Who can give me an example of a field extension that has the Galois group isomorphic with Z_3, Z_p (p=prime number), S_p?
In general any finite group can be made isomorphic to a Galois group. However, you might be interested in finding the inverse Galois problem for $\displaystyle \mathbb{Z}_p$ and $\displaystyle S_p$. Here is a start. If $\displaystyle f(x)$ is an irreducible polynomial over $\displaystyle \mathbb{Q}$ with prime degree such that it has exactly two non-real zeros then form the splitting field $\displaystyle F$. We argue that the group $\displaystyle \mbox{Gal}(F/\mathbb{Q})\simeq S_p$. First, note that $\displaystyle \mbox{Gal}(F/\mathbb{Q})$ is a subgroup of the permutation group of the zeros. Second, let $\displaystyle \theta$ be complex-conjugation, then $\displaystyle \theta \in \mbox{Gal}(F/K)$ and it leaves all real zeros intact, therefore, $\displaystyle \theta$ can be viewed as a transposition. Third, let $\displaystyle \phi$ be an automorphism which premutes the zeros in a cyclic fashion then $\displaystyle \phi$ can be regarded as a cycle of length $\displaystyle p$. Thus, we have that $\displaystyle \text{Gal}(F/\mathbb{Q})\leq S_p$ has a cycle of length $\displaystyle p$ and a transpotion, thus in fact, $\displaystyle \text{Gal}(F/\mathbb{Q}) = S_p$.
(The link provides how to get $\displaystyle \mathbb{Z}_n$ as an inverse Galois problem).
This is Mine 97th Post!!!
you should know that every finite extension of a finite field is Galois. so, for example, if $\displaystyle q=2^n,$ then $\displaystyle \text{Gal}(\mathbb{F}_q/\mathbb{F}_2) \simeq \mathbb{Z}_n.$
to answer your last question, i just answered the question for case p = 2, because $\displaystyle S_2 \simeq \mathbb{Z}_2,$ or another example for this
case is $\displaystyle \mathbb{Q}(\sqrt{2})/\mathbb{Q}.$ so i will assume, from now on, that $\displaystyle p > 2.$
recall that if $\displaystyle f \in \mathbb{Q}[x]$ is irreducible and $\displaystyle \deg f = p,$ where p is prime, and if $\displaystyle f$ has exactly two non-real roots, then the
Galois group of $\displaystyle f$ is $\displaystyle S_p.$ so we just need to find an example of such polynomials. here's one example:
let $\displaystyle f_{\alpha}(x)=x^p + \alpha p^2(x -1)(x-2) \ ... \ (x - p + 2) - p,$ with the integer $\displaystyle \alpha > 0$ to be determined! obviously, by Eisenstein
lemma, $\displaystyle f_{\alpha}$ is irreducible. now let $\displaystyle x_j, \ 1 \leq j \leq p,$ be the roots of $\displaystyle f_{\alpha}.$ then $\displaystyle \sum x_j^2 = -2 \alpha p^2 < 0,$ and thus $\displaystyle f_{\alpha}$ has at least
two non-real roots. so the only thing we need to show now is that there exists $\displaystyle \alpha > 0$ such that $\displaystyle f_{\alpha}$ has at least $\displaystyle p-2$
real roots. to prove this, first see that $\displaystyle f_{\alpha}(1) < 0$ and $\displaystyle f_{\alpha}(2) > 0.$ so $\displaystyle f_{\alpha}$ has a root in the interval (1,2), which proves the
claim for p = 3. so i will assume that p > 3. now let $\displaystyle 1 \leq j \leq \frac{p-3}{2},$ and fix $\displaystyle 2j < t_j < 2j+1.$ choose $\displaystyle \alpha_j > 0$ large enough
so that $\displaystyle f_{\alpha}(t_j) < 0, \ \forall \alpha \geq \alpha_j.$ (why is it possible?) thus if $\displaystyle \alpha \geq \alpha_j,$ then $\displaystyle f_{\alpha}$ has at least 2 roots in the interval $\displaystyle (2j, 2j+1),$
since $\displaystyle f_{\alpha}(2j) > 0, \ f_{\alpha}(2j+1) > 0,$ and $\displaystyle f_{\alpha}(t_j) < 0.$ now let $\displaystyle \alpha \geq \max _j \alpha_j.$ then $\displaystyle f_{\alpha}$ will have at least $\displaystyle p-2$ real roots. Q.E.D.
Here is a way to show any (finite) group can be thought of as a Galois group. Let $\displaystyle F$ be a field and $\displaystyle E=F(x_1,...,x_n)$ be the rational function field in $\displaystyle n$ variables. Let $\displaystyle K = F(f_1,...,f_n)$ where $\displaystyle f_1,...,f_n$ are the elementary symettric polynomials. Now $\displaystyle \mbox{Gal}(E/K)\simeq S_n$ because any permutation of $\displaystyle x_1,...,x_n$ will leave $\displaystyle K$ fixed. By Cayley's theorem every group can be regarded as a subgroup of the symettric group. Using the fundamental theorem we simply pick the appropriate subgroup $\displaystyle H$ of $\displaystyle S_n$ isomorphic to $\displaystyle G$ and take the corresponding fixed field, i.e. take $\displaystyle E^H$ and form the group $\displaystyle N = \text{Gal}(F/E^H)$ then $\displaystyle N\simeq G$.
Here is another way to get $\displaystyle \mathbb{Z}_n$. Let $\displaystyle F=\mathbb{C}(t)$ the rational function field. Let $\displaystyle x^n - t\in F[x]$ then $\displaystyle x^n - t$ is irreducible by Eisenstein (since $\displaystyle \mathbb{C}(t)$ is a quotient field of $\displaystyle \mathbb{C}[t]$). Let $\displaystyle \alpha$ solve the equation $\displaystyle x^n - t = 0$ then $\displaystyle \alpha , \zeta \alpha , ... , \zeta^{n-1}\alpha$ all solve this equation where $\displaystyle \zeta = e^{2\pi i/n}$. Thus, $\displaystyle K = F(\alpha)$ and furthermore $\displaystyle \mbox{Gal}(K/F) \simeq \mathbb{Z}_n$ since we can map $\displaystyle \alpha \mapsto \zeta^k \alpha$ and so $\displaystyle \alpha$ generates this group with order $\displaystyle n$.