TITLE: Why is $x^4-3x^2+18$ irreducible over the $3$-adics?
QUESTION [6 upvotes]: Let $f = x^4 - 3x^2 + 18 \in \mathbb{Q}_3[x]$. Since there is an LMFDB page of an extension defined by this polynomial, I assume that $f$ is irreducible. Could you verify if this is true and also why that is (not) the case?
It is not an Eisenstein polynomial (as $3^2$ divides $18$, the constant coefficient of $f$). I also tried the substitution $y = x^2$ and asked myself if $g = y^2 - 3y + 18$ is irreducible. Its reduction mod $3$ is $y^2$ which is not irreducible, so I cannot say whether $g$ is irreducible or not.

REPLY [3 votes]: Well, here’s an argument that uses a somewhat non-elementary fact about $p$-adic field extensions, namely that if the residue-field extension degree is $f$ and the ramification degree is $e$, then the total degree $n$ is $n=ef$. This $f$ is also the degree over the base field of the maximal unramified intermediate extension.
I’m about to show that if $\alpha$ is a root of $F(x)=x^4-3x^2+18$, then the extension $\Bbb Q_3(\alpha)\supset\Bbb Q_3$ has $f$ and $e$ both equal to $2$, so that $[\Bbb Q_3(\alpha):\Bbb Q_3]=4$. This shows that your polynomial $F$ is irreducible.
You’ve already seen that $G(x)=x^2-3x+18$ has $\alpha^2=\gamma$ for a root, and as I’ve pointed out in a comment, $\gamma/3$ is a root of $x^2-x+2=g(x)$. Use the Quadratic Formula to see that to get roots of $g$, you need $\sqrt{-7}$, and since $x^2+7\equiv x^2+1\pmod3$, adjoining $\sqrt{-7}$ is same as adjoining $i$ to $\Bbb Q_3$, giving the quadratic unramified extension of $\Bbb Q_3$. So $f=2$ for the extension $\Bbb Q_3(\gamma)\supset\Bbb Q_3$.
Since the roots of $g(x)$ are $3$-adic units (the constant term of the minimal polynomial is a unit), we have $v_3(\gamma)=1$, as @RobertIsrael has shown by different methods. This means that $v_3(\alpha)=v_3(\sqrt\gamma\,)=1/2$, which in turn means that the ramification degree $e$ of $\Bbb Q_3$ over $\Bbb Q_3$ is at least two. Thus we have $f\ge2$, $g\ge2$, and their product $n$, the degree of the minimal polynomial for $\alpha$ over $\Bbb Q_3$, is $\le4$. It follows that all three inequalities are equalities, and $F$ is irreducible.