6

I'm trying to show that $x^3-21x+35$ has Galois group $Z_3$ over $\mathbb Q$. I started out by taking $\alpha$ to be a root. Supposing for the moment that the polynomial is irreducible, it'd then be enough if I can show that a polynomial in $\alpha$ also satisfies this polynomial. The question gives a hint that if $\alpha$ is a root, then so is $\alpha^2+2\alpha-14$. Once I show this, then since the sum of the roots is 0, I've that all roots are in $\mathbb Q(\alpha)$ and since the polynomial has degree 3, the Galois group has order 3 and so we're done. The real problem is how do I show $\alpha^2+2\alpha-14$ is a root. The brute force method (just plugging it in) is prohibitively cumbersome. Can someone provide me a simple way to do this? I mean, even if I do really need to plug in and compute, is there an efficient way to do it?

adrija
  • 3,027

1 Answers1

5

  • You have an irreducible polynomial $P(x) = x^3-21x+35 = (x-\alpha)(x-\beta)(x-\gamma)$

    so that $0x^2 = (-\alpha-\beta-\gamma)x^2$ and $ \beta+\gamma=-\alpha$.

  • The discriminant of $P(x) = x^3+bx+c$ is $$\Delta = (\alpha-\beta)^2(\alpha-\gamma)^2(\beta-\gamma)^2=-4b^3-27c^2=4.21^3-27.35^2=63^2$$

  • Also $P'(x) = 3x^2-21=(x-\beta)(x-\gamma)+(x-\alpha)(x-\gamma)+(x-\alpha)(x-\beta)$ so that $$P'(\alpha) = (\alpha-\beta)(\alpha-\gamma) \qquad \text{and}\qquad (\beta-\gamma)P'(\alpha) = \Delta^{1/2}=63$$

  • $\Delta \ne 0$ means there is no double root so that $gcd(P(x),P'(x)) = 1$ and there exists $U,V$ such that $PU+P'V = 1$. The Euclidean algorithm gives $V(x) = \frac{2}{63}x^2 + \frac{5}{63}x - \frac{4}{9}$

  • You have a field isomorphism $\varphi:\mathbb{Q}[x]/(P)\to \mathbb{Q}(\alpha) $ given by $\varphi(f) = f(\alpha)$. Since $ P' V= 1$ in $\mathbb{Q}[x]/(P)$ it means $ \frac{1}{P'(\alpha)}=V(\alpha)=\frac{2}{63}\alpha^2 + \frac{5}{63}\alpha - \frac{4}{9}$ i.e. $$\beta-\gamma = \frac{\Delta^{1/2}}{P'(\alpha)} =\Delta^{1/2}V(\alpha)= 2\alpha^2 + 5\alpha - 28$$

  • and $$\boxed{\beta = \frac{1}{2}(\Delta^{1/2}V(\alpha) - \alpha)=\alpha^2 + 2\alpha - 14, \qquad \gamma = -\alpha-\beta=-\alpha^2 -3\alpha + 14}$$ and hence $\mathbb{Q}(\alpha) \simeq \mathbb{Q}(\beta) \simeq \mathbb{Q}(\gamma)$ and $Gal(\mathbb{Q}(\alpha)/\mathbb{Q})=\mathbb{Z}_3$

For an arbitrary irreducible polynomial degree $3$, checking if $\Delta^{1/2} \in \mathbb{Q}$ (more generally if $\Delta^{1/2} \in\mathbb{Q}(\alpha)$) is a fundamental step. And if $P$ is not irreducible, you'll have some troubles but I'm not sure where.

reuns
  • 77,999