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 $\mathbb{Z}_3?$ 
 A: 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 $8x^3-6x+1$.
A: 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. 
A: 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: 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. 
