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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.