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?
inverse Galois problem for and . Here is a start. If is an irreducible polynomial over with prime degree such that it has exactly two non-real zeros then form the splitting field . We argue that the group . First, note that is a subgroup of the permutation group of the zeros. Second, let be complex-conjugation, then and it leaves all real zeros intact, therefore, can be viewed as a transposition. Third, let be an automorphism which premutes the zeros in a cyclic fashion then can be regarded as a cycle of length . Thus, we have that has a cycle of length and a transpotion, thus in fact, .
(The link provides how to get as an inverse Galois problem).
This is Mine 97th Post!!!
to answer your last question, i just answered the question for case p = 2, because or another example for this
case is so i will assume, from now on, that
recall that if is irreducible and where p is prime, and if has exactly two non-real roots, then the
Galois group of is so we just need to find an example of such polynomials. here's one example:
let with the integer to be determined! obviously, by Eisenstein
lemma, is irreducible. now let be the roots of then and thus has at least
two non-real roots. so the only thing we need to show now is that there exists such that has at least
real roots. to prove this, first see that and so has a root in the interval (1,2), which proves the
claim for p = 3. so i will assume that p > 3. now let and fix choose large enough
so that (why is it possible?) thus if then has at least 2 roots in the interval
since and now let then will have at least real roots. Q.E.D.
Here is a way to show any (finite) group can be thought of as a Galois group. Let be a field and be the rational function field in variables. Let where are the elementary symettric polynomials. Now because any permutation of will leave 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 of isomorphic to and take the corresponding fixed field, i.e. take and form the group then .
Here is another way to get . Let the rational function field. Let then is irreducible by Eisenstein (since is a quotient field of ). Let solve the equation then all solve this equation where . Thus, and furthermore since we can map and so generates this group with order .