Galois Theory – Understanding the Galois Group of an Irreducible Polynomial

Question

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.

Answer 1

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!