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I'm trying to prove the following equivalence

Let $p\neq 3$, then $f(X)=X^3-1$ splits completely in $\mathbb{Z}_p$ ($p$-adic integers) iff $p\equiv1 \bmod 3$.

This is my attempt: first I noticed that $\mathbb{Z}/p\mathbb{Z}$ contains a primitive 3rd root of unity iff $p\equiv1\bmod3$.

Then if $p\equiv1\bmod3$ I have 3 different roots $x_1,x_2,x_3$ of $f(X)$ in $\mathbb{Z}/p\mathbb{Z}$ s.t. $f'(x_i)\neq0$ and by Hensel Lemma I can lift them to 3 different roots in $\mathbb{Z}_p$, so $f(X)$ splits completely.

I have a doubt on the reverse implication: if $f(X)$ splits completely in $\mathbb{Z}_p$ then it splits completely in $\mathbb{Z}/p\mathbb{Z}$. I want to deduce that $\mathbb{Z}/p\mathbb{Z}$ contains a primitive root of unity and I think this is true since if $p\neq 3$ then $f(X)\neq (X-1)^3$ and so if it splits is has 3 different roots. Is it true?

user26857
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Oromis
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2 Answers2

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Suppose that $\omega$ is a non-trivial cube root of unity in $\Bbb Z_p$. If $p\equiv2\pmod3$ then $\omega\equiv1\pmod p$. But this is impossible. For some $n$, $\omega\equiv1+rp^n\pmod{p^{n+1}}$ with $p\nmid r$. Then $\omega^3\equiv 1+3rp^n\pmod{p^{n+1}}$ and so $\omega^3\ne1$.

Angina Seng
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It is more fun to use some arithmetic to show that the polynomial $X^3 -1$ splits completely in $\mathbf F_p$ iff $p \equiv 1$ mod $3$. Oviously $X^3 - 1$ can be replace by $X^2 + X + 1$, whose discriminant is $-3$, so the problem is equivalent to the computation of the quadratic residue symbol $(\frac {-3}p)$ . The case $p=2$ is obvious. For $p$ odd, the quadratic reciprocity law implies : $(\frac {-3}p)=(\frac {-1}p)(\frac {3}p)=(-1)^{\frac {2(p-1)}4}(\frac {-1}p)(\frac p 3)$ . In the extreme RHS, the first two symbols are the same by Euler's criterion, so finally, $-3$ is a square in $\mathbf F_p$ iff $(\frac p 3)=1$, i.e. $p \equiv 1$ mod $3$ (again by Euler's criterion).

nguyen quang do
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