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So the only possible roots in $\mathbb{Q}$ are $1$ or $-1$ and neither are roots. So all I rely know is that the group is isomorphic to $S_3$ or $A_3$ and the polynomial has no rational roots.

Also this problem is very similar to this problem Galois groups of $x^3-3x+1$ and $(x^3-2)(x^2+3)$ over $\mathbb{Q}$ So I am trying to apply the answer for this question, but to be honest I am not sure what the discriminate means?

Mathmo123
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Samantha Wyler
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  • Google will find you lots of information about the rational root theorem and about discriminants, particularly if you spell the latter correctly. – Rob Arthan Jul 18 '20 at 21:35
  • @RobArthan looked up the rational root theorem and realized that was just the name of the theorem I used to find that there were no rational roots. – Samantha Wyler Jul 18 '20 at 21:46
  • So I found this about the discriminate https://www.johndcook.com/blog/2019/07/14/discriminant-of-a-cubic/ – Samantha Wyler Jul 18 '20 at 21:52
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    The point is that the splitting field of $f$ contains, as a quadratic subfield, the field $\mathbb{Q}(\sqrt{\Delta(f)})$ where $\Delta(f)$ is the discriminant of $f$. Can you figure out why it suffices for you to show that $\Delta(f)$ is not a square? – Mummy the turkey Jul 18 '20 at 21:56
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    So in my case would the discriminate be $0 + 108 + 0 - 0 - 27 = 81$ So since $81$ is a square in $\mathbb{Q} $, the Galois group is $A_3$? – Samantha Wyler Jul 18 '20 at 22:00
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    Sorry to be pedantic, but it will save you trouble if you learn it's called and spelled "discriminant". – Torsten Schoeneberg Jul 18 '20 at 23:22

2 Answers2

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Substituting $x=y-1$ for $x$ yields the well-known equation $y^3-3y+1=0$, whose roots are $2cos(2\pi n/9)$ for $n=1, 4, 7$. This equation has Galois Group $A_3$; in fact, if a root is $\alpha$, then the other two roots are $\alpha ^2-2$ and $2-\alpha-\alpha ^2$, showing that adding $\alpha$ to the rationals also adds the other two roots.

jimvb13
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Let $$P(x) = x^3 + 3x^2 - 1$$

First of all lets check that $P(x)$ is irreducible. If it factors at all it will have a linear factor. So we can show it is irreducible by showing it has no rational roots, this is done with the rational root theorem.

This is a degree 3 irreducible polynomial so the Galois group will be $S_3$ or $C_3$.

One tool that we can use to differentiate between these two Galois groups is the discriminant: $$\operatorname{disc}(x^3 + b x^2 + c x + d) = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$$. The Galois group will be $C_3$ if and only if the discriminant is a square.

In our case the discriminant is $81 = 9^2$ so the Galois group is $C_3$.