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In continuation to my last post:

In class we saw an example that says: $n=[\mathbb{F}_{p^n}:\mathbb{F}_{p}]=|\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_{p})|$ ; ($p$ is prime).

My thoughts are that if I look at any Galois extension then it is the splitting field of separable polynomials in the field I started with. I know that any automorphism of the extension field that fixes the field I started with sends each root of an irreducible factor (of one of the polynimials that the Galois expenstion is their splitting field), and also that for every such permutation I have an automorphism .

But since if an irreducable factor have $k$ roots then I can permute on them in $k!$ options I deduced that the size of Galois group is of the form $k_1!k_2!\cdots k_t!$, in particular it is either $1$ or devisable by $2$.

But this contradicts what I wrote in my first paragraph if for example we take $p=2,n=5$ since $5$ is odd.

Can someone please point out the mistake ?

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Again, you have mistakenly assumed that all permutations are possible. This is not true, even if your polynomial is irreducible. – Zhen Lin May 10 '12 at 9:40
You're not missing any roots. You're missing permutations! – Zhen Lin May 10 '12 at 9:49
@Belgi, as an example of what Zhen Lin is saying consider the following. If $\zeta=e^{2\pi i/5}$, then all the roots of $x^4+x^3+x^2+x+1$ are $\zeta,\zeta^2,\zeta^3,\zeta^4$. But if an automorphism $\sigma$ of $\mathbf{Q}(\zeta)$ sends $\zeta$ to $\zeta^j$, for some $j=1,2,3,4$, then we must have $$ \sigma(\zeta^2)=\sigma(\zeta)^2=\zeta^{2j}, $$ and similarly the image of $\zeta$ uniquely determines the images of $\zeta^3$ and $\zeta^4$. IOW, $\sigma$ cannot just permute those roots any which way we want. It has to respect the "hidden relations" among the roots. – Jyrki Lahtonen May 10 '12 at 10:16
@Belgi: You are right in that you can send any root to any other root. This merely means the Galois group has size $\geq k$. – Hurkyl May 10 '12 at 10:35
@Belgi: In the last couple of days you have asked a number of questions related to field extensions or Galois groups which are all treated in books. What are you reading as you are learning this material? These questions would often be addressed in course textbooks, as the questions are quite basic (e.g., why a Galois group of a polynomial of degree $n$ does not have size $n!$). That doesn't mean the questions are unimportant, but they ought to be better understood by reading about worked examples in books. – KCd May 10 '12 at 16:09
up vote 5 down vote accepted

Since you don't seem to be convinced by your own example, let's look at another polynomial whose splitting field does not have the full symmetric group as its Galois group.

We take $\mathbb{Q}$ as our base field. Let $P(x) = x^3 - 3 x + 1$. For various reasons, $P$ is irreducible over $\mathbb{Q}$. Let $K$ be its splitting field over $\mathbb{Q}$, with three zeros $\alpha, \beta, \gamma$. Then, $$(x - \alpha)(x - \beta)(x - \gamma) = x^3 - 3 x + 1$$ so we can deduce various equations like $\alpha + \beta + \gamma = -1$. This can be done generically for any polynomial. But $P$ has a special property which gives an extra equation: $$(\alpha - \beta)(\beta - \gamma)(\gamma - \alpha) = 9$$ This is a polynomial equation with integer coefficients that the zeros of $P$ satisfy, so any automorphism of $K$ over $\mathbb{Q}$ must preserve this equation. But that means we cannot transpose just two of the roots, because that would invalidate the equation. So $\textrm{Gal}(K \mid \mathbb{Q})$ is the cyclic group of order 3.

Now, what is true is for any two zeros $\alpha$ and $\beta$ of a irreducible polynomial $P$, there is an automorphism $\sigma$ such that $\sigma (\alpha) = \beta$. But $\sigma$ could do anything else to any other zero – including $\alpha$. In slightly more sophisticated terms, the Galois group of the splitting field of $P$ is always a transitive subgroup of the symmetric group and acts transitively on the set of zeros of $P$. But it is not guaranteed to act freely.

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