Three basic questions about class field theory over $\mathbb{Q}$, splitting versus merely having a zero, and finite fields

Question

(Questions 1, 2 and 3 have been answered, but not the extension of 2 to all finite fields. This is more or less formal, and easy to check given the other statements here.)

I've been reading about class field theory specifically to help me answer a question in finite group theory, so I'm trying to pin down whether what I think is correct actually is, and get at very specific statements. Furthermore, I have no local algebraic number theorist (meaning close to me, not local fields) to pester with questions, hence am turning to the Internet.

Let $f$ be an irreducible polynomial with coefficients in $\mathbb Z$. Let $\mathrm{Spl}(f)$ denote the set of unramified primes for which the reduction of $f$ modulo $p$ splits, and let $\mathrm{HasaRoot}(f)$ denote the set of unramified primes for which the reduction of $f$ modulo $p$ has a zero.

1. Suppose that $f$ has abelian Galois group, let $m$ be a defining modulus of $f$. Then $\mathrm{Spl}(f)$ consists of the primes congruent modulo $m$ to $i$ for some subset $I$ of the set $\{0,\dots,m-1\}$. Is it also true that if $q$ is a prime power then $f$ splits over $\mathbb{F}_q$ if and only if $q\bmod m$ lies in $I$? (i.e., can I extend the statement about splitting to all finite fields, rather than just prime fields?) Answer 'yes', see Mindlack's answer below.

2. Suppose that $f$ has abelian Galois group (and is a Galois extension). Is $\mathrm{Spl}(f)$ equal to $\mathrm{HasaRoot}(f)$? If so, does this also hold for all finite fields, as with 1? Answer to the first part is 'Yes', see KCd's answer.

3. If 2. is true, does the statement $\mathrm{Spl}(f)=\mathrm{HasaRoot}(f)$ hold if and only if $f$ has abelian Galois group? Answer is 'No', the condition is Galois, not abelian. See KCd's answer.

My experiments with a few cases suggest that all of these are true, but 1. is especially important for applications to matrix groups, which occur for all finite fields, not just prime fields. 2. would be really useful as it would solve a difficult question about choices of roots of unity when reducing representations of finite groups from rings of integers to finite fields. 3. is just for my education.

Answer 1

Edit: Here’s a simpler argument for question 1. I’m still assuming the same things (that is, $p$ is unramified for $f$, does not divide $m$, and it does not divide the leading coefficient of $f$). I’m showing a result essentially similar but easier to formulate, namely that whether $f$ splits or not in $\mathbb{F}_q$ depends only on the class of $q$ mod $m$.

Let $K$ be the splitting field of $f$ (as $K/\mathbb{Q}$ is abelian, every subextension is Galois, and thus $K$ is generated by any root of $f$) $L=\mathbb{Q}(e^{2i\pi/m’})$ where $m’$ is the lcm of $m$ and $q-1$.

Let $\mathfrak{p}$ be some prime ideal of $L$ above $p$, $\mathcal{O}_{\mathfrak{p}}$ be the local ring and $k$ be the residual field. Write $f(x)=c(x-x_1)\ldots (x-x_n)$, where $x_i \in \mathcal{O}_{\mathfrak{p}} \cap K$, $c \in \mathbb{Z}_{(p)}^{\times}$.

So $f$ splits in $\mathbb{F}_q$ iff the image of each $x_i$ in $k$ is in $\mathbb{F}_q$, iff for each $1 \leq i \leq n$, the images in $k$ of $x_i$ and $x_i^q$ are the same. Write $q=p^r$, let $\psi \in Gal(L/\mathbb{Q})$ be the Frobenius at $p$, then $f$ splits in $\mathbb{F}_q$ iff for each $1 \leq i \leq n$, the images in $k$ of $x_i$ and $\psi^r(x_i)$ are the same (by definition of $\psi$).

Now, if $1 \leq i \leq n$, $\psi^r(x_i)$ is another root of $f$, hence some $x_j$; $f$ is separable mod $p$ so the images of the $x_i$ in $k$ are pairwise distinct. Therefore, $f$ splits in $\mathbb{F}_q$ iff $\psi^r$ has all the $x_i$ as fixed points, iff $\psi^r$ is in $Gal(L/K)$.

But what is $\psi^r$ in $Gal(L/\mathbb{Q})$ depends only on $\psi^r(e^{2i\pi/m’})=e^{2i\pi p^r/m’}=e^{2i\pi q /m’}$. But $e^{2i\pi q/m’}$ depends only on $q$ mod $m’$. By the Chinese remainder theorem, what $q$ mod $m’$ is depends only on what $q$ mod $m$ is. QED.

---

Original answer: For question 1, I think the answer is yes as long as $p$ does not divide $m$.

Let $p$ be an unramified prime for $f$, and $q=p^r$. Let $K$ be the splitting field of $f$, so that $K \subset K_0=\mathbb{Q}(e^{2i\pi/m})$.

Then $f$ splits in $\mathbb{F}_q$ iff the residual field of $K$ at a prime over $p$ is contained in $\mathbb{F}_q$. This occurs iff the residual degree at $p$ of $K/\mathbb{Q}$ divides $r$. As $K/\mathbb{Q}$ is unramified at $p$, said residual degree $\delta$ is the cardinality of the decomposition subgroup at $p$ of $Gal(K/\mathbb{Q})$, thus the order of the Frobenius at $p$ in $Gal(K/\mathbb{Q})$.

Thus $\delta$ is the smallest positive integer such that $Frob_p^{\delta} \in Gal(K_0/K)$ (where this Galois group is seen as a subgroup of $Gal(K_0/\mathbb{Q})$ and we are speaking of a Frobenius above $\mathbb{Q}$). But under the identification $Gal(K_0/\mathbb{Q}) \cong (\mathbb{Z}/(m))^{\times}$, $Frob_{\ell}$ is mapped to $\ell$ and $Gal(K_0/K)$ is mapped to the image $H$ of $I$ in $(\mathbb{Z}/(m))^{\times}$, which is thus a subgroup. Therefore, $f$ splits in $\mathbb{F}_q$ iff $r$ is a multiple of the smallest integer $\delta>0$ such that $p^{\delta} \in H$, iff ($H$ is a subgroup) $q \in H$ iff $q \text{ mod } m \in I$.

Answer 2

I will address question $3$.

Let $K = \mathbf Q(\alpha)$ where $\alpha$ is a root of $f(x)$ and $K'$ be the splitting field of $f(x)$ over $\mathbf Q$ (equivalently, $K'$ is the Galois closure of $K$ over $\mathbf Q$). Let $G = {\rm Gal}(K'/\mathbf Q)$ and $H = {\rm Gal}(K'/K)$. A prime number is unramified in $K$ and only if it is unramified in $K'$ and it splits completely in $K$ if and only if it splits completely in $K'$. The set of primes that split completely in $K'$ has natural density $1/|G|= 1/[K':\mathbf Q]$ (special case of the Chebotarev density theorem). The set of primes that are unramified in $K$ and have at least one prime ideal factor in $\mathcal O_K$ of residue field degree $1$ has natural density $|\bigcup_{g \in G} gHg^{-1}|/|G|$.

The primes $p$ not dividing ${\rm disc}(f)$ (this excludes just a finite set of primes) are unramified in $K$ and thus also in $K'$, and the way $f(x) \bmod p$ factors into irreducible parts (number of monic irreducible factors and their degrees, necessarily all factors have multiplicity $1$ since $f(x) \bmod p$ is separable from $p \nmid {\rm disc}(f)$) matches the way $p\mathcal O_K$ factors into prime ideals in $\mathcal O_K$. In particular, we can interpret the density formulas for prime ideal factorizations in the previous paragraph as density formulas for the way $f(x) \bmod p$ factors over $\mathbf F_p$: the set of $p$ such that $f(x) \bmod p$ splits completely has natural density $1/|G| = 1/[K':\mathbf Q]$ and the set of $p$ not dividing ${\rm disc}(f)$ such that $f(x) \bmod p$ has a root in $\mathbf F_p$ has natural density $|\bigcup_{g \in G} gHg^{-1}|/|G|$. So in your notation, ${\rm Spl}(f)$ has natural density $1/|G|$ and ${\rm HasaRoot}(f)$ has natural density $|\bigcup_{g \in G} gHg^{-1}|/|G|$.

In particular, if ${\rm Spl}(f) = {\rm HasaRoot}(f)$ then $1/|G| = |\bigcup_{g \in G} gHg^{-1}|/|G|$, so $|\bigcup_{g \in G} gHg^{-1}| = 1$. That is equivalent to saying $H$ is trivial, which is equivalent to $K' = K$, which is equivalent to saying $K/\mathbf Q$ is Galois. Conversely, if $K/\mathbf Q$ is Galois then of course $K' = K$, $H$ is trivial, and ${\rm Spl}(f) = {\rm HasaRoot}(f)$ (in a Galois extension, all residue field degrees at prime ideals over a prime are equal). Thus ${\rm Spl}(f) = {\rm HasaRoot}(f)$ (even up to an a priori finite set of exceptions) if and only if $K/\mathbf Q$ is Galois, meaning $f(x)$ is normal. It is not necessary that $f(x)$ have an abelian Galois group. What we need (necessary and sufficient) is that $\mathbf Q(\alpha)$ is Galois over $\mathbf Q$. If $f(x)$ has an abelian Galois group then all subfields of $K'$ are Galois over $\mathbf Q$, so $K' = K$, but you can have $K' = K$ without ${\rm Gal}(K'/\mathbf Q)$ being abelian. For example, $\mathbf Q(\sqrt[4]{2},i)$ is Galois over $\mathbf Q$ with Galois group $D_4$ (dihedral group of order $8$) and a primitive element for the extension is $\sqrt[4]{2}+i$, which has minimal polynomial $f(x) = x^8 + 4x^6 + 2x^4 + 28x^2 + 1$. For this $f(x)$, ${\rm Spl}(f) = {\rm HasaRoot}(f)$ but the Galois group of $f(x)$ over $\mathbf Q$ is not abelian.