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Let $k$ be an arbitrary field. Suppose that $C/k$ is an integral curve which is birationally equivalent to a projective line. Is it true that $C$ is geometrically reduced and irreducible?

This is of course true if $C$ is a normal, since then $C$ is just isomorphic to some open subset of the projective line. So I guess the above question can be generalized to the following one:

Let $X$ and $Y$ be two integral (so both reduced and irreducible) schemes of finite type over $\text{Spec }k$ and let $X\longrightarrow Y$ be a finite, birational $k$-morphism. Suppose that $X$ is geometrically reduced (irreducible). Is then $Y$ also geometrically reduced (irreducible)?

rschwieb
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Tomasz Lenarcik
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    When $X$ is integral, being geometrically reduced or geometrically irreducible is a property of the field extension $k(X)/k$, so the property is invariant by birational morphisms. – Cantlog Dec 17 '13 at 21:08
  • Good catch! I haven't looked at the problem this way. Would you mind turning your comment into answer? – Tomasz Lenarcik Dec 18 '13 at 01:59
  • From this point of view, the answer is yes when $X\to Y$ is just assumed to be dominant. If you agree with my comments, you could copy them into an answer. I stopped to write answers. – Cantlog Dec 19 '13 at 16:35

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