How to compute $I(Y)$ for the curve $Y$ defined parametrically by $x=t^{3}$, $y=t^{4}$, $z=t^{5}$?

Let $Y\subset\mathbb{A}^{3}$ defined parametrically by $x=t^{3}$, $y=t^{4}$, and $z=t^{5}$. I want to compute $I(Y)$.

I think that
$$I(Y)=(x^{20}-z^{12},x^{20}-y^{15},y^{15}-z^{12}),$$
but I can only get inclusion $\supset$. Any ideas on how to get the other inclusion or is this even true?

I am not really interested in the ideal itself but more the height.

I know that it is a prime ideal. So another way I thought about getting at the height was trying to compute the dimension of $\mathbb{C}[t^{3},t^{4},t^{5}]$. I’m hoping it is 2. Any ideas on either of these approaches?
Thanks.

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If you are only interested in computing the height of $I(Y)$, then you can argue as follows.

First, since $\dim k[x,y,z]=3$ it follows immediately that $\operatorname{ht}I(Y) \leq 3$.
The existence of the chain

\begin{split}
(0) \subsetneq (xz-y^2) \subsetneq I(Y)
\end{split}

of prime ideals in $k[x,y,z]$ implies moreover that $\operatorname{ht}I(Y) \geq 2$, so that either $\operatorname{ht}I(Y)=2$, or $\operatorname{ht}I(Y)=3$.

But $\operatorname{ht}(I(Y))=3$ would imply that $I(Y) \subset k[x,y,z]$ is a maximal ideal and hence $Y$ would have to consist of a single point, which is false. Therefore $\operatorname{ht}I(Y)=2$.


As you indicated, you can also compute the height of $I(Y)$ by calculating the dimension of $A(Y):=k[x,y,z]/I(Y)$, the affine coordinate ring of $Y$, thanks to the formula
\begin{split}
\operatorname{ht}I(Y)+\dim A(Y)=\dim k[x,y,z]=3
\end{split}
which is a special case of the dimension-formula. (cf., e.g. Hartshorne, Theorem I.1.8A). Thus $\operatorname{ht} I(Y)=2$ is equivalent to $\dim A(Y)=1$.

Now if you already know that $A(Y)=\Bbb{C}[t^3,t^4,t^5]$, then it is easy to show that $\dim A(Y)=1$ (since the dimension of $A(Y)$ is equal to the transcendence degree of the field of fractions of $A(Y)$ over $\Bbb{C}$, which is $\Bbb{C}(t)$, and this has transcendence degree $1$ over $\Bbb{C}$).

But proving that $A(Y)=\Bbb{C}[t^3,t^4,t^5]$ you need to know what $I(Y)$ looks like, and this seems to be a little more complicated than the above method for computing the height of $I(Y)$.

In fact you should have:
$$I(Y) = (x^3 – yz, y^2 – xz, z^2 – x^2y)$$
One inclusion is easy. For the other the proof I have is a little messy, maybe someone else will post a better one:

We need to prove that the ideal
$$I(Y)/(x^3 – yz, y^2 – xz, z^2 – x^2y) \subseteq k[x, y, z]/(x^3 – yz, y^2 – xz, z^2 – x^2y)$$
is zero. First observe that in the quotient ring, the second two relations imply that every element is represented by a polynomial with at most one $y$ and at most one $z$ in any monomial and from the first relation we can also ask that $yz$ doesn’t appear in any monomial. This means every $\overline{f} \in I(Y)/(x^3 – yz, y^2 – xz, z^2 – x^2y)$ can be represented by a polynomial $f \in k[x, y, z]$ of the form
$$f = f_0(x) + f_1(x)y + f_2(x)z.\tag{$\ast$}$$

Now assume $I(Y)/(x^3 – yz, y^2 – xz, z^2 – x^2y)$ is nonzero and choose $f \in k[x, y, z]$ to be of minimal degree among those polynomials of the form $(\ast)$ which lie in $I(Y)$. Note $f$ has no constant term because $(0, 0, 0) \in Y$ so in fact we have
$$f = f_0′(x)x + f_1(x)y + f_2(x)z$$
where $f_0 = xf_0’$. The polynomial $f(t^3, t^4, t^5) \in k[t]$ is identically zero by hypothesis. Note the smallest possible monomial in $f_0′(t^3)t^3$ has degree $3$ and can’t be cancelled by anything in $f_1(t^3)t^4 + f_2(t^3)t^4$ so it’s coefficient must be zero, thus $x$ divides $f_0’$. Let $f_0′ = xf_0”$, now we have
$$f = f_0”(x)x^2 + f_1(x)y + f_2(x)z.$$
A similar argument gives that x divides $f_1$ and that $x$ divides $f_2$ so
$$f = f_0”(x)x^2 + f_1′(x)xy + f_2′(x)xz.$$
But now $x$ divides $f$. Write $g = f_0”(x)x + f_1′(x)y + f_2′(x)z$ so that $f = xg$. The polynomial $g(t^3, t^4, t^5)t^3$ must be identically zero so $g(t^3, t^4, t^5)$ is zero. This gives $g \in I(Y)$, contradicting the minimality of $f$.

Thus $I(Y)/(x^3 – yz, y^2 – xz, z^2 – x^2y)$ is zero.