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This is inspired by https://math.stackexchange.com/questions/705979/purely-inseparable-morphisms-and-factorizations-of-a-morphism-of-finite-type that I asked previously .

Let $k$ be an algebraically closed field. Let $f: X \rightarrow Y$ be a finite surjective morphism of integral algebraic varieties over k. Suppose we know the following:

If $K(X)/K(Y)$ is separable of degree n, then there exists a non-empty open subset V of Y such that for any $y \in V(k)$, $f^{-1}(y)$ consists of at least n points.

Then, with this I want to show that if f induces an injective map $X(k) \rightarrow Y(k)$ then it is purely inseparable. I see that one should try to apply the above, but I can't quite make it work - what if the extension $K(X)/K(Y)$ is a mix between being separable and inseparable for example?

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  • Suppose that $f:X\to Y$ is not purely inseparable. Then $f$ factors as $g\circ h$ where $g:X\to Z$ is purely inseparable and $h:Z\to Y$ is separable. Note that the degree of $h$ is at least two. (Else $h$ would be an isomorphism and $f$ would be purely inseparable.) By your remark, there exists a point $y$ in $Y$ such that $h^{-1}(y)$ has at least two points. In particular, $f^{-1}(y)$ has at least two points. We conclude that $X(k) \to Y(k)$ is not injective (because the fiber over $y$ has more than two elements). This is what you want to show, right? – Ariyan Javanpeykar Mar 13 '14 at 12:10
  • @Ari Why does f factor as such? – user101036 Mar 13 '14 at 12:31
  • Assuming $X$ and $Y$ to be normal, I think you can see this by factoring $K(Y) \subset K(X)$ appropriately as $K(Y)\subset L\subset K(X)$, and normalizing $Y$ in $L$ to get $Z$. This only gives a rational map from $X$ to $Z$ though. I will think about it a bit more. – Ariyan Javanpeykar Mar 14 '14 at 10:51
  • @Ari please do! I thought if we could somehow normalize $X$ andtake the normalization of $Y$in L would give us something, but I didn't get anywhere with that approach. – user101036 Mar 14 '14 at 13:22
  • Assume $X$ and $Y$ are normal. Normalize $Y$ in $L$ to get $Z$ and normalize $Y$ in $K(X)$ to get $X^\prime$. Since $X$ is normal, the morphism $X\to Y$ factors through $X^\prime$. (This is the universal property of normalization, and can be found in Liu's book in Chapter 4.1 I think.) So now you get $X\to X^\prime\to Z\to Y$, where we use that the normalization of $Y$ in $K(X)$ is the same as the normalization of $Z$ in $K(X)$. Now apply what I said to $X^\prime\to Y$ to see that $X^\prime(k)\to Y(k)$ is not injective. If you think a bit more, you can use this (in a slightly stronger form).. – Ariyan Javanpeykar Mar 14 '14 at 15:46
  • ..to conclude that $X(k)\to Y(k)$ is not injective. If $X$ and $Y$ are not normal, you can probably work on the normalizations to get what you want. In fact, the normalization of $X$ maps to the normalization of $Y$ (by the universal property of normalization I mentioned before). – Ariyan Javanpeykar Mar 14 '14 at 15:47

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