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By a result of Klein-Nagata rings of the form $A_Q=K[x_1,...,x_n]/(Q)$ are factorial when $K$ is a field, $n \geq 5$ and $Q$ is a non-degenerate quadratic form.

Question 1: When is $A_Q$ a principal ideal domain or even an euclidean ring for $n \geq 2$ for a general quadratic polynomial $Q$?

Question 2: What is the global dimension of $A_Q$ when $Q$ is a quadratic form?

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    $\begingroup$ I don't think they will ever be. The ideal $(x_1,...,x_n)$ will not be principal unless $n=1$, in which case it will not be domain. $\endgroup$ Apr 16, 2021 at 0:43

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PID's have Krull dimension $1$ (or $0$, if you call a field a PID); $A_Q$ will have Krull dimension $n-1$. So the only option is $n=2$ (the case $n=1$ doesn't apply since $k[x]/x^2$ is not a domain).

However, the $n=2$ case will also not give a PID. The ideal $\langle x_1, x_2, \ldots, x_n \rangle$ is not even locally principal (compute the dimension of the Zariski tangent space).

When I say the title of this question, I thought it would be asking about $k[x,y]/(ax^2+bxy+cy^2+dx+ey+f)$. The answer is that this is a PID if the conic $ax^2+bxy+cy^2+dx+ey+f=0$ is smooth$^*$ and either

(1) $a x^2+b xy + c y^2$ has roots in $\mathbb{P}^1_k$ or

(2) the conic $ax^2+bxy+cy^2+dx+ey+f=0$ has no roots in $k$.

For example, $\mathbb{R}[x,y]/\langle x^2-y \rangle$, $\mathbb{R}[x,y]/\langle xy-1 \rangle$ and $\mathbb{R}[x,y]/\langle x^2+y^2+1 \rangle$ are all PID's, but $\mathbb{R}[x,y]/\langle x^2+y^2-1 \rangle$ is not.

$^*$ In characteristic $2$, I should say regular instead of smooth. For example, I think that $\mathbb{F}_2(t,u)[x,y]/(t x^2+u y^2+1)$ is a PID. I'm not completely confident in this.

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  • $\begingroup$ Thanks, I was consufed what a quadratic form (it was after midnight when I posed it...) is for my question and thought it is more general as in the second form of your answer. I expanded my question to avoid it being closed. $\endgroup$
    – Mare
    Apr 16, 2021 at 8:52

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