Compute integral closure of $F[x,y,z]/(x^2-y^2z)$.

I want to compute integral closure of $R:=F[x,y,z]/(x^2-y^2z)$. Let $S$ the integral closure. I have proved that $\frac{\bar x}{\bar y}$ and $\bar y$ are integral over $R$ and that $\{\bar y,\frac{\bar x}{\bar y}\}$ is a transcendence basis of $\text{Frac}(R)$. My questions are the following one :

Q1) Why $S\supset F[\bar y,\frac{\bar x}{\bar y}]$ ? I agree that if $S\supset R[\bar y,\frac{\bar x}{\bar y}]\subset S$, but why $F[\bar y,\frac{\bar x}{\bar y}]\subset S$ ?

Q2) Why $$\text{trdeg}_F(\text{Frac}(R))\leq \text{trdeg}_F(\text{Frac}(F[\bar y,\frac{\bar x}{\bar y}])) \ \ ?$$

With those answer, I can conclude.

Solution 1:

I believe that you mean the integral closure of $R$ in its field of fractions $K$ (notice that $x^2-y^2z \in F[y,z][x]$ is irreducible by Eisenstein at $p=z$ so $(x^2-y^2z)$ is a prime ideal of the UFD $F[x,y,z]$).

1) You proved that $\bar y,\frac{\bar x}{\bar y}$ are integral over $R$. But any element $a$ of $F$ is integral over $R$ (take the polynomial $P(t) = t-a \in R[t]$, considering $F$ as a subring of $R$). Then any polynomial $P(x,y)$ with coefficients in $F$, evaluated at $\bar y,\frac{\bar x}{\bar y}$ will give you an integral element of $R$, i.e. $$S \subset F\left[\bar y,\frac{\bar x}{\bar y} \right].$$

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2) We have $$ F \;\subset\; R = F[\bar x, \bar y, \bar z] \;\subset\; S \;\subset\; F\left[\bar y,\frac{\bar x}{\bar y} \right] \;\subset\; K = \mathrm{Frac}(R) $$

and applying $\mathrm{Frac}(\cdot)$, we get

$$ F \;\subset\; \mathrm{Frac}(R)=K \;\subset\; L := \mathrm{Frac}\left( F\left[\bar y,\frac{\bar x}{\bar y} \right] \right) \;\subset\; K $$