the action of galois group in prime ideals

Question

Take $L=\mathbb Q(\omega,\sqrt[3]{5})$ the splitting field of the polynomial $f=X^3-5$, with Galois group $\text{Gal}(f/\mathbb Q)\approx S_3$.

We consider the intermediate extension $$\mathbb Q\subseteq \mathbb Q(\sqrt[3]{5})$$ of degree 3, and the algebraic integers of this extension is $\mathbb Z[\sqrt[3]{5}]$. By Kummer theorem the prime number $13$ have 3 factor distinct in $\mathbb F_{13}[X]$ because the minimum polynomial of $\sqrt[3]{5}$ satisfies $$X^3-5\equiv (X+2)(X+5)(X+6)\mod 13.$$

My conclusion is that the number of prime factors of $13$ in the ring $\mathcal O_L$ is $3$ or $6$ by the formula $\sum_ie_if_i=6$.

It is possible to know which is the number of the prime factors of 13 in $\mathcal O_L$?

Viktor Vaughn · Accepted Answer · 2025-02-02T06:39:24.747

4

By this post or this post, we have that a prime $\newcommand{\frakp}{\mathfrak{p}} \frakp$ splits completely in a compositum $KF$ iff $\frakp$ splits completely in both $K$ and $F$. Let $\newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}} K = \Q(\sqrt[3]{5})$ and $F = \Q(\omega)$. You have shown that $(13)$ splits completely in $K$. Since the splitting field $L = KF$, then it suffices to check that $(13)$ splits completely in $F$ as well.

Since $\newcommand{\calO}{\mathcal{O}} \calO_F = \Z[\omega]$ and $[F : \Q] = 2$, by Dedekind-Kummer this occurs iff there is a primitive $3^\text{rd}$ root of unity in $\Z/13\Z$, i.e., if there is an element of order $3$ in $(\Z/13\Z)^\times \cong \Z/12\Z$. Since $(\Z/13\Z)^\times$ is cyclic and $3 \mid 12 = \#(\Z/13\Z)^\times$, then there is such an element. Indeed, $3$ has order $3$ since $3^3 = 27 \equiv 1 \bmod{13}$ and no lower power gives $1$.

Thus $(13)$ splits completely in $F$, so $13$ splits completely in the compositum $KF = L$. Thus $(13)$ splits into $6$ distinct prime factors in $L$.

Also note that you can easily compute the factorizations of ideals using a computer algebra system. For instance, using Magma, the following code outputs the 6 prime factors of $(13)$.

R<x> := PolynomialRing(Rationals());
f := x^3 - 5;
L<b> := SplittingField(f);
L;
OL := Integers(L);
Factorization(13*OL);

edited Feb 02 '25 at 06:39

answered Feb 02 '25 at 06:11

Viktor Vaughn

20,897

It is possible to find a extension $\mathbb Q\subseteq \mathbb Q(\sqrt[3]{5})\subseteq L$ such that $\text{Gal}(L/\mathbb Q)\approx S_3$ and (13) have only three prime factors in $\mathcal O_L$? Thanks – Luis Antonio Sanchez Feb 02 '25 at 06:22
@LuisAntonioSanchez If you want $L$ to be Galois of degree $6$ and to contain $\mathbb{Q}(\sqrt[3]{5})$, I think that forces $L$ to be the splitting field $\mathbb{Q}(\sqrt[3]{5}, \omega)$. – Viktor Vaughn Feb 02 '25 at 06:27
yees me too.... – Luis Antonio Sanchez Feb 02 '25 at 06:28
or maybe another prime number $p\neq 13$ and a extension of degree 6 $L/\mathbb Q$ with Galois group $S_3$ such that $p$ have only three factor in $\mathcal O_L$?? This problem is important for me, any ideas please – Luis Antonio Sanchez Feb 02 '25 at 06:33
1

@LuisAntonioSanchez Sure, if you're willing to choose a different $p$, there are many possibilities. Take the same field $L$ as before, and take $p = 2$. (Or $11$, or $17$, or any prime whose Frobenius cycle type is $2^3$ at this page.) You'll see that for $p=2$, $2$ is inert in $F$, since $x^2 + x + 1$ has no roots mod $2$. – Viktor Vaughn Feb 02 '25 at 06:38
@LuisAntonioSanchez See this computation – Viktor Vaughn Feb 02 '25 at 07:00
Thank you, all is clear now – Luis Antonio Sanchez Feb 02 '25 at 07:23

the action of galois group in prime ideals

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