Let $I \subset \mathbb{F}[x_1, \dots, x_n]$ be an ideal containing (among possibly other generators) the polynomials $x_1^2 - x_1,\dots,x_n^2 - x_n$. We can call such an ideal multilinear because every element in the ideal is a degree 1 polynomial in every variable. Is it true that $I$ must be a radical ideal?
I have done some experimenting with Sage, and I can't easily come up with a counterexample. For instance, the ideal $\langle x^2 - x, y^2 - y, (x+y)^2 \rangle$ is radical, and in particular, you can get $x+y$ by expanding \begin{equation} (x+y)^2(2xy - 3x - 3y + 5), \end{equation} reducing exponents on variables to 1 as necessary, and dividing by 2. This kind of strange equation also gives me no leads towards a constructive proof.
Yes. Recall that $I$ is radical iff the quotient $F[x_1, \dots x_n]/I$ has no nontrivial nilpotents. Since $I$ contains the polynomials $x_i^2 - x_i$, this quotient factors through the quotient $F[x_1, \dots x_n]/(x_i^2 - x_i)$. This quotient is the tensor product $\bigotimes_{i=1}^n \mathbb{F}[x_i]/(x_i^2 - x_i)$, and applying the Chinese remainder theorem in each factor gives $F[x]/(x^2 - x) \cong F^2$ and hence
$$F[x_1, \dots x_n]/(x_i^2 - x_i) \cong \bigotimes_{i=1}^n F^2 \cong F^{2^n}$$
where the isomorphism is given by evaluating a polynomial $f \in F[x_1, \dots x_n]$ at the $2^n$ points with coordinates either $0$ or $1$. Every quotient of $F^{2^n}$ has the form $F^k$ for some $k \le 2^n$ and all of these have no nontrivial nilpotents.