Find Galois Group of cubic polynomial $f(x)=x^3+6x^2+9x+3$ over $\mathbb{Q}$

Question

I want to find the Galois Group of the following cubic polynomial: $f(x)=x^3+6x^2+9x+3$ over $\mathbb{Q}$

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By Eisentein's criterion with $p=3$, I know that $f(x)$ is irreducible, since $3$ divides all coefficients except the first one, and $9$ does not divide $3$

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So since it is irreducible, I know that the Galois group of f(x) over $\mathbb{Q}$ is either $S_3$ or $A_3$

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My trouble is in how to distinguish between the two cases. I think it has something to do with splitting fields. Any help would be much appreciated

Answer 1

It is $A_3$. You could show this for example by any of the following criteria:

• The discriminant of $f$ (which is 81) is a square.

• If $\alpha$ is a root of $f$, that is $f$ defines the field $K=Q(\alpha)$, then $f$ splits over $K$ into linear factors (namely $f=(x-\alpha)\cdot(x-\alpha^2-4\alpha)\cdot(x+\alpha^2+5\alpha+6)$.

• The polynomial has a root (and thus -- as the field is normal all roots) over the 9-th cyclotomic field: $\zeta+\bar\zeta-2$ is a root of $f$ where $\zeta$ is a primitive 9th root of unity. Thus, as a subfield of a cyclotomic field, the Galois groups must be abelian.