No prime is inert in non-cyclic Galois extension

Question

It is well-known that in a non-cyclic Galois extension $K/\mathbf{Q}$ no prime $p$ is inert.

Say $\zeta$ is a primitive $n$-th root of unity such that $(\mathbf{Z}/n\mathbf{Z})^\times$ is non-cyclic and set $K=\mathbf{Q}(\zeta)$. The minimal polynomial of $\zeta$ is the $n$-th cyclotomic polynomial $\Phi_n$. By the Kummer-Dedekind Theorem, the factorisation of $\Phi_n$ modulo $p$ determines the decomposition of $p$ in $\mathbf{Z}[\zeta]=\mathcal{O}_K$.

Now say we have a $(\mathbf{Z}/2\mathbf{Z})^2$-extension of $\mathbf{Q}$. These are of the form $K=\mathbf{Q}(\sqrt{n},\sqrt{m})$ with $n,m\in \mathbf{Z}$ such that $n,m,nm\not\in \mathbf{Q}^{\times 2}$. A primitive element is $\sqrt{n}+\sqrt{m}$, with minimal polynomial $f=x^4+(2(n-m)-4n)x^2+(n-m)^2$. I want to now conclude that such polynomials are always reducible modulo $p$.

For instance, when $n=-1$ and $m=2$, we get $f=X^4-2X^2+9$. The only prime factors of its discriminant are $2$ and $3$. Hence, the decomposition of $p$ in $K$ is the factorisation of $f$ mod $p$ by the Kummer-Dedekind Theorem for all $p\neq 2,3$. Hence, we know by the above that $f$ is reducible mod $p$ for all $p\neq 2,3$. We can treat the cases $2$ and $3$ separately, these are easy.

But in general, Qiaochu Yuan seems to use that the minimal polynomial of a primitive element (chosen integral, of course) of a non-cyclic Galois extension is reducible mod $p$ for all $p$ here.

In the cyclotomic case, the ring of integers is monogenic and generated by a primitive element of $K$. However, in general, this need not be the case. How can we then connect the decomposition of $p$ in $\mathcal{O}_K$ to the factorisation of the minimal polynomial $f^{\alpha}_{\mathbf{Q}}$ of a primitive element $\alpha$ of $K$? Since, a priori, the factorisation of $f^{\alpha}_{\mathbf{Q}}$ modulo $p$ only gives the decomposition of $p$ in $\mathcal{O}_K$ when $p$ does not divide the index $[\mathcal{O}_K:\mathbf{Z}[\alpha]]$.

There must be something I am not seeing or forgot. Any help is welcome. :)

Answer 1

The Kummer-Dedekind theorem tells you how to determine how $p$ splits in terms of the splitting of $f$ modulo $p$ only if $p\nmid [\mathcal O_K : \mathbb Z[\alpha]]$.

However, what you're after is the converse: you want to understand how $f$ splits modulo $p$ given that you already know how $p$ splits. This direction doesn't require $p\nmid [\mathcal O_K : \mathbb Z[\alpha]]$.

So fix your polynomial $f$ and a prime $p$. By assumption, $p$ cannot be inert in $K = \mathbb Q(x)/f$. So we can write $p\mathcal O_K = \mathfrak{p}_1^{a_1}\cdots \mathfrak{p}_n^{a_n}$ as a product of primes such that $n >1$ or $a_1 > 1$.

So if $\alpha$ is a root of $f$ $$\frac{\mathbb F_p[x]}{f(x)} \cong \frac{\mathbb Z[x]}{(p, f(x))} \cong \frac{\mathbb Z[\alpha]}{p}\subset \frac{\mathcal O_K}{p} = \prod_{i}\frac{\mathcal O_K}{\mathfrak p_i^{a_i}}.$$

We see that $\mathbb F_p[x]/f$ cannot be a field extension of $\deg f$. So $f$ must be reducible.

For the usual Kummer-Dedekind theorem, you'd need to know that $\frac{\mathbb Z[\alpha]}{p}=\frac{\mathcal O_K}{p}$, which is what requires $p\nmid [\mathcal O_K : \mathbb Z[\alpha]]$.