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I am presently learning the Galois theory, and is often confused when putting what I have learned to applications to polynomials. The following are some examples.

(a) Every polynomial has a Galois group, and this group in some sense enables us to look into the polynomials. Natural it is then to ask what happens if two polynomials have the isomorphic Galois groups? If one denotes this equivalent relation by ~, then what is F[x]/~?

(b) Albeit aware already of the main theorem of Galois theory, I can barely directly compute the Galois group of polynomials. For example,
$f(x)=x^3-x-1$, over $Q$.

Thanks and best regards.

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Galois groups of irreducible monic cubics can be found by verifying whether the discriminant is a square of a rational or not, if it is a square then the galois group is cyclic of order 3, if it is not then it is just $S_3$. Note that Galois group of a polynomial of degree n is always a subgroup of $S_n$ and if the polynomial is irreducible then it is transitive. –  Dinesh Nov 3 '11 at 9:27
    
Dinesh is right of course - in my answer I was trying to outline a bare-hands approach from first principles. –  Alon Amit Nov 4 '11 at 0:37
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1 Answer 1

up vote 2 down vote accepted

(I think it's generally better to ask one question per post. I'll take a stab at your first two).

(a) I don't think this equivalence relation is very informative. Over $\mathbb{Q}$, for example, there are lots of degree-5 polynomials with Galois group $S_5$, and they are not related to each other in any particularly illuminating way as far as I can tell. Also, the equivalence relation doesn't respect any algebraic operation, so $F[x]/\sim$ is just a set, which can be described by listing in some way one polynomial for each isomorphism class of Galois groups. This is a computationally hard task - we don't even know what all the Galois groups of $\mathbb{Q}$ are (see "inverse Galois problem").

(b) Consider:

  • Is your polynomial irreducible over $\mathbb{Q}$? (in your case, degree 3, show that it's irreducible iff it has no rational root. Does it?).

  • How many real and how many non-real complex roots does it have? (in your case, show that it has exactly one real root).

  • If there are non-real roots, what can you conclude from "complex conjugation" being an automorphism of the splitting field? (it's an element of order 2 in the Galois group).

  • Is there an automorphism mapping the real root to one of the non-real ones?

Calculating Galois groups in general is not a straightforward task, but for polynomials of reasonably small degree you can usually get by with considerations such as the above, plus a few more you'll learn with practice.

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@Alon Amit:Thanks for your answer; and I think I get some idea of computing the Galois groups of polynomials of degree at most 4 now, by means of reading through the articles on this topic. Still thanks for the judgement of the above equivalence relation; the sentence "It respects not to any algebraic operation" is somewhat illuminating, and I think this equivalence relation makes indeed barely any sense with respect to polynomials, except for the trivial ones, if any. Thanks again for the time to response. –  awllower Nov 3 '11 at 13:51
    
I think this answer really answered the question now. Therefore it is accepted. Still, the third problem is left open. –  awllower Nov 16 '11 at 16:22
    
Therefore I would suggest asking the third question separately, and separating questions in general. –  Alon Amit Nov 16 '11 at 21:22
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