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How would I go about determining the singular locus of the hypersurface ring $R=\mathbb{F}_p[x,y,z]/(xy-z^2)$? I conjecture that the ring is regular at every maximal ideal except $(x,y,z)$. The trouble I am having is that $\mathbb{F}_p$ is not algebraically closed.

I was able to determine that $R$ is regular at every maximal ideal of the form $(x-a,y-b,z-c)$ for which $(a,b,c)\neq(0,0,0)$ by showing that the partial derivatives of $xy-z^2$ are all $0$ only when $x=y=z=0$. But of course, there are maximal ideals not of the form $(x-a,y-b,z-c)$ since $\mathbb{F}_p$ is not algebraically closed. How would I go about determining if $R$ is regular at those maximal ideals? And how about prime ideals which are not maximal?

Anon
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  • Every maximal ideal is the kernel of a homomorphism to $\overline{\mathbb{F}_p}$ (by the Nullstellensatz); in other words, you can treat them as evaluation homomorphisms but at the points over $\overline{\mathbb{F}_p}$. – Qiaochu Yuan Jul 09 '24 at 03:07
  • @QiaochuYuan Thanks! I can see that much, but I am still a little stuck on finding the singular locus. Can I just say that every maximal ideal other than $(x,y,z)$ is in the regular locus since at least one partial derivative of $xy-z^2$ is nonzero when plugging in any point of $\overline{\mathbb{F}_p}^3\setminus{(0,0,0)}$? And then can I conclude that $(x,y,z)$ is the only prime ideal in the singular locus? – Anon Jul 09 '24 at 03:37
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    The way I like to approach these is to compute the non-smooth locus by looking at the gradient. (Since non-regular points are non-smooth.) In your case this is just a point and you can then check directly that this point is non-regular by computing the Zariski-tangent space. – Daniel Jul 09 '24 at 04:52
  • @Daniel I think I am okay with $(x,y,z)$ being in the singular locus. The issue I am having is why this is the only prime ideal in the singular locus. E.g. if $f(x)$ is any irreducible polynomial over $\mathbb{F}_p$, how do I show that $(f(x))$ is not in the singular locus? – Anon Jul 09 '24 at 08:51
  • Is this not basically about checking that $(0,0,0)$ is the only point where $F=xy-z^2$ as well as its partial derivatives $$\nabla F=(y,x,-2z)$$ vanish? Even in the case $p=2$, when $-2z$ vanishes everywhere, does not create any other singular points. – Jyrki Lahtonen Jul 11 '24 at 08:34
  • ^ the only point $(x_0,y_0,z_0)\in\overline{\Bbb{F}_p}\ldots$ – Jyrki Lahtonen Jul 11 '24 at 10:53
  • @JyrkiLahtonen That’s what I’m trying to understand. QiaochuYuan suggested we can use $\overline{\mathbb{F}_p}$ somehow but didn’t fully expand on that answer. I’m $\textit{guessing}$ his claim is that if we have some maximal ideal $m$ then we can choose some point $(a,b,c)\in\overline{\mathbb{F}_p}$ which is a root of the polynomials in $m$ and then check that if this point isn’t $(0,0,0)$ then it’s not a common zero of $xy-z^2$ and it’s partial derivatives, and if this is the case, then the ring must be regular at $m$. Is this true? Also, what about prime ideals that aren’t maximal? – Anon Jul 11 '24 at 15:38
  • I am not an expert on this, but I do think that the following related threads will clear some of the fog :-) 1, 2, 3, 4, 5. But I'm afraid they won't give you everything. May be more can be found with different buzzwords? – Jyrki Lahtonen Jul 11 '24 at 16:45

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Recall that morphism $f:X\to Y$ is smooth iff $f$ is flat, locally of finite presentation, and for every geometric point $\overline{y}$ of $Y$ the geometric fiber $X\times_Y \overline{y}$ is regular. In the case when $X\to Y$ is a variety over a field $k$, this gives that $X$ is smooth over $k$ exactly when $X_{\overline{k}}$ is regular. Let's recall another helpful result:

Lemma. Suppose $x\in X$ is a point of a scheme of finite type over a field $k$ and $k\subset K$ is a field extension. If $x'\in X_K$ lies over $x$ and $x'$ is regular, then $x$ is also regular.

Proof. Since $\operatorname{Spec} K\to\operatorname{Spec} k$ is flat, so is $X_K\to X$, and we may apply 00OF to $\mathcal{O}_{X,x}\to\mathcal{O}_{X_K,x'}$. $\blacksquare$

In particular, if $X_\overline{k}$ is regular, $X$ must also be regular, so the non-smooth locus contains the non-regular locus. In our case, the non-smooth locus is the origin, and we can verify that the origin is not regular: the maximal ideal of $\Bbb F_p[x,y,z]_{(x,y,z)}/(xy-z^2)$ has minimal generating set of size 3, but it is a 2-dimensional ring.

KReiser
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