If \(X\) is a complete intersection of surfaces of degrees \(a, b\) in \(\PP^3\), then \(X\) does not lie on any surface of degree \(<\min (a, b)\).
Solution
Let \[ S=k[x_0,x_1,x_2,x_3]. \]
1. The saturated homogeneous ideal of \(X\) is\[ I(X)=(F,G), \]where \(F\) and \(G\) are homogeneous forms of degrees \(a\) and \(b\).
By hypothesis, \(X\) is the scheme-theoretic complete intersection of two surfaces of degrees \(a\) and \(b\). Under the complete-intersection convention of Exercise II.8.4, its saturated homogeneous ideal is generated by the defining equations of those surfaces. Thus we may choose \[ \deg F=a, \qquad \deg G=b, \] with \[ I(X)=(F,G). \]
2. For every integer\[ m<\min(a,b), \]the degree-\(m\) component of \(I(X)\) is zero:\[ I(X)_m=0. \]
Because \((F,G)\) is a homogeneous ideal, \[ I(X)_m = F\,S_{m-a}+G\,S_{m-b}. \] If \(m<a\) and \(m<b\), then \[ S_{m-a}=S_{m-b}=0, \] since a polynomial ring has no nonzero homogeneous forms of negative degree. Hence \[ I(X)_m=0. \]
3. The curve \(X\) lies on no surface of degree\[ m<\min(a,b). \]
Suppose a surface \(T\subseteq\PP^3\) of degree \(m<\min(a,b)\) contained \(X\). Let \[ 0\ne H\in S_m \] be a homogeneous equation for \(T\). Since \(T\) contains \(X\), the form \(H\) lies in the saturated homogeneous ideal of \(X\), so \[ H\in I(X)_m. \] Step 2 gives \(I(X)_m=0\), contradicting \(H\ne0\). Therefore no such surface exists.
4. Q.E.D.
Step 3 is exactly the required assertion.