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?
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 $\operatorname{ht} I = 2$.
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 coincide with the symbolic powers.
Here is a (hopefully correct) elementary proof of the minimality of the generating set. Suppose $I = (f,g)$. Then $I = (f + rg,g)$ for any polynomial $r$, and so we may assume that $f$ and $g$ do not have the monomial term $z^2$ in common. Also, note that since $f,g \in I$, the following terms cannot be nonzero in $f,g: 1,x,y,z,xy,x^2.$ Since we must have polynomials $a,b$ such that $z^2 - x^2y = af + bg$, somehow a nonzero $z^2$ must appear. It cannot come about from a $1$ or $z$ in $f,g$, and so one of these must contain the term $z^2$. WLOG $f$ does, and so $a$ must have nonzero constant term.
Now choose polynomials $p,q,r$ with $f = p(z^2 - x^2y) + q(x^3 - yz) + r(y^2 - xz)$. Since $z^2$ appears in $f$ as a nonzero term, the constant term of $p$ must be nonzero, and the $-x^2y$ term from $p(z^2 - x^2y)$ cannot be cancelled from any other term from the RHS of the equation for $f$ above. Hence $x^2y$ is a nonzero term of $f$.
Now, write $x^3 - yz = cf + dg$ for some polynomials $c,d$. Then one of the constant terms of $c$ or $d$ must be nonzero and the term $yz$ appear in $f$ or $g$. Suppose that the constant term of $c$ is nonzero. Then the $x^2y$ term of $f$ appears and must be cancelled by some term of $dg$. Hence $g$ must contain one of the following terms: $1,x,y,x^2,xy,x^2y$. The first of these are forbidden since $g$ lies in $I$, while if $g =p'(z^2 - x^2y) + q'(x^3 - yz) + r'(y^2 - xz)$ contains the term $x^2y$, then the constant term of $p'$ must be nonzero, and so $g$ must contain a nonzero $z^2$, contradicting our assumption. So the constant term of $c$ is zero, and the constant term of $d$ is not zero. That case ruled out, we must have $yz$ nonzero in $g$. Note that $x^3$ must also be nonzero in $g$. So the constant term of $q'$ is not zero.
Now write $y^2 - xz = jf + kg$. As before, we know that one of the constant terms of $j,k$ must be nonzero to yield a $y^2$ in the LHS. If the constant term of $j$ is nonzero, then a $z^2$ appears from the $jf$ term and so $1,z,z^2$ must appear in $g$, impossible. So the constant terms of $k,r'$ are nonzero. But then one of $1,x,x^2,x^3$ must appear in $jf$. But the constant term of $j$ is zero, so one of $1,x,x^2$ appear in $f$, impossible.
Firstly, it is clear that $f$ has not constant coefficient. I claim that for any $f\in \mathfrak{p}$, $f$ has not terms $X^1, X^2, Y^1, Z^1$. Since, for example, $$f(X,Y,Z)=Y^1+\textrm{(other terms)}=Y^1+Y^2(\ldots)+X(\ldots)+Z(\ldots)$$ then $$0=f(T^3,T^4,T^5)=T^4+\underbrace{T^8(\ldots)+T^3(\ldots)+T^4(\ldots)}_{\textrm{has not term $T^4$}}$$ since $3a+4b+5c\neq 1$ for $a,b,c\in \mathbb{Z}_{\geq 0}$. But, $X^3-YZ, Y^2-XZ, Z^2-X^2Y\in \mathfrak{p}$ have terms $X^3, Y^2,Z^2$ respectively. If $\left<f,g\right>=\mathfrak{p}$, assume $a,b,p,q,n,m\in k[X,Y,Z]$ such that $$af+bg=X^3-YZ\qquad pf+qg=Y^2-XZ\qquad nf+mg=Z^2-X^2Y$$ Assume that $$\left(\begin{matrix}f\\ g\end{matrix}\right) =\left(\begin{matrix}x_f& y_f & z_f\\\ x_g& y_g & z_g\end{matrix}\right) \left(\begin{matrix}X^3\\ Y^2\\ Z^2\end{matrix}\right)+ \left(\begin{matrix}\textrm{other terms}\\ \textrm{other terms}\end{matrix}\right)$$ and that $a_0=a(0,0,0)$ and so on. Then by looking at the term $X^3,Y^2,Z^2$, one has $$\left(\begin{matrix}a_0 & b_0\\ p_0& q_0\\ n_0&m_0\end{matrix}\right) \left(\begin{matrix}x_f& y_f & z_f\\\ x_g& y_g & z_g\end{matrix}\right) =\left(\begin{matrix}1 & 0 &0 \\ 0& 1 & 0\\ 0&0 &1\end{matrix}\right)$$ which is a contradiction by standard linear algebra. $\square$
I think it is a solution in flavor of Gröbner basis. Maybe it is the proof "which Hartshorne wanted readers to give".
