Galois group of an irreducible polynomial Find the Galois group of the polynomial $x^5-9x+3$ over $\mathbb{Q}$. 
since $3$ cannot divide $a_5$, $3$ divide other coefficients, $3^2$ cannot divide $a_0$, we see that the polynomial is irreducible. 
Is the Galois group of an irreducible polynomial always $S_n$, the permutation group of all roots?
I can find all roots of $x^5-1$. But I cannot find out any roots of $x^5-9x+3$ so I do not konw what is the splitting field. Since it is impossible to know roots of the polynomial, and impossible to know the splitting field, we see that to find the Galois group (the group of splitting field automorphisms) is impossible. 
 A: IF that polynomial has exactly two non real roots, then the Galois group is $S_5$. Since it has degree 5, its group is a subgroup of $S_5$, containing an element of order 5; if it has exactly two non real roots, then complex conjugation is a transposition; any subgroup of $S_5$ containing a 5-cycle and a transposition is all of $S_5$. 
A: The Galois group of an irreducible polynomial is not always $S_n$.  A good example that LASV seems to be getting at in the comments is the polynomial $x^4 + x^3 + x^2 + x + 1$.  This is irreducible over $\mathbb{Q}$, but the Galois group is $(\mathbb{Z}/5\mathbb{Z})^\times$, not $S_4$.  (In general the Galois group of the nth Cyclotomic Field over $\mathbb{Q}$ is $(\mathbb{Z} / n\mathbb{Z})^\times$.)
The basic method for finding the Galois group of a field extension is:


*

*First, verify, the field extension is a Galois extension.

*Then, find the degree of the extension.  This will give you the size of the Galois group.

*Then, find a table of all the finite groups of that order, and apply some sort of logical reasoning to decide which group you have.  (E.g., if you know the Galois group is abelian, this will help narrow things down.)


So let's apply this strategy.  We are looking for the Galois group of $K / \mathbb{Q}$, where $K$ is the splitting field of $x^5 - 9x + 3$.


*

*The extension is a Galois extension because it is the splitting field of a separable polynomial.  (Why is $x^5 - 9x + 3$ separable?)

*You observed that $x^5 - 9x + 3$ is irreducible.  That means $\mathbb{Q}(\alpha)$ has degree $5$ over $\mathbb{Q}$, where $\alpha$ is a root of the polynomial.  Then the goal is to step up to $K$ through intermediate extensions of various degrees; the degree of $K / \mathbb{Q}$ will be the product of all the individual degrees.  This is probably the trickiest part; I don't have time to do it right now but let me know if you're stuck.  This might help: if $\alpha$ is a root of the polynomial,
$$
x^5 - 9x + 3 = (x - \alpha) (x^4 + \alpha x^3 + \alpha^2 x^2 + \alpha^3 x + \alpha^4 - 9)
$$

*Finally, look up finite groups (assuming you don't know them by heart).  When I was doing Galois theory I found this wikipedia page and this list to be very helpful.


There are also hundreds of questions on mathSE already about finding the Galois groups of polynomials; use the search bar at the top of the screen and you will probably find a lot of relevant help.  Good luck!
A: (see Thomas W. Hungerford, Abstract Algebra: An Introduction, Saunders College Publing, 1990., page 366, eample)
Claim: The Galois group of $f(x)=x^5-9x+3$ over $\Bbb{Q}$ is $S_5$.
Note that the derivative of $f(x)$ is $5x^4-9$,
whose only real roots are $\pm \sqrt[4]{\frac{9}{5}}$ (the others being $\pm i\sqrt[4]{\frac{9}{5}}$).
Then $f''(x)=20x^3$,
and the second-derivative test of elementary calculus shows that $f(x)$ has exactly one relative maximum at $x=-\sqrt[4]{\frac{9}{5}}$,
one relative minimum at $x=\sqrt[4]{\frac{9}{5}}$,
and one point of inflection at $x=0$.
So its graph must have the general shape shown below.
In particular,
$f(x)$ has exactly three real roots.

Note that $f(x)$ is irreducible in $\Bbb{Q}[x]$ by Eisenstein's Criterion (with $p=3$).
If $K$ is a splitting field of $f(x)$ in $\Bbb{C}$,
since $char\Bbb{Q}=0$, $K$ is a Galois extension field of $\Bbb{Q}$,
($char F=0$ and $K$ is a splitting field of some polynomial over $F$ implies
$K$ is a finite, normal, separable extension field of $F$,
that is, Galois extension)
then $|Gal_{\Bbb{Q}} K|=[K:\Bbb{Q}]$ by the Fundamental Theorem of Galois Theory.
If $r$ is a root of $f(x)$,
then $\Bbb{Q}\leq \Bbb{Q}(r)\leq K$
and $[K:\Bbb{Q}]=[K:\Bbb{Q}(r)][\Bbb{Q}(r):\Bbb{Q}]$.
Since $r$ is a root of the irreduclbe polynomial $f(x)$ over $\Bbb{Q}$,
hence $r$ is algebraic over $\Bbb{Q}$ with minimal polynomial $f(x)$,
then $[\Bbb{Q}(r):\Bbb{Q}]=deg f(x)=5$.
Thus, $|Gal_{\Bbb{Q}}K|=[K:\Bbb{Q}]=[K:\Bbb{Q}(r)][\Bbb{Q}(r):\Bbb{Q}]=5\cdot [K:\Bbb{Q}(r)]$,
$|Gal_{\Bbb{Q}}K|$ is divisible by 5.
By Cauchy Theorem, $Gal_{\Bbb{Q}}K$ contains an element of order $5$.
The group $Gal_{\Bbb{Q}}K$,
considered as a group of permutations of the roots of $f(x)$,
is a subgroup of $S_5$ (see Corollary 11.5 in the textbook).
But the only elements of order 5 in $S_5$ are the 5-cycles.
So $Gal_{\Bbb{Q}}K$ contains a 5-cycle.
Since $K$ is a normal extension of $\Bbb{Q}$,
$\Bbb{Q}\leq K\leq \Bbb{C}$.
Complex conjugation $\phi$ is an automorphism on $\Bbb{C}$ fixed $\Bbb{Q}$,
restrict the domain of $\phi$ induces an automorphism $\phi'$ on $K$ fixed $\Bbb{Q}$ (Corollary 11.13).
This automorphism $\phi'$ interchanges the two NONreal roots of $f(x)$
and fixes the three real ones.
Thus $Gal_{\Bbb{Q}}K$ contains a transposition (2-cycle).
The Exercise 8 in the section 11.3 shows that the only subgroup of $S_5$
that contains both a 5-cycle and a transposition is $S_5$ itself.
Therefore $Gal_{\Bbb{Q}}K=S_5$.
A: Nowadays one can test the OP's question  using a computer. For example, using Magma one sees

P<x>:=PolynomialAlgebra(Rationals());
f:=x^5-9*x+3;
print IsIrreducible(f), Order(GaloisGroup(f));

true 120

f:=x^4+x^3+x^2+x+1;
print IsIrreducible(f), Order(GaloisGroup(f));

true 4
So there is a negative answer to the question "Is the Galois group of an irreducible polynomial of degree $n$ always $S_n$?".
