Prove that an ideal has height 2

Question

I know this question has been asked before, and I throughly looked at all the answers, but can't find one that matches my math knowledge, or the contents of the course the exercise is derived from.

In $A=\mathbb{C}[x,y,z]$ I have the ideal $I=(xz-y^2,x^3-yz,z^2-x^2y)$. I need to find the height of this ideal. There have been questions about this, but I don't understand them. I showed that $I$ is prime by proving that the related algebraic variety $X=V(I)$ is irreducible, and I did it using the parametrization of the variety $t \mapsto (t^3,t^4,t^5)$. So to find the height I'd need to find some primes inside this one. I know the height needs to be 2, so I'd only have to find another prime, and prove there are no others between $I$ and $(0)$.

I saw someone proving the fact that $A/I$ is isomorphic to $\mathbb{C}[t^3,t^4,t^5]$ and, even though I don't get how $I$ should be the kernel of the immersion $\Phi$ of $A$ in $\mathbb{C}[t^3,t^4,t^5]$ where we map $x \mapsto t^3, y\mapsto t^4, z \mapsto t^5$ (of course, $I \subseteq \ker\Phi$, but how do I prove the opposite?), I still don't have the knowledge about $\mathbb{C}$-algebras to conclude that the height is $2$. I know that in general $\mathrm{ht}(I) + \dim(A/I) \leq \dim A$ but I don't know if and when the equality holds. There are other answers like these here, but there's no one that seems clear enough for me. Would you help me understand? Thank you!

Answer 1

The outline you set up in your post is of course something that can be followed, but there is a major result which will make your life much easier:

Key fact 1. Let $A$ be a finitely generated algebra over a field $k$, and let $P\subset A$ be a prime ideal. Then $$\dim A = \dim A/P + \operatorname{height} P.$$ See here for discussion about when this is true for more general classes of rings, and a reference to Eisenbud's Commutative Algebra for the statement in your case.

In your case, $A=\Bbb C[x,y,z]$, and $P=I$. As $\dim\Bbb C[x,y,z]=3$, if we can show $\dim A/P=1$, then we'll have $\operatorname{height} I =2$.

Key fact 2. If $A\subset B$ is an integral extension of rings, then $\dim A = \dim B$. See for instance here.

Since $A/P\cong\Bbb C[t^3,t^4,t^5]$ has integral extension $\Bbb C[t]$ which is of dimension one, we have $\dim A/P=1$.

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All that's left is to show $I$ is the kernel of the map $\Bbb C[x,y,z]\to\Bbb C[t^3,t^4,t^5]$. I claim that any element $f\in\Bbb C[x,y,z]$ can be written as $p(x)+yq(x)+y^2r(x)+\lambda z + s(x,y,z)$ for $s\in I$.

• If $f$ has a term of the form $kx^ay^bz^c$ with $k\neq 0$ for $c>1$, then $kx^ay^bz^c=kx^ay^bz^{c-2}(z^2-x^2y+x^2y)$, so we can write $f'=f-kx^ay^bz^c+kx^{a+2}y^{b+1}z^{c-2}$, and $f$ is of the given form iff $f'$ is. Inductively applying this rule, we may assume that $f=\sum_{a,b,c} k_{a,b,c}x^ay^bz^c$ with $0\leq c \leq 1$.
• If $f$ has a term of the form $kx^ay^bz^c$ with $k\neq 0$, $c=1$, and $a>0$ or $b>0$, then we can use the same process in the last step with $xz-y^2$ or $x^3-yz$ to turn this term in to one with $c=0$. So we may assume that $f=g(x,y)+\lambda z$.
• Next, we note that $x^4-y^3=y(xz-y^2)+x(x^3-yz)$, so by the same logic as in the first step we may assume that $\deg_y f < 3$, and this gives the claim.

Now look at what happens when we evaluate $p(x)+yq(x)+y^2r(x)+\lambda z + s(x,y,z)$ for $s\in I$ at $x=t^3$, $y=t^4$, and $z=t^5$. We get $p(t^3)+t^4q(t^3)+t^8r(t^3)+\lambda t^5$, and this is zero iff $p=q=r=\lambda=0$: there cannot be any cancellation! This shows that $\ker\Phi\subset I$, and we've wrapped up the problem.