Let \(k\) be algebraically closed and let \(Y \subseteq \AA_k^3\) be the curve given parametrically by \(x = t^3\), \(y = t^4\), \(z = t^5\). Show that \(I(Y)\) is a prime ideal of height \(2\) in \(k[x,y,z]\) which cannot be generated by \(2\) elements. We say \(Y\) is not a local complete intersection; cf. (Ex. 2.17).
Solution
Put \(S=k[x,y,z]\), \(\mathfrak m=(x,y,z)\), and define \[ \phi:S\longrightarrow k[t],\qquad x\longmapsto t^3,\quad y\longmapsto t^4,\quad z\longmapsto t^5. \] Since \(k\) is infinite, a polynomial in \(k[t]\) vanishing at every element of \(k\) is zero. Thus \(I(Y)=\ker\phi\).
1. The defining ideal is\[ \boxed{I(Y)=(y^2-xz,\ yz-x^3,\ z^2-x^2y).} \]
Let \(J\) be the ideal displayed in the claim. Substitution in each generator gives zero, so \(J\subseteq\ker\phi\). Modulo \(J\), the replacements \[ y^2=xz,\qquad yz=x^3,\qquad z^2=x^2y \] reduce the total exponent of \(y\) and \(z\) in a monomial whenever that exponent is at least two. Repeating these replacements therefore expresses every element of \(S/J\) as \[ p(x)+q(x)y+r(x)z\qquad(p,q,r\in k[x]). \] Its image under \(\phi\) is \[ p(t^3)+t^4q(t^3)+t^5r(t^3). \] The exponents in these three summands are congruent respectively to \(0,1,2\) modulo \(3\). They cannot cancel between summands, and distinct powers within each summand are also linearly independent over \(k\). Hence a zero image forces \(p=q=r=0\). It follows that \(S/J\to k[t]\) is injective, giving \(J=\ker\phi\) as claimed.
The parametrized image is itself closed. Indeed, at a common zero of these equations with \(x\ne0\), set \(t=y/x\). The equations give \(t^3=x\), then \(t^4=y\) and \(t^5=z\). If \(x=0\), the equations \(y^2=0\) and \(z^2=0\) give the origin, corresponding to \(t=0\). Thus the image is exactly \(V(J)\).
2. The ideal \(I(Y)\) is prime and has height \(2\).
Step 1 identifies the quotient with the subring \[ S/I(Y)\cong k[t^3,t^4,t^5]\subseteq k[t]. \] This is a domain, so \(I(Y)\) is prime. Its fraction field is \(k(t)\): it contains \(t=t^4/t^3\) and is contained in \(k(t)\). Its transcendence degree over \(k\) is therefore one. The dimension formula for a prime ideal in a polynomial ring gives \[ \operatorname{height}I(Y) =3-\operatorname{trdeg}_k\operatorname{Frac}(S/I(Y))=2 \] [@Har10a, Chapter I, §1].
3. The ideal \(I(Y)S_{\mathfrak m}\) needs exactly three generators in the local ring \(S_{\mathfrak m}\).
1.They are linearly independent in that space.Indeed, \(I(Y)\subseteq\mathfrak m^2\), so \(\mathfrak m I(Y)\subseteq\mathfrak m^3\).If \(aF+bG+cH\in\mathfrak m I(Y)\) for \(a,b,c\in k\), its homogeneous quadratic part must vanish:\[ a(y^2-xz)+b yz+c z^2=0. \]The coefficients of \(y^2\), \(yz\) and \(z^2\) give \(a=b=c=0\).Consequently \(\dim_k I(Y)/\mathfrak m I(Y)=3\).
Localization gives \[ I(Y)S_{\mathfrak m}/\mathfrak m S_{\mathfrak m}\,I(Y)S_{\mathfrak m} \cong I(Y)/\mathfrak m I(Y), \] because the right-hand module is a \(k=S/\mathfrak m\)-vector space and every element outside \(\mathfrak m\) acts on it by a nonzero scalar. Any generating family for the localized ideal must span this three-dimensional quotient, so it has at least three elements. The three displayed polynomials already generate the ideal, proving the assertion. In particular, a two-element global generating family is impossible, since it would remain a generating family after localization.
4. Q.E.D.
Step 2 proves primality and height two. Step 3 proves that the ideal cannot be generated by two elements even at the origin. Its codimension there is two, so this also proves failure of the local complete-intersection condition at that point.