Is $p$ irreducible in $\Bbb Z[(-1 + i\sqrt{3})/2]$ if $p \equiv 2\pmod{3}$?

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Here is this week's POTW:

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Let $p$ be a prime integer. Prove $p$ is irreducible in $\Bbb Z[(-1 + i\sqrt{3})/2]$ if and only if $p \equiv 2\pmod{3}$.
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No one answered this week's problem. You can read my solution below.

Spoiler

Let $d = (-1 + i\sqrt{3})/2$. The isomorphism $$\Bbb Z[d]/p\Bbb Z[d] \approx (\Bbb Z/p\Bbb Z)[x]/\langle x^2 + x + 1 \rangle$$ induced by the ring morphism $f : \Bbb Z[d] \to (\Bbb Z/p\Bbb Z)[x]/\langle x^2 + x + 1\rangle$, $f(a + bd) = [a]_p + _px + \langle x^2 + x + 1\rangle$ shows that $p$ is irreducible in $\Bbb Z[d]$ if and only if $x^2 + x + 1$ is irreducible in $(\Bbb Z/p\Bbb Z)[x]$. Indeed, since $\Bbb Z[d]$ is a PID, $p$ is irreducible in $\Bbb Z[d]$ if and only if $p\Bbb Z[d]$ is maximal in $\Bbb Z[p]$, which occurs if and only if $\Bbb Z[d]/p\Bbb Z[d]$ is a field. The isomorphism above then gives the desired equivalence.

Now it suffices to show that $x^2 + x + 1$ is irreducible in $\Bbb Z/p\Bbb Z[x]$ if and only if $p \equiv 2\pmod{3}$. The result is clear when $p = 2$, and over $\Bbb Z_3$, $x^2 + x + 1 = x^2 - 2x + 1 = (x - 1)^2$ is reducible. Now consider $p > 3$. We show that $x^2 + x + 1$ is reducible in $\Bbb Z/p\Bbb Z[x]$ if and only if $p \equiv 1\pmod{3}$. Due to the factorization $x^3 - 1 = (x - 1)(x^2 + x + 1)$ and the fact that $[1]_p$ is not a zero of $x^2 + x + 1$ when $p > 3$, we have that $x^2 + x + 1$ is reducible over $\Bbb Z/p\Bbb Z$ if and only if it has a zero of $x^3 - 1$ that is not equal to $[1]_p$. But then the zeros of $x^2 + x + 1$ give elements of order $3$ in $(\Bbb Z/p\Bbb Z)^*$, which occurs if and only if $3\, |\, (p - 1)$, i.e., $p \equiv 1\pmod{3}$.