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  • $x^3+x^2+1$
  • $x^3-12x+8$
  • $x^4+4x^2+9$
  • $x^4+3x^3-3x+3$

I started taking class in Galois theory this fall and here are some problems I was told to solve. The problem is I feel huge lack of practice and methods I could use. On previous practice class we tried to solve one problem of this kind using resolvent, symmetric polynomials and basic group theory. I hope it will help you to guess the main Idea (but I understood nothing).

I was told I should start with my ideas. As I know, if I have to compute Galois group of polynomial $f$ and $\deg f = n$, it will be subgroup of $S_n.$ What should I do next?

P.S. I'm BEGGING: when you use any fact, describe what is it (theorem, lemma, something empirical, etc.) cause I really struggle with applying lecture material even to elementary problems.

P.P.S. Ok, now I also know that if $f(x)$ is irreducible and separable (no roos in $\mathbb C$), than $G$ is transitive. Hopefully there's a page https://people.maths.bris.ac.uk/~matyd/GroupNames/T31.html

So, for example: $x^3+x^2+1$ is irreducible and has no roots, so its Galois group equals $S_3$ or $C_3.$ What could I do next?

Big Coconut
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    Have you reviewed https://math.stackexchange.com/questions/45893/how-to-find-the-galois-group-of-a-polynomial? – Moo Sep 23 '22 at 16:57
  • No, thanks for link! – Big Coconut Sep 23 '22 at 17:06
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    Artin’s algebra covers the cubic case and leaves as an exercise how to classify quartic Galois groups. When I have time - and if you are still stuck - I will show the exercises’ solutions to you and apply it to your problem: in the meantime, their book is freely available on the internet as a pdf. – FShrike Sep 23 '22 at 17:19
  • Ok, I'll check the book and I would edit question in case I solved quartic part. Thanks! – Big Coconut Sep 23 '22 at 17:25
  • what book is your class using??????? – Will Jagy Sep 23 '22 at 17:35

1 Answers1

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I will give the cubic Galois groups - if the quartic ones don't make sense, let me know and I can do those too.

The discriminant of a cubic $ax^3 + b x^2 + c x + d$ is given by $\Delta^2 = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$. If $\Delta$ is a square, the Galois group is a subgroup of the alternating group $A_3 \cong C_3$.

In the case b=1, c=0, d=1, we have discriminant 0 - 4 - 27 = -31. Clearly, -31 is not a square, so the Galois group isn't $C_3$, and hence must be $S_3$.

Similarly, $x^3-12x+8$ is irreducible, and the discriminant is $-4(-12)^3-27(8)^2 = 2^8 \cdot 3^3 - 3^3 \cdot 2^6 = 2^6 \cdot 3^4$, which is clearly a square, and so this cubic has Galois group $C_3$.

Fred T
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