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I'm trying to do the following:

Let $R = K[X,Y,Z]$ and $\mathfrak{p}$ = $(X+Y,Z^{2}-X)$. Show that $\mathfrak{p}$ is prime and find the transcendence degree of $R/\mathfrak{p}$.

If I prove that $\mathfrak{p}$ is prime the question is over just using the fact that the quotient is integral over $K[y]$. But my problem is to show that $\mathfrak{p}$ is prime. I know so far that $\mathfrak{p}$ is generated by a regular sequence in $R$, i.e., $X+Y$ is regular in $R$ and $Z^{2}-X$ is regular in $R/(X+Y)$, and thus every minimal prime of $\mathfrak{p}$ has height $2$. I tried do it by contradction but I get nothing.

Thank you for any help.

User43029
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3 Answers3

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Recall that $\mathfrak{p}$ is prime if and only if $R/\mathfrak{p}$ is an integral domain. I suggest that this is the easiest way to approach this question.

bradhd
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Hint: Note that in $R/\mathfrak p$, we have $f(X,Y,Z) = f(X,-X, Z) = f(Z^2, -Z^2, Z)$. Now it might be interesting to look at the natural inclusion followed by projeciton $$K[Z] \to K[X,Y,Z] \to R/\mathfrak p$$ as well as the map $$K[X,Y,Z]\to K[Z]$$ given by $f(X,Y,Z) \mapsto f(Z^2, - Z^2, Z)$. Show that the latter descends to a homomorphism $R/\mathfrak p \to K[Z]$.

Sam
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Consider the following map: $T: K[x,y,z]/(x+y, z^2-x) \to K[z]$ where $T$ is defined by $1 \mapsto 1$, $x \mapsto z^2, y \mapsto -z^2, z \mapsto z$. Then $x+y$ and $z^2-x$ both map to $0$ and so $T$ is well defined. $T$ is surjective and if $Tf=0$, then $f(z^2, -z^2, z)=0$. But in the domain space, $x=z^2, y=-z^2$ and so then $f(x,y,z)=0$. Thus, the map is an isomorphism and we notice the image is definitely integral.

Alexander
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