Let $p$ be a prime number and $$I:=(x_1+x_2+x_3+x_4+x_1 x_2+x_1 x_3+x_1 x_4+x_2 x_3+x_2 x_4+x_3 x_4,x_1 x_2 x_3+x_1 x_2 x_4+x_1 x_3 x_4+x_2 x_3 x_4+x_1 x_2 x_3 x_4)$$ be an ideal of $F_p[x_1,x_2,x_3,x_4]$. Is this true $I$ is a prime ideal for any $p$?
I know this is a prime ideal of ring $\mathbb{Z}[x_1,x_2,x_3,x_4]$. I use the Macaulay2 to test the small prime number (say, the prime number less than 100), they are all prime ideal. But how can we show this is prime for any $p$?
The goal is to show that, in $k[a,b,c,d]$ (for any field $k$), the ideal $I$ generated by $\sigma_1+\sigma_2$ and $\sigma_3+\sigma_4$ is prime (where $\sigma_i$ is the elementary symmetric polynomial of degree $i$).
Note that $I$ is generated by $(1+a)(1+b)(1+c)(1+d)-1$ and by $(d+1)(a+b+c+1)+ab+bc+ca-1$.
So $k[a,b,c,d]/I$ is isomorphic to $(k[a,b,c]/(P))[d]/((1+d)(1+c)(1+b)(1+a)-1),$ where $$P=a+b+c+1+(1+a)(1+b)(1+c)(ab+bc+ca-1).$$
Thus, we only need to prove that $P \in k[a,b,c]$ is irreducible.
Note that $P$ has degree $2$ with respect to $a$ and is symmetric. Moreover, the leading and constant coefficients of $P$ (wrt $a$) are $(1+b)(1+c)(b+c)$ and $(1+b)(1+c)(bc-1)+b+c+1=bc(b+c+bc)$, so they are coprime in $k[b,c]$.
Thus, if $P$ is reducible, then we can write $P=QR$, where $Q$ and $R$ have degree one with respect to $a$ and are irreducible. The same reasoning shows that $Q,R$ must have degree one with respect to $b$ and $c$ as well.
Since $P$ has total degree $5$, $Q$ and $R$ cannot have the same total degree. Since permuting $a,b,c$ leaves $P$ invariant, it must preserve (up to scalar) the decomposition in irreducibles. Since this action also preserves the total degree, $Q$ and $R$ are $S_3$-invariant (up to scalar).
Since $S_3^{ab}=\{\pm 1\}$, $Q$ and $R$ are at least cyclic. One of them has degree at most two, so it is symmetric, hence $Q,R$ are both symmetric.
If $Q$ has total degree one, then $R$ has total degree four, but degree one with respect to every variable, so it is impossible.
Thus $Q$ has total degree two, so it is of the form $Q(a,b,c)=(ab+bc+ca)+v(a+b+c)+w$ with $v,w \in k$ (up to scalar).
The leading coefficient of $Q$ wrt $a$ must divide $(1+b)(1+c)(b+c)$, and it is $(b+c+v)$: hence $v=0$.
The constant coefficient of $Q$ wrt $a$ must divide $bc(b+c+bc)$, but it is $bc+w$: hence $w=0$.
Therefore $ab+bc+ca \mid P$. This implies $ab+bc+ca \mid abc$, which is clearly impossible, and we are done.