Height and minimal number of generators of an ideal

Question

How can I determine the height and the least number of generators of the ideal $ I=(xz-y^2,x^3-yz,z^2-x^2y) \subset K[x,y,z] $?

I tried to calculate the dimension of the vector space $I/I\mathfrak m$ with $\mathfrak m=(x,y,z)$ but I'm not able to find it.

I think the height is 2, because the affine variety should be a curve, and also I know that the ideal is prime, so I need a chain $0\subset \mathfrak p \subset I$. Maybe $(xz-y^2)?$ Is it prime?

Answer 1

For simplicity I will assume $\operatorname{char} k = 0$, though probably $\operatorname{char} k \notin \{ 2, 3, 5 \}$ will be enough.

First, one makes an inspired guess and observes that $$\begin{align} x & = t^3 \\ y & = t^4 \\ z & = t^5 \end{align}$$ parametrises all the solutions to the equations. (One is led to guess this by inspecting the Gröbner bases of the ideal with respect to various monomial orderings.) This implies that the variety is the continuous image of an irreducible variety, so must itself be irreducible. (Thus, the ideal in question is prime.) Note also that this parametrisation is singular at $t = 0$, so one suspects that the behaviour of the variety at $(0, 0, 0)$ may not be generic.

Let us instead look at the point $(1, 1, 1)$. Let $A = k[x, y, z] / I$. The cotangent space of $X$ at $(1, 1, 1)$ is the $A / \mathfrak{m}$-module $\mathfrak{m} / \mathfrak{m}^2$, where $\mathfrak{m}$ is the maximal ideal of $A$ (not the polynomial ring!) corresponding to the point. In order to use Gröbner basis techniques, we must lift this definition to the polynomial ring. If $\tilde{\mathfrak{m}} = (x, y, z)$ is the maximal ideal of $k[x, y, z]$ above $\mathfrak{m}$, then we have $$\frac{\mathfrak{m}}{\mathfrak{m}^2} \cong \frac{(\tilde{\mathfrak{m}} + I) / I}{(\tilde{\mathfrak{m}}^2 + I) / I} \cong \frac{\tilde{\mathfrak{m}} + I}{\tilde{\mathfrak{m}}^2 + I}$$ Mathematica informs me that $$\tilde{\mathfrak{m}}^2 + I = (1 - 2z + z^2, -1 + 5y - 4z, -2 + 5x - 3z)$$ and so $\mathfrak{m} / \mathfrak{m}^2$ is indeed $1$-dimensional, as expected. On the other hand, the cotangent space of $X$ at $(0, 0, 0)$ is $3$-dimensional!

Now, by general facts about $k$-algebras, we have $$\dim A + \operatorname{ht} I = \dim k[x, y, z] = 3$$ where $\dim$ here refers to Krull dimension. Since the variety is indeed a curve, $\dim A = 1$. So $\operatorname{ht} I = 2$.

Answer 2

Let $R=K[X,Y,Z]$ and $P=(XZ-Y^2,X^3-YZ,Z^2-X^2Y)$. Then $P$ is a prime ideal, $\operatorname{ht}P=2$ and $\mu(P)=3$, where $\mu(I)$ denotes the minimal number of generators of the ideal $I$.

Zhen Lin already proved that $P$ is prime and $\operatorname{ht}P=2$ showing that $R/P\simeq K[T^3,T^4,T^5]$.

It remains to show that $\mu(P)=3$.

Since $\operatorname{ht}P=2$, $P$ can't be principal, so $\mu(P)\ge 2$.

Suppose $\mu(P)=2$. Using again that $\operatorname{ht}P=2$ and the well known fact that $R$ is Cohen-Macaulay we get $\operatorname{grade}P=2$. Then $P$ can be generated by an $R$-sequence of length $2$. The powers of ideals generated by $R$-sequences are perfect (see Bruns and Herzog, Cohen-Macaulay Rings, exercise 1.4.27), in particular, grade unmixed (see the same book, Proposition 1.4.16). Since $P^m$ is grade unmixed it follows that the only associated prime of $R/P^m$ is $P$ (otherwise there exists an associated prime $P'$ with $P\subset P'$, and then we have $\operatorname{ht}P'>\operatorname{ht}P$, that is, $\operatorname{grade} P'>\operatorname{grade} P$, false). This shows that $P^m$ is a $P$-primary ideal for every $m\ge 1$. But $P^2$ is not primary as show the following: $$X(X^5-3X^2YZ+XY^3+Z^3)=(X^3-YZ)^2+(XZ-Y^2)(Z^2-X^2Y)\in P^2,$$ but $X\notin P$ and $X^5-3X^2YZ+XY^3+Z^3\notin P^2$ (for every polynomial in $P^2$ contains no monomials of total degree less than $4$).

Edit. Actually I've proved above that in a Cohen-Macaulay ring the prime ideals of principal class, that is, ideals whose minimal number of generators equals their height, have the following property: the ordinary powers concide with the symbolic powers.