realizable over A simple example: cyclic groups It is possible, using classical results, to construct explicitly a polynomial whose Galois group over is the cyclic group for any positive integer . To do this, choose a prime such that ; this is possible by Dirichlet's theorem. Let be the cyclotomic extension of generated by , where is a primitive -th root of unity; the Galois group of is cyclic of order . Since divides , the Galois group has a cyclic subgroup of order . The fundamental theorem of Galois theory implies that the corresponding fixed field, , has Galois group over . By taking appropriate sums of conjugates of , following the construction of Gaussian periods, one can find an element of that generates over and compute its minimal polynomial. This method can be extended to cover all finite abelian groups, since every such group appears in fact as a quotient of the Galois group of some cyclotomic extension of . (This statement should not though be confused with the Kronecker–Weber theorem, which lies significantly deeper.) Worked example: the cyclic group of order three For , we may take . Then is cyclic of order six. Let us take the generator of this group which sends to . We are interested in the subgroup } of order two. Consider the element . By construction, is fixed by , and only has three conjugates over : , , . Using the identity: , one finds that , , . Therefore is a root of the polynomial , which consequently has Galois group over . Symmetric and alternating groups Hilbert showed that all symmetric and alternating groups are represented as Galois groups of polynomials with rational coefficients. The polynomial has discriminant We take the special case . Substituting a prime integer for in gives a polynomial (called a specialization of ) that by Eisenstein's criterion is irreducible. Then must be irreducible over . Furthermore, can be written and can be factored to: whose second factor is irreducible (but not by Eisenstein's criterion). Only the reciprocal polynomial is irreducible by Eisenstein's criterion. We have now shown that the group is doubly transitive. We can then find that this Galois group has a transposition. Use the scaling to get and with we arrive at: which can be arranged to . Then has as a double zero and its other zeros are simple, and