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Here's exercise 9.5G from Ravi Vakil's Foundation of algebraic geometry

Recall that the quartic curves in $\mathbb{P}^2_k$ are parametrized by a $\mathbb{P}^{14}_k$. Show that the points of $\mathbb{P}^{14}_k$ corresponding to geometrically irreducible curves form an open subset. Explain the necessity of the modifier “geometrically” (even if $k$ is algebraically closed).

It is also remarked that the dimension $2$ and degree $4$ can be replaced by other numbers. I think irreducible curves here mean irreducible and reduced curves.

  1. My first question is why the abverb "geometrically" is needed?

A homogeneous polynomial of degree corresponds corresponds to an reducible or non-reduced closed subscheme if and only if the polynomial is reducible, which means that the polynomial is the product of two quadratic polynomial or a product of a linear polynomial and a cubic, and these should correspond to two closed subset of $\mathbb{P}^{14}_k$ since the map between two projective variety is proper (but this fact has not been proved before this exercise, so is there any other way to see this?), and thus the complement of their union, which corresponds to irreducible quartic curves, is open. It seems that this argument works without the term "geometrically".

  1. My second question is how to do the question when the term "geometrically" is added.

Let $K$ be the algebraic closure of $k$, by using the arguments above (if it is correct), I can find an open subset $W$ of $\mathbb{P}^{14}_K$ which corresponds to irreducible quartic curves in $\mathbb{P}^2_K$. Let $\phi:\mathbb{P}^{14}_K\to\mathbb{P}^{14}_k$ be the pullback of the map $specK \to speck$. I would want to show that $\phi$ maps $W$ to an open set of $\mathbb{P}^{14}_k$. I want to show this by showing $\phi$ is both closed and surjective and hence open.

I can show that $\phi$ is surjective on closed set: Let $p\in \mathbb{A}^{14}_k \subset \mathbb{P}^{14}_k$ be a closed point, then $p$ also corresponds to an ideal of $K[x_1,\dots,x_{14}]$, let $m$ be any maximal ideal containing this ideal, then the image of $m$ would be $p$, but I am not sure if the map is surjective or not since $\mathbb{A}^{14}_K$ is not a $k$ variety and we cannot use Chevalley's theorem on constructible sets. To show that $\phi$ is closed, I find the same difficulties in showing that the map $specK[x_1,\dots,x_{14}]/I \to speck[x_1,\dots,x_{14}]/I\cap k[x_1,\dots,x_{14}]$ is surjective.

Any helps are appreciated, thank you.

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As you say, we can find the loci where the polynomials factor. Let's consider the case where it factors into a line and a cubic. So lines are parameterized by $\mathbb{P}^2$ and cubics are parameterized by $\mathbb{P}^9$, and we have the multiplication map $\mathbb{P}^2\times \mathbb{P}^9\rightarrow \mathbb{P}^{14}$. As you say, the image is closed by properness.

The issue is that, if ${\rm Spec}(\ell)\rightarrow \mathbb{P}^{14}$ is a $\ell$-point in the image, it might not be hit by a $\ell$-point of $\mathbb{P}^2\times \mathbb{P}^9$. Maybe it is hit by an $L$-point of $\mathbb{P}^2\times \mathbb{P}^9$ for $L$ some finite extension of $\ell$. This will happen if we have a quartic that doesn't factor into a line and a cubic over $\ell$ but it does over the algebraic closure.

As a toy example of such a thing happening, we can consider the family of subchemes of $\mathbb{P}^1_{\mathbb{R}}$ over $\mathbb{A}^1_{\mathbb{R}}$ given by the equation $X^2+tY^2=0$. We see that this is irreducible when $t>0$, which is not an open subset.

As for how to finish the argument, I think there are no further changes needed. Maybe there is a way to base change to the algebraic closure, prove the result there, and then deduce it for the original base field, but I think it's cleaner as is. Besides, even if we do base change to the algebraic closure, we might still have the same issue happening at scheme-theoretic points (as the problem hints at).

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    I think by the points in $\mathbb{P}^{14}_k$, it means the points as a topological space/set. If it means the $l$-valued points (for some extension $l/k$), how can we talk about the corresponding subset in $\mathbb{P}^{14}_k$ ? – 2017-01-21
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    I'm just thinking of $\ell$ as the residue field of a point – 2017-01-21
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    I see what you mean. And when $k$ is algebraically closed, why the same issue would happen? Do we have an example of this? – 2017-01-21
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    Do you mean we may have something like a polynomial $f(t,x,y,z) \in k[t,x,y,z]$ which is reducible in $k(t)[x,y,z]$, but $f(a,x,y,z)$ is irreducible in $k[x,y,z]$ for any $a\in k$? – 2017-01-21
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    Yeah, I think we're on the same page. For an example of such a thing happening, we can consider the polynomial $X^2+tY^2\in k(t)[X,Y]$. It only factors if we adjoin a squareroot of $t$. – 2017-01-21
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    Should we find an example which is reducible over $k(t)$ but irreducible over $k$? your example seems to be irreducible over $k(t)$ but reducible over $k$ (after substituting $t=a$) – 2017-01-21
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    Oh, sorry such example would not exist, for if $f=gh$ in $k(t)[x,y,z]$, by Gauss lemma, we may assume $g,h$ in $k[t,x,y,z]$, then if we evaluate $t=a$ for some suitable $a \in k$, we would have a factorization of $f(a,x,y,z)$ in $k[x,y,z]$. The problem would not arise when $k$ is algebraically closed, right? – 2017-01-21
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    What problem would not arise? – 2017-01-22
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    I see what you mean now, I had some misunderstanding before. So I do not need to do base change at all, thank you so much! – 2017-01-22