Galois group

**Veve** · May 17th 2008, 11:57 AM

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?

**ThePerfectHacker** · May 17th 2008, 05:26 PM

Originally Posted by Veve

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 $\mathbb{Z}_p$ and $S_p$ . Here is a start. If $f(x)$ is an irreducible polynomial over $\mathbb{Q}$ with prime degree such that it has exactly two non-real zeros then form the splitting field $F$ . We argue that the group $\mbox{Gal}(F/\mathbb{Q})\simeq S_p$ . First, note that $\mbox{Gal}(F/\mathbb{Q})$ is a subgroup of the permutation group of the zeros. Second, let $\theta$ be complex-conjugation, then $\theta \in \mbox{Gal}(F/K)$ and it leaves all real zeros intact, therefore, $\theta$ can be viewed as a transposition. Third, let $\phi$ be an automorphism which premutes the zeros in a cyclic fashion then $\phi$ can be regarded as a cycle of length $p$ . Thus, we have that $\text{Gal}(F/\mathbb{Q})\leq S_p$ has a cycle of length $p$ and a transpotion, thus in fact, $\text{Gal}(F/\mathbb{Q}) = S_p$ .

(The link provides how to get $\mathbb{Z}_n$ as an inverse Galois problem).

This is Mine 97

th Post!!!

**NonCommAlg** · May 17th 2008, 06:43 PM

Originally Posted by Veve

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?

you should know that every finite extension of a finite field is Galois. so, for example, if $q=2^n,$ then $\text{Gal}(\mathbb{F}_q/\mathbb{F}_2) \simeq \mathbb{Z}_n.$

to answer your last question, i just answered the question for case p = 2, because $S_2 \simeq \mathbb{Z}_2,$ or another example for this

case is $\mathbb{Q}(\sqrt{2})/\mathbb{Q}.$ so i will assume, from now on, that $p > 2.$

recall that if $f \in \mathbb{Q}[x]$ is irreducible and $\deg f = p,$ where p is prime, and if $f$ has exactly two non-real roots, then the

Galois group of $f$ is $S_p.$ so we just need to find an example of such polynomials. here's one example:

let $f_{\alpha}(x)=x^p + \alpha p^2(x -1)(x-2) \ ... \ (x - p + 2) - p,$ with the integer $\alpha > 0$ to be determined! obviously, by Eisenstein

lemma, $f_{\alpha}$ is irreducible. now let $x_j, \ 1 \leq j \leq p,$ be the roots of $f_{\alpha}.$ then $\sum x_j^2 = -2 \alpha p^2 < 0,$ and thus $f_{\alpha}$ has at least

two non-real roots. so the only thing we need to show now is that there exists $\alpha > 0$ such that $f_{\alpha}$ has at least $p-2$

real roots. to prove this, first see that $f_{\alpha}(1) < 0$ and $f_{\alpha}(2) > 0.$ so $f_{\alpha}$ has a root in the interval (1,2), which proves the

claim for p = 3. so i will assume that p > 3. now let $1 \leq j \leq \frac{p-3}{2},$ and fix $2j < t_j < 2j+1.$ choose $\alpha_j > 0$ large enough

so that $f_{\alpha}(t_j) < 0, \ \forall \alpha \geq \alpha_j.$ (why is it possible?) thus if $\alpha \geq \alpha_j,$ then $f_{\alpha}$ has at least 2 roots in the interval $(2j, 2j+1),$

since $f_{\alpha}(2j) > 0, \ f_{\alpha}(2j+1) > 0,$ and $f_{\alpha}(t_j) < 0.$ now let $\alpha \geq \max _j \alpha_j.$ then $f_{\alpha}$ will have at least $p-2$ real roots. Q.E.D.

**ThePerfectHacker** · May 17th 2008, 07:03 PM

Here is a way to show any (finite) group can be thought of as a Galois group. Let $F$ be a field and $E=F(x_1,...,x_n)$ be the rational function field in $n$ variables. Let $K = F(f_1,...,f_n)$ where $f_1,...,f_n$ are the elementary symettric polynomials. Now $\mbox{Gal}(E/K)\simeq S_n$ because any permutation of $x_1,...,x_n$ will leave $K$ 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 $H$ of $S_n$ isomorphic to $G$ and take the corresponding fixed field, i.e. take $E^H$ and form the group $N = \text{Gal}(F/E^H)$ then $N\simeq G$ .

**ThePerfectHacker** · May 31st 2008, 08:06 PM

Here is another way to get $\mathbb{Z}_n$ . Let $F=\mathbb{C}(t)$ the rational function field. Let $x^n - t\in F[x]$ then $x^n - t$ is irreducible by Eisenstein (since $\mathbb{C}(t)$ is a quotient field of $\mathbb{C}[t]$ ). Let $\alpha$ solve the equation $x^n - t = 0$ then $\alpha , \zeta \alpha , ... , \zeta^{n-1}\alpha$ all solve this equation where $\zeta = e^{2\pi i/n}$ . Thus, $K = F(\alpha)$ and furthermore $\mbox{Gal}(K/F) \simeq \mathbb{Z}_n$ since we can map $\alpha \mapsto \zeta^k \alpha$ and so $\alpha$ generates this group with order $n$ .

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