[Math] Counter example of a radical extension that is not Galois/normal over $\mathbb{Q}(\omega)$

algebraic-number-theoryfield-extensionsgalois-theory

Most proofs of Galois theorem stating that "an equation is solvable in radicals if and only if its Galois group is solvable," show the left to right direction by induction on the height of the radical extension. Starting with a cyclotomic field, adjoining "enough" roots of the unity (e.g. adjoining the product of all the ) and continuing with successive simple radical extensions, which are all Galois and Abelian over the previous one (because we ensured that there were enough roots of unity to start with). The key element in proving that all these extensions are solvable over the base field is then to define a solvable extension as an extension which normal closure has solvable Galois group (equivalently such that there exist an extension which Galois group is solvable) (def (a)), this makes "being a solvable extension" transitive (it is distinguished) as proved in Lang Algebra. Doing this makes it possible to complete the proof by induction, while "being Galois and solvable" (def (b)) does not go through (since Galois extensions are not distinguished).

Many texts also state that the (radical) extensions arising in the above mentioned induction are not Galois in general. While this is indeed not the case in general (as with the traditional example $\mathbb{Q} \subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt[4]{2})$ since it's missing $i\sqrt[4]{2}$), one might wonder if this is true or not for radical extensions where the base field contains enough roots of unity.

More precisely, the question is to determine whether the following statement is true.

Let $\omega_n$ be a primitive $n^\textrm{th}$ root of the unity.

Let $x = (x_1, \ldots, x_h)$ be sequence of elements in the algebraic closure of $\mathbb{Q}$, such that $x_{i+1}^{m_{i+1}} \in \mathbb{Q}(x_1, \ldots, x_{i})$ for some minimal $m_i > 1$ for all $i < h$.

Is there a natural number $N$, such that in the radical extension series
$$
\mathbb{Q}(\omega_{N}) = K \subset K(x_1) \subset \ldots \subset K(x_1, \ldots, x_n)
$$
each $K(x_1, \ldots, x_{i+1})/K$ is Galois ?

Definition
Let's call $N_x$ the smallest of such natural numbers if it exists, and set $N_x = \infty$ if there is no such natural number.
The above question amounts to «Is $N_x$ always finite?»

In a typical proof of Galois theorem above, since the primitive root of the unity that is chosen is typically a divisor of $d_x = \prod_i m_i$ (one can also pick the lcm afaik).

If it were true that $N_x | d_x$ for all $x$ then the proof above could be carried out with def (b) instead of (a). (The answers seem to indicate this is not true though.)
However if $N_x$ is finite for all $x$, then for each $x$ one can use def (b) by starting the radical series with $\mathbb{Q}(\omega_{N_x})$ (instead of $\mathbb{Q}(\omega_{d_x})$ as in the proof above).

I conjectured the above statement is false, i.e. that there are $x$s such that $N_x > d_x$ and even such that $N_x = \infty$.

Hence I am searching for a (minimal?) counter example, i.e. a radical extension
$$
\mathbb{Q}(\omega_N) = K \subset K(x_1) \subset \ldots \subset K(x_1, \ldots, x_h)
$$
such that each for all $N$ such that each $K(x_1, \ldots, x_{i+1})/K(x_1, \ldots, x_i)$ is Galois, $K(x_1, \ldots, x_n)/K$ is not Galois.

(Another way to state it is to start with $K$ being an extension of $\mathbb{Q}$ containing all roots of unit.)

I think the following is a radical extensions of total degree $6$ over $\mathbb{Q}(\omega)$, where $\omega = \frac{\sqrt{3}}{2} + \frac{i}{2}$ is a primitive 12th root of unity and where each intermediate extension is Galois over the previous one:
$$K = \mathbb{Q}(\omega) \subset F = K\left(\sqrt[3]{2}\right) \subset K\left(\sqrt[3]{2}, \sqrt{1+ \sqrt[3]{2}}\right) = E.$$
However, I would like to claim $E$ is not Galois over $K$, even though all 12th roots of unity are in $K$.
Indeed a conjugate of $\sqrt{1+ \sqrt[3]{2}}$ over $K$ is a square root $\alpha$ of $1 + j\sqrt[3]{2}$ — their minimal polynomial is $X^6 – 3 X^4 + 3 X^2 – 3$ — but $\alpha$ would not be in $E = \textrm{span}_F\left(1, \sqrt{1+ \sqrt[3]{2}}\right).$

I reduced the problem to checking that $\frac{1 + j\sqrt[3]{2}}{1+ \sqrt[3]{2}}$ is not a square in $F$ (unless I made a mistake) and some famous online computer algebra system tells that if I compare it to $(x+y\sqrt[3]{2}+z\sqrt[3]{2}^2)^2$ then $z$ needs to be expressed using a square root of an element of $F$ that does not look like it's in $F$, but I am still not 100% sure this is correct…

Does someone has indications on better ways to find a counter example, check this one or prove the theorem that I believe is wrong in the first place? (i.e. that all radical extension over a field with enough roots of unity are Galois.)

EDIT 25/01/2021

I made my question more precise by showing the statement for which I am searching for a counter example.

Unless I am mistaken, the answers and their comments below show that both

$x_1 ^ 2 = 3, \quad x_2 ^ 2= 1 + x_1$
$y_1 ^ 3 = 2, \quad y_2 ^ 2= 1 + y_1$

are indeed counter examples of $N | d$ since

$d_x = 4$ and $N_x = 8$
$d_y = 6$ and $N_y > 12$.

They also show that $x$ is not a counter example that $N_x < \infty$, and I think the finiteness status of $N_y$ is unknown.

Best Answer

The polynomial $f:=(x^2-1)^3-2$ has Galois group $G \simeq S_4 \times C_2$ over $\mathbb{Q}$, so the commutator subgroup $[G,G] \simeq A_4$. Let $L$ be the splitting field (over $\mathbb{Q}$) of $f$. This contains the $12$-th roots of unity, i.e. contains your $K$. Since $[K:\mathbb{Q}]=4=[G:[G,G]]$, $L$ contains no other roots of unity and $\operatorname{Gal}(L/K)\simeq A_4$. Since $A_4$ has no normal subgroups of order $2$ the extension $E/K$ is not Galois.

Certifying that $G \simeq S_4 \times C_2$ could e.g. be done using (an exact version of a small part of) Algorithm 6.3.10 in Cohen's "A Course in Computational Algebraic Number Theory" or using more basic Galois resolvents, or...

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[Math] Extensions obtained adding torsion points of an elliptic curve

This question is discussed very carefully in Section 3 of the paper Mod $p$ representations on elliptic curves, by Frank Calegari (available here).

In particular, after noting that the answer is positive when $p \leq 5$ (as was already observed in the comments above), he proves (in Theorems 3.3 and 3.4) that if $p \geq 7$ then there exists a continuous representation $\rho: Gal(\overline{\mathbb Q}/ \mathbb Q) \to GL_2(\mathbb F_p)$ with cyclotomic determinant which does not come from the $p$-torsion of an elliptic curve. (He also notes that the same result was established by Dieulefait, in the paper Existence of non-elliptic mod $\ell$ Galois representations for every $\ell > 5$.)

Solvable Octic Equation – Analysis of x^8-x^7+29x^2+29=0

To answer your second question, there are soluble octics with the same Galois group involving other $p$th roots of unity. Take $p\equiv 1\mod 7$, and $K={\mathbb Q}(\alpha)$ the unique degree 7 extension of ${\mathbb Q}$ in ${\mathbb Q}(\zeta_p)$. E.g. take $\alpha=\sum_i \zeta_p^i$ where $i$ ranges over all seventh powers in ${\mathbb F_p}$.

If $f(x)$ is the minimal polynomial of $\alpha$ (degree 7), then $f(x^2)$ is the minimal polynomial of $\sqrt\alpha$, which defines, generally, a 'random' quadratic extension of $K$. That is, its Galois group over ${\mathbb Q}$ is $$ G=C_2\wr C_7\cong C_2^7:C_7. $$ Viewing $C_2^7$ as a 7-dimensional representation of $C_7$ over ${\mathbb F}_2$, it decomposes as a 1-dimensional (trivial) representation plus two distinct 3-dimensional ones. (The reason for this is that $2^3\equiv 1\mod 7$.) Factor out $C_2^4\triangleleft G$, which is one of those plus the trivial one. This gives a Galois group $C_2^3:C_7$ that you are after, and a subgroup $C_7$ in it cuts out the required octic field.

Here is a Magma code that can be used in the Magma online calculator to get such an octic:

p:=43;        // or some other p = 1 mod 7

K<z>:=CyclotomicField(p);
alpha:=&+[z^i: i in [1..p] | IsPower(GF(p)!i,7)];    

R<x>:=PolynomialRing(Rationals());
f:=Evaluate(MinimalPolynomial(alpha),x^2);
K:=NumberField(f);
assert exists(a){a: a in ArtinRepresentations(K) | #Kernel(Character(a)) eq 16};
F:=Field(Minimize(a));
DefiningPolynomial(F);

You can also stick in a Tschirnhaus transformation, say,

alpha:=alpha^3+alpha+1;

in the 5th line to vary the generator of $K$ - in this way you get all possible $C_2^3:C_7$-extensions involving $p$th roots of unity.

For your questions in the comments, the roots might be real or complex, and the constant term may or may not be a square - this depends on whether $\alpha$ is chosen to be totally real, and on the way 'Minimize' works; you can always use an additional Tschirnhaus transformation to modify the final output or Pari's 'polredabs' function to try and get the coefficients smaller.

I do not know the reason why for $p=29$ there is such an elegant octic, this is very curious. It is a bit like the Trinks polynomial $x^7-7x+3$ with Galois group PSL(2,7), and I wonder whether simple polynomials having interesting Galois group is such a statistical blip, or there is a reason behind it.

Best Answer

Related Solutions

[Math] Extensions obtained adding torsion points of an elliptic curve

Solvable Octic Equation – Analysis of x^8-x^7+29x^2+29=0

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