complex multiplication - A problem of Shimura and its relation to class field theory

I would like to write some kind of summary of the above two answers. There is nothing new here.

Consider $mathbb{Q} subset mathbb{Q}(zeta_7)$,(and we fix $zeta_7=e^{frac{2pi i}{7}}$) this is an abelian Galois extension, and the Galois group is $(Z/7Z)^{times}$. Frobenius element over $p$ for $p neq 7$ (7 is the ramification) acts on $zeta_7$ by sending it to its $p$-th power.

Now consider $alpha=zeta_7+zeta_7^{-1}$, which is in $mathbb{Q}(zeta_7)$, consider all its Galois conjugates, which are precisely $alpha_2 = zeta^2 +zeta^{-2}$, $alpha_3=zeta^3+zeta^{-3}$. So we have $mathbb{Q} subset mathbb{Q}(alpha)$ is Galois.

Now $Frob_p$ maps $alpha$ to $zeta^p + zeta^{-p}$, which is equal to $alpha$ if and only if $p equiv 1, -1 (mod 7)$. And they are precisely those primes that are totally split in $mathbb{Q}(alpha)$. For those primes, $Z/p = mathbb{O}_{mathbb{Q}(alpha)}/p$, thus we can always find some $n$, s.t., $alpha_i equiv n(mod p)$, which is equivalent to say that $F(x)$ totally split over $Z/p =F_p$.

I am glad to know this problem and answer since I finally found an explicit example of cyclic Galois extension of $mathbb{Q}$...(which I had been wondering for a while...)

We can also do the similar things for p=13. just take $beta=theta +theta^5+theta^8+theta^{12}$,where $theta$ is the 13-th root of unity, then $Q(beta)$ is again a cyclic Galois extension of $Q$.

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when I said cyclic above, I meant for cyclic of order 3. Thank Peter for pointing out.