0
$\begingroup$

Let $F$ be a field. I know that $F[x,y,z]$ is a UFD, that $x^2-y^2z\in F[x,y,z]$ is prime. Let $R:=F[x,y,z]/(x^2-y^2z)$.

Q1) Why $R$ is an integral domain ? It's clear that $R$ is a field, and thus it's a domain, but why is it integral ? (maybe a field is always integral).

Let $S$ be the integral closure of $R$. We have that $\bar x^2=\bar y^2\bar z\implies \left(\frac{\bar x}{\bar y}\right)^2=\bar z$, we have that $\frac{\bar x}{\bar y}$ is integral over $R$.

Q2) Why now we have that $S\supset F[\frac{\bar x}{\bar y},\bar y]$ ?

After, they say that $\frac{\bar x}{\bar y}$ and $\bar y$ are algebraically independent, otherwise,

$$2>\operatorname{trdeg}_F \operatorname{Frac} F\left[\frac{\bar x}{\bar y},\bar y\right]\geq \operatorname{trdeg}_F R=2,$$ but I don't understand each inequality.

Q3) I mean, why $2>\operatorname{trdeg}_F \operatorname{Frac} F\left[\frac{\bar x}{\bar y},\bar y\right]\geq \operatorname{trdeg}_F R=2$, why $\operatorname{trdeg}_F \text{Frac}( F\left[\frac{\bar x}{\bar y},\bar y\right])\geq\operatorname{trdeg}_F R$, and why $\operatorname{trdeg}_F R=2$ ?

  • 1
    (1) You need to check the definition of integral domain. Google it up! $R$ is an integral domain exactly because $x^2-y^2z$ is prime. (2) $R$ is not a field. – 2017-01-22
  • 0
    @MathChat: You right, it would be a field if $F[x,y,z]$ would be a PID, because in PID, prime are maximal, right ? – 2017-01-22
  • 0
    $\operatorname{trdeg}_FR=\dim R=2$, while $u,v$ algebraically dependent over $F$ means $\operatorname{trdeg}_FF(u,v)\le 1$. – 2017-01-22
  • 0
    @user386627 : $F[x_1,\dots,x_n]$ is never a field, for $x_1$ is not invertible. And in a PID, only non-zero primes are maximal (PID $\implies$ Krull dimension is at most 1 ; converse is not true, there are Dedekind non-UFDs, or $k[x_1,...]/(x_1,...)^2$ which has dimension 0 but is not even Noetherian). – 2017-01-29
  • 0
    By the way, Q2) is answered [here](http://math.stackexchange.com/questions/2119160) (as well as Q1, actually). – 2017-01-29

0 Answers 0