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


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 $3$


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


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

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  • $\begingroup$ Call $t=x+2$. Now, your polynomial is $t^3-3t+1$, and the discriminant is easier to compute. $\endgroup$ – Crostul Mar 14 '16 at 0:33
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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.

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  • $\begingroup$ Thanks for the response. Just to clarify: If f(x) is an cubic reducible polynomial of three linear factors, then we calculate its discrimnant to determine the Galois Group. If it is square it will be $A_3$ and if not, $S_3$. Is this correct? Thanks $\endgroup$ – thinker Mar 14 '16 at 1:47
  • $\begingroup$ Yes. If the polynomial is irreducible of degree $n$ over $Q$ then the Galois group lies in $A_n$ if and only if the discriminant is a square. For $n=3$ then $A_3$ is the only option. $\endgroup$ – ahulpke Mar 14 '16 at 14:19

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