphism $\operatorname{Spec}(K)\to X$ does the job.

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Ricky
Ricky
  • 3.7k
  • 4
  • 35
  • 53

In general the answer is no. Take $k \subset K$, a finite extension of field (so the morphism $\operatorname{Spec}(K)\to\operatorname{Spec}(k)$ is proper). Let $X$ be an affine variety over $k$. Let now $x$ be a $K$-point of $X$ that is not defined over $k$. The corresponding morphism $\operatorname{Spec}(K)\to X$ does the job.