# Galois group that is isomorphic to $\mathbb{Z}_3?$

I am curious if we can construct arbitrary number fields whose Galois group over $$\mathbb{Q}$$ is $$G,$$ for an arbitrary group $$G.$$ However, I cannot do this for $$\mathbb{Z}_3.$$ I think that if the Galois group is cyclic, the group would have to permute roots of unity. However, $$3 \neq \phi(n)$$ for any $$n$$ so I don't think this is possible. Is there such an extension over $$\mathbb{Q}$$ such that its Galois group is $$\mathbb{Z}_3?$$

• The general case is known as the Inverse Galois Problem and it is still open. Hilbert settled the problem for symmetric groups (the article contains the construction). – lulu Jan 29 at 11:09
• just take $\mathbb{Q}[-\frac{1}{2} + \frac{\sqrt{3}}{2} i]$ – Enkidu Jan 29 at 11:12
• @Enkidu, that field has degree $2$. – lhf Jan 29 at 11:15
• You may well already know this, but I wanted to add that while the inverse Galois problem is open, the question of whether there is a Galois extension of some field with finite Galois group $G$ not so hard and follows from the fundamental theorem of symmetric polynomials and Cayley's embeddability theorem for finite groups (see this question). – csprun Jan 29 at 14:33
• @Mindlack The cubic $p(x)=x^3-4x+1$ is a semistandard counterexample. It has three real roots, in the intervals $(-3,-2)$, $(0,1)$ and $(1,2)$. Modulo $3$ it is irreducible, so the Galois group has a 3-cycle. Modulo $2$ it factors as $(x^2+x+1)(x-1)$ so the Galois group also has a permutation with cycle type $(2,1)$, i.e. a 2-cycle. Therefore the Galois group is $S_3$. The discriminant is the tool here. $\Delta(p)=229$ is a non-square, so the Galois group is all of $S_3$. – Jyrki Lahtonen Jan 29 at 14:39

Take an irreducible cubic polynomial with rational coefficients that has three real roots and a square discriminant. Then its splitting field has degree $$3$$ over $$\mathbb Q$$ and so is a Galois extension of $$\mathbb Q$$ with Galois group isomorphic to $$\mathbb{Z}_3$$.

One example is $$8x^3-6x+1$$.

• – lhf Jan 29 at 11:17
• I see. However, does this follow because all groups of size 3 are cyclic? That is, can we generalize this method to arbitrary cyclic groups of arbitrary size? – 伽罗瓦 Jan 29 at 11:20
• @伽罗瓦, see en.wikipedia.org/wiki/… – lhf Jan 29 at 11:21
• Very helpful. Thanks! – 伽罗瓦 Jan 29 at 11:22
• Ihf , you are partially wrong. The polynomial $x^3-4x+1$ has $3$ real distinct roots and, yet, has $S_3$ as Galois group. Your polynomial works because its discriminant is a square in $\mathbb{Q}$. – loup blanc Jan 29 at 12:07

The field generated over the rationals by $$e^{2\pi i/7}+e^{-2\pi i/7}$$ has Galois group cyclic of order 3.

A number of the other answerers of your question are certainly aware of it, but did not mention the solution to your problem that is in some sense universal.

It’s that every abelian extension of $$\Bbb Q$$ is contained in a cyclotomic field $$\Bbb Q(\zeta_n)$$, where $$\zeta_n$$ is a primitive $$n$$-th root of unity. These are the cyclotomic fields, and since $$\text{Gal}^{\Bbb Q(\zeta_n)}_{\Bbb Q}\cong(\Bbb Z/n\Bbb Z)^\times$$, a completely understood group, there is, in principle, no difficulty in constructing an extension of $$\Bbb Q$$ with any particular abelian Galois group $$G$$, and indeed in describing all extensions of $$\Bbb Q$$ with Galois group $$G$$.

By taking $$n=7$$ and $$\Bbb Q(\zeta_7)$$, Gerry Meyerson gave you the example that I most often use for offering a field of the type that you’re asking for. He could have just as well offered $$\Bbb Q(\zeta_p)$$ for any $$p\equiv1\pmod3$$, or $$\Bbb Q(\zeta_9)$$. These certainly do not exhaust the possibilities, though.

Things get to be much more fun if you look for Galois cubic extensions of such a field as $$\Bbb Q(i)$$, but that starts to get into very deep mathematics. You have opened a door into very beautiful phenomena.

Let $$p(x)=x^3+ax^2+bx+c\in \mathbb{Q}[x]$$ be the general monic polynomial. Putting $$X=x+a/3$$, we can reduce the problem to the polynomials in the form $$P(x)=x^3+px+q$$ (note that $$p$$ irreducible iff $$P$$ irreducible).

Then $$P$$ has $$C_3$$ as Galois polynomial iff

$$P$$ has no rational roots and $$P$$ satisfies $$-4p^3-27q^2=r^2$$ where $$r\in\mathbb{Q}$$.

EDIT 1. Without any calculation, we can see that the example given by @Gerry Myerson works; indeed the conjugates of $$\cos(2\pi/7)$$ are $$\cos(4\pi/7),\cos(8\pi/7)=\cos(6\pi/7)$$ and, therefore, the function $$x\mapsto 2x^2-1$$ realizes a permutation of the set of these $$3$$ roots.

EDIT 2. For the lhf's polynomial, the function $$\phi:x\mapsto 1-x-2x^2$$ realizes a permutation of the set of the roots.

Such a polynomial $$\phi\in\mathbb{Q}_2[x]$$ exists when the considered polynomial has $$3$$ real roots and has $$C_3$$ as Galois group.

Conversely, I don't know what are the conditions that the functions $$\phi$$ must fulfill; for example, $$\phi=1+2x-x^2$$ is not convenient.