The intersection of two varieties, each of which is a complete intersection

[naturemath]

Consider f_1,..., f_k, g_1,..., g_l in a polynomial ring C[x_1,...,x_M], where f_i's are homogeneous of degree 2 while g_j's are linear polynomials.

Suppose the codim of the variety cut out by S_1 = {f_1,..., f_k} is k while the codim of the variety cut out by S_2 = {g_1,..., g_l} is l.

Assume any or all of the following:
* {f_i, g_j} is a complete intersection for some i and for some j;
* {f_i, g_j} is a complete intersection for any i and for any j;
* S_1 union {g_j} is a complete intersection for any g_j in S_2;
* {f_i} union S_2 is a complete intersection for any f_i in S_1;

Isn't there a result somewhere in a commutative algebra book or in some paper which says that the variety cut out by all of S_1 union S_2 is a complete intersection?

 

[fresh_42]

Maybe an answer can be found in https://www.jmilne.org/math/CourseNotes/AG.pdf

 

[mathwonk]

This rather old question has the answer "no". In P^4 take a general pair of quadric hyper surfaces, which intersect in a quartic "segre" surface containing 16 lines. Now take two general hyperplanes each containing that line. Then the two quadrics cut a complete intersection, the segre surface, and the two hyperplanes cut a complete intersection, a plane, and each hyperplane cuts the segre surface in a complete intersection, a curve with two components, one a cubic and one the given line. Each quadric moreover also cuts the plane in a curve, since no smooth quadric in P^4 contains a plane. The 4 varieties taken together however do not define a complete intersection, since they meet in more than a finite set of points, namely they all contain the given line.