# Summing complete intersections

Suppose we have polynomials $$f_1,\dots,f_r\in k[X_1,\dots,X_N]$$ defining a complete intersection in $$\mathbb{A}^N$$. I suspect that it is then true that $$f_1(X)+f_1(Y),\dots,f_r(X)+f_r(Y)\in k[X_1,\dots,X_N,Y_1,\dots,Y_N]$$ defines a complete intersections in $$\mathbb{A}^{2N}$$. Is this true? If so, is there a generalization of this to commutative $$k$$-algebras and regular sequences?

This is not true. We can have $$N=r+1$$ and $$f_i = X_{i+1} X_1 - 1$$ for all $$i$$. Then the intersection is the locus where $$X_i = X_1^{-1}$$ for all $$i$$ from $$2$$ to $$r+1$$, thus has dimension $$1$$ and is a complete intersection. However $$f_i(X_1,\dots,X_n) + f_i(Y_1,\dots,Y_n)$$ will vanish as soon as $$X_1=Y_1=0$$ and has dimension $$2N-2$$ so i snot a complete intersection as long as $$r>2$$.
However, it is true for homogeneous polynomials. The reason is that by a degeneration argument, if the vanishing set of $$f_1,\dots,f_r$$ is a complete intersection, then the vanishing set of $$f_1-c_1,\dots, f_r-c_r$$ is a complete intersection for any $$c_1,\dots,c_r$$. Because of this, we can see that the vanishing locus of $$f_i(X) + f_i(Y)$$ has dimension $$r+ (N-r) + (N-r) = 2N-r$$ and thus is a complete intersection.