Summing complete intersections

Question

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?

Answer 1

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.