How many elements are in $V((x^3-1,p))\subset\mathrm{Spec}(\mathbb{Z}[x])$?

Question

I am trying to solve the following problem:

Consider $\mathrm{Spec}(\mathbb{Z}[x])$. List the first four primes $p$ such that $$|V((x^3-1))\cap V((p))|=3.$$

However, I doubt there exist such prime numbers. It is well-known that the prime ideal in $\mathbb{Z}[x]$ is one of the followings

1. $(0)$.

2. $(f(x))$, where $f(x)$ is an irreducible polynomial.

3. $(p)$, where $p$ is a prime number.

4. $(p, f(x))$, where $p$ is a prime number and $f(x)$ is an irreducible polynomial in $\mathbb{Z}_p[x]$.

So $V((x^3-1))\cap V((p))=V((x^3-1,p))$ should only have two elements: the prime ideals $$(x-1,p)\quad \quad (x^2+x+1,p)$$ Where is the third prime ideal? I am really confused.

Answer 1

Only prime ideal of type 4 in your list can contain the ideal $(x^3-1,p)$. So in order to to find three such prime ideals you must find three distinct irreducible factors of $x^3-1$ modulo $p$. In that case the said factors are necessarily linear, so $|V((p,x^3-1))|=3$ is equivalent to $x^3-1$ having three pairwise non-congruent zeros modulo $p$.

Claim. This happens if and only if $p\equiv1\pmod3$.

If $a\in \Bbb{F}_p^*$ is a solution of $a^3=1$, then either $a=1$ (one solutions), or $a$ has order three. By Lagrange's theorem in the latter case we must have $3\mid p-1$. On the other hand, if $3\mid p-1$ then, by cyclicity of $\Bbb{F}_p^*$ there exists and element $a$ of order three. In that case $1,a,a^2$ are all solutions of $x^3\equiv1$. As $a\neq1$ they are pairwise distinct.

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For example when $p=7$ we see that $1,2,4$ are all solutions of $x^3\equiv1\pmod7$. These yield the three prime ideals $(x-1,7)$, $(x-2,7)$ and $(x-4,7)$ all containing the polynomial $x^3-1$.

Answer 2

The third element is $(p)$.

On the other hand, $V\bigl((x^3-1,p)\bigr)$ has $3$ elements if $x^2+x+1$ is irreducible in $\mathbf Z/p\mathbf Z$, i.e. if and only if its discriminant $-3$ is not a square mod. $p$.

Now, if $p=2$, $x^2+x+1$ is irreducible, so $2$ is the first prime in the list.

Suppose $p$ is odd. By the law of quadratic reciprocity, we have $$\biggl(\frac{-3}p\biggr)=\biggl(\frac{-1}p\biggr)\biggl(\frac{3}p\biggr)=\bigl(-1\bigr)^{\tfrac{p-1}2}\cdot\biggl(\frac p{3}\biggr)\bigl(-1\bigr)^{\tfrac{p-1}2}=\biggl(\frac p{3}\biggr),$$ which shows $x^2+x+1$ is irreducible mod. $p$ if and only if $p$ is not a square mod. $3$, i.e. if $$p\equiv 2\mod 3.$$