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I want to determine the Galois group of $x^4+8x+12$. For any polynomial of degree four over $Q$; $f(x)=x^4+bx^3+cx^2+dx+e$, the resolvent cubic of $f$ is given by $R_3(f)=x^3-cx^2+(bd-4e)x-b^2e+4ce-d^2$ and the discriminant $disc(R_3)=-4p^3-27q^2$ considering $R_3$ on the form $R_3(f)=x^3+px+q$ (which always can be obtained by a change of variables).

I know:

  • Gal$(f)\cong A_4$ $\Longleftrightarrow$ $R_3(f)$ irreducible and $disc(R_3)$ is a square
  • Gal$(f)\cong S_4$ $\Longleftrightarrow$ $R_3(f)$ irreducible and $disc(R_3)$ is not square
  • Gal$(f)\cong V_4$ $\Longleftrightarrow$ $R_3(f)$ reducible and $disc(R_3)$ is a square
  • Gal$(f)\cong D_4$ or $Z/4Z$ $\Longleftrightarrow$ $R_3(f)$ reducible and $disc(R_3)$ is not square

So first I calculate $R_3(f)=x^3-12x-8^2=x^3-3\cdot 2^2x-2^6$. But how do I determine whether $R_3$ is irreducible? I cant use Eisenstein, so maybe it's reducible?

Next I calculate $disc(R_3)=-4(-12)^3-27(8^2)^2=-103700$. I see that $\sqrt{-103700}\notin Q$, hence not a squre in $Q$. So we are in case 2 or case 4, but yeah struggle to figure the rest out.

(As a side note we have $disc(R_3)=disc(f)$, which is why we just consider $disc(R_3)$. The details of this is not interesting for my question.)

slowpoke
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2 Answers2

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For a polynomial of degree $3$, being reducible is the same as having a root. And since the polynomial is integral and monic, any rational root must be an integer, furthermore dividing the constant coefficient $2^6$. That should be enough to test irreducibility pretty quickly.

Captain Lama
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"But how do I determine whether $R_3$ is irreducible? "

This follows by the Rational Root Theorem. It yields that $x^3-12x-64$ has no rational root and therefore is irreducible over $\Bbb Q$.

Dietrich Burde
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