3
$\begingroup$

I was reading galois theory and galois group from Dummit Foote and while reading Galois groups of polynomial a sudden question came into my mind that if $f(x)$ is an irreducible separable polynomial of degree $n$ and $K$ be a splitting field of $f(x)$ then is it true that $Gal(K/F) \cong S_n$?

Moreover, can we generalize this question that if irreducible factorization of $f(x)=f_1(x)....f_k(x)$ where $f_i(x)$ is an irreducible separable polynomial of degree $n_i$ and $K$ be a splitting field of $f(x)$ then is it true that $Gal(K/F)\cong S_{n_1}\times \cdots S_{n_k}$? If so then what is the arguement?

For $\Bbb Q(\sqrt 2, \sqrt 3)$ we see that $f(x)=(x^2-2)(x^2-3)$ and the $Gal(K/F)\cong S_2 \times S_2 $ again for Galois group of $x^3-2$ we see that $Gal(K/F)\cong S_3$.

$\endgroup$
  • $\begingroup$ $X^4-2$ is the minimal polynomial of $2^{1/4}$ over $F=\mathbb{Q}(i)$ and $F(2^{1/4})/F$ is Galois with Galois group $C_4$ (sending $ 2^{1/4}$ to $i^m2^{1/4}$). In general for $K/F$ the splitting field of $f \in F[X]$ then $Gal(K/F)$ is a subgroup of $S_n, n = \deg(f)$ and iff $f$ is irreducible separable then $Gal(K/F)$ is a transitive subgroup of $S_n$ $\endgroup$ – reuns Jan 30 at 0:11
  • $\begingroup$ What is true is that the Galois group of a splitting field of a degree $n$ polynomial is a subgroup of $S_n$. Better, if the polynomial is irreducible, then the Galois group is a transitive subgroup of $S_n$. $\endgroup$ – Mathmo123 Feb 25 at 15:29
3
$\begingroup$

This is not always true as shows the answer of this question for $X^n-2$, the order of the Galois group is $n\phi(n)$ or $n\phi(n)/2$.

https://mathoverflow.net/questions/143739/galois-group-of-xn-2

$\endgroup$
1
$\begingroup$

After reading one general answer given by @Tsemo Aristide one trivial answer came into my mind which is for the splitting field of $x^p-1$ over $\Bbb Q$ we have $\Bbb Q(\psi_p)$ which is Galois and of the degree of extension $p-1$. Basically, the element of Galois group will be completely determined by the image of $\psi_p$ so arbitrary permutation is not possible at all.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.