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I'm trying to prove that the quadric surfaces $Q_1:xy-zw=0$ and $Q_2:z^2-xy=0$ are normal (exercise 3.17(b) from Hartshorne's Algebraic Geometry).

According to the exercise, $Y$ is normal when $\mathcal{O}_P$ is integrally closed for all $P\in Y$. I have a feeling that using the definition directly is not a good plan,now I'm stuck because I can't think of anything else.

rmdmc89
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    If you are working over a field where the Jacobian criterion works, I believe it is the simplest way to show the normality. For a hypersurface, it is normal iff codim (to the hypersurface) of the Jacobian is at least 2. – Youngsu Oct 11 '17 at 17:44
  • @Youngsu, in section $3$ he hasn't mentioned jacobian criterion. – rmdmc89 Oct 11 '17 at 17:52
  • I think he may want you to do some unpleasant equation hacking at this stage in the text... – Tabes Bridges Oct 11 '17 at 18:28
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    I see. What you can do and now I think it is better for your example is checking the affine charts of $Q_1$ and $Q_2$. I believe, except for one chart for $Q_2$ everything is a polynomial ring. The chart on $D(z)$ for $Q_2$ is isomorphic to k[t, 1/t] which is a UFD so normal. – Youngsu Oct 11 '17 at 18:41
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    The example of the cone is done on p. $124$ ($\S5.1$) of Shafarevich's Basic Algebraic Geometry Vol. 1 just using basic properties of integrality. The answer here gives more or less the same argument. There is also a shorter answer using a bit more machinery here. – Viktor Vaughn Oct 11 '17 at 19:18
  • Here is an overkilled argument : they are toric varieties and the cones $\sigma_1, \sigma_2$ corresponding $to Q_1$ and $Q_2$ are strongly convex and rational, so $Q_1,Q_2$ are normal. In fact I'm even sure that it would be possible to translate it to a nice argument without refering to toric geometry, it would be a nice computation. – Nicolas Hemelsoet Oct 11 '17 at 19:46
  • @youngsu can you please explain why the chart on D(w) for $ Q_2 $ is a polynomial ring? – sss89 Oct 11 '17 at 21:57
  • @sss89 $Q_2$ does not have a $D(w)$ chart. I'd be nice if you can clarify this and let me know what you are getting for the chart you are asking. – Youngsu Oct 14 '17 at 03:33
  • @yongsu $ Q_2 $ is embedded in $ \mathbb{P}^3 $ where the coordinates are $ x,y,z,w $ – sss89 Oct 14 '17 at 08:32
  • @sss89: You are right. $D(w)$ is not a polynomial ring. I do not think my approach works for $Q_2$ as it has a singularity. – Youngsu Oct 14 '17 at 20:10

1 Answers1

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Using the hint Hartshorne provides for this exercise I was finally able to solve it. Hope it will be helpful for you too.

Let $P\in Q_{1}$. Since all the coordinates are symmetric, we can assume, WLOG, that $w\neq0$. Then $\mathcal{O}_{Q_{1},P}\cong\mathcal{O}_{D\left(w\right),P}$ so it is sufficient to show that $Y=V\left(xy-z\right)\subseteq\mathbb{A}^{3}$ is normal. The morphism $\varphi:\mathbb{A}^{2}\rightarrow Y$ given by $\left\langle x,y\right\rangle \mapsto\left\langle x,y,xy\right\rangle $ is an isomorphism (the inverse is given by $\left\langle x,y,z\right\rangle \mapsto\left\langle x,y\right\rangle $) and thus $Y$ is normal since $A\left(\mathbb{A}^{2}\right)\cong k\left[x,y\right]$ is UFD.

Let $P\in Q_{2}$. If $x\neq0$ or $y\neq0$, $P$ has a neighborhood isomorphic to $V\left(z^{2}-x\right)\subseteq\mathbb{A}^{3}$ which is normal since it is isomorphic to $\mathbb{A}^{2}$. For $z\neq0$, $P$ has a neighborhood isomorphic to $V\left(xy-1\right)\subseteq\mathbb{A}^{3}$ which is isomorphic to $\mathbb{A}^{2}\backslash V\left(x_{1}\right)$ which is an open subset of a normal variety and thus obviouslly normal. Finally, suppose $w\neq0$, so $P$ has a neighborhood isomorphic to $V\left(z^{2}-xy\right)\subseteq\mathbb{A}^{3}$ so we need to show that $R=k\left[x,y,z\right]/\left(z^{2}-xy\right)$ is integrally closed. Let $\alpha\in Quot\left(R\right)$ be integral over $R$. $ Quot\left(R\right)/k\left(x,y\right)$ is finite field extension of degree 2, so we can write $\alpha=f+g\cdot z$ for $f,g\in k\left(x,y\right)$. The minimal polynomial of $\alpha$ over $k\left(x,y\right)$ is $\alpha^{2}-2f\cdot\alpha+\left(f^{2}-g^{2}xy\right)$ and since $k\left[x,y\right]$ is an UFD, by Gauss lemma, it is also the minimal poynomial of $\alpha$ over $k\left[x,y\right]$ thus $f\in k\left[x,y\right]$ and $f^{2}-g^{2}xy\in k\left[x,y\right]$ for which follows that $g^{2}xy\in k\left[x,y\right]$ and since $xy$ is square free, $g\in k\left[x,y\right]$ so $\alpha\in R$.

sss89
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  • Although $k[x,y,z]/(z^{2}-xy)$ is integrally closed,why $Q_{2}$ is normal? We just know that for affine varieties, normal iff the coordinate ring is integrally closed. Here $Q_{2}$ is a projective variety,right? – Jiabin Du Nov 15 '17 at 06:48
  • I've covered $Q_{2}$ with affine varieties and showed that each of them is integrally closed. Since a point being normal is a local property, every point of $Q_{2}$ has a normal neighbourhood and thus normal. – sss89 Nov 15 '17 at 06:53
  • you are right! Got it! Normality is a local property. I have trouble in proving this problem. I don’t take an open normal affine cover before.Since $xy$ is square free, I know $k[x,y,z]/(z^{2}-xy)$ is integrally closed . But I can’t get the desired conclusion that $Q_{2}$ is normal because for projective varieties, normal is not necessarily equivalent to the coordinate ring integrally closed .It seems that taking an open affine normal cover is necessary.Thanks – Jiabin Du Nov 15 '17 at 07:06
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    At the very end you conclude $f \in k[x, y]$ because $2f \in k[x, y]$, but what if the characteristic of $k$ is $2$? – Jim Mar 23 '20 at 02:38
  • Jim, you're right, I did implicitly assumed the characteristic is other than 2. I'm yet to figure out what happens in characteristic 2. – sss89 Mar 23 '20 at 06:42
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    In last part, how to apply Gauss lemma? When polynomial is already element of $k[x,y]$ , we can use Gauss lemma. For example, $f(x)=x^2 + \frac{1}{4}$ is irrreducible in $\mathbb{Q}[x]$ , but it is not in $\mathbb{Z}[x]$. ($\mathbb{Z}$ is UFD) – hew Jun 15 '20 at 12:33
  • @hew, this follows from the fact that $\alpha$ is integral over $k[x,y]$ (since its integral over $R$ which, in turn, is integral over $k[x,y]$). So exist some monic irreducible polynomial with coefficients in $k[x,y]$ that vanishes on $\alpha$ and Gauss lemma ensures it is equal to the minimal polynomial of $\alpha$ over $k(x,y)$. – sss89 Jun 15 '20 at 14:21
  • @sss89 Do you means that "$\alpha$ is integral over $R$" imply "$\alpha$ is integral over $k[x,y]$" by regarding $^{k[x,y,z]} / _{(z^2 -xy)} \leq k[x,y]$ ? (mapping $x \mapsto x^2$ , $y \mapsto y^2$ , $z \mapsto xy$). – hew Jun 16 '20 at 01:05
  • @hew I mean that $k[x,y]\subset R$ by $x\mapsto x$ and $y\mapsto y$. – sss89 Jun 18 '20 at 10:27