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I want to compute the Galois group of some polynomials, but I want to see some examples first. For example this proposition could be helpful. I don't know how to prove it <.<

Let's consider a polynomial $f(x)= x^4+ax^2+b \in \mathbb{Z} $. Let $ \pm\alpha , \pm\beta$ denote the roots of $f(x)$. Note that $f$ is irreducible iff $\alpha^2, \alpha\pm\beta \notin \mathbb{Q}$. So let's suppose that $f(x)$ is irreducible. Let's denote $G$ the galois group of $f$ (the Galois group of the splitting field $\mathbb{Q}(\alpha,\beta)/\mathbb{Q}$)

$i)$ $G\cong V$ the Klein 4-group iff $\alpha\beta \in \mathbb{Q}$

$ii)$ $G\cong \mathbb{Z}_4$ iff $\mathbb{Q}(\alpha\beta)=\mathbb{Q}(\alpha^2)$

$iii)$ $G\cong D_8$ the dihedral group of order 8, iff $\alpha\beta \notin \mathbb{Q}(\alpha^2)$

I need help with the proofs of this propositions... Well I'm very lost with the computation of Galois group, some care must be exercised, I don't have it. For example I know that Galois group acts transitively, in the sense that $u\in K/F$ a galois extension, the other roots of the minimal polynomial of $m_u(x)\in F[x]$ are precisely the elements $ \sigma(u)$ , where $\sigma \in Gal(K/F)$. And clearly any automorphism, is determined by only the value of a basis. In this case we are working on $\mathbb{Q}(\alpha,\beta)$, so I have to know the values of both. But I have to respect the algebraic relations involving $\alpha , \beta$ I proved that in $i),ii)$ $\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\alpha) $ So I have to know the values on $\alpha$.

My first question is: If in both cases $i),ii)$, I have an extension of degree 4 (so $|G|=4$) I have to map $\alpha$ to determine the automorphism, but I have also to map $\alpha$ to all of it's four roots. Why the galois group in this case are different?

My second question is about $iii)$ I have no idea How to attack it.

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1 Answer 1

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For i) and ii), let the conjugates of $\alpha$ be $\gamma$, $\delta$, and $\epsilon$. Consider the automorphism that maps $\alpha$ to $\gamma$. If it maps $\gamma$ to $\delta$ (and then, as you can show, $\delta$ to $\epsilon$, and $\epsilon$ to $\alpha$), then this automorphism is of order 4, and the group is the cyclic group of order 4.

But it could be that every automorphism is its own inverse; if it maps $\alpha$ to some other conjugate, say, to $\eta$, then it maps $\eta$ to $\alpha$, and also interchanges the other two conjugates. Then the group is Klein-4.

In other words, just knowing where $\alpha$ goes doesn't tell you where the other conjugates go, and there are two essentially different possibilities, corresponding to the two groups of order 4.

To get you started on iii), try to find an automorphism $\sigma$ of order 4, and an automorphism $\tau\ne\sigma^2$ of order 2, and try to show they satisfying the defining relation of the dihedral group, $\tau\sigma=\sigma^3\tau$.

EDIT: Maybe an example of ii) will help to see what happens in this case. Take the polynomial $x^4-4x^2+2$. The zeros are $\pm\alpha$ and $\pm\beta$ where $$\alpha=\sqrt{4+2\sqrt2},\quad\beta=\sqrt{4-2\sqrt2}$$ We have $$\alpha^2=4+2\sqrt2,\quad\alpha\beta=2\sqrt2$$ Let $\sigma(\alpha)\beta$. Then $\sigma(\alpha^2)=\beta^2$, which implies $\sigma(\sqrt2)=-\sqrt2$. Then $$\sigma(\beta)=\sigma(2\sqrt2/\alpha)=-2\sqrt2/\beta=-\alpha$$ and we see that $\sigma$ has order 4.

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I'm a little confused right now.In $ii)$ the hypothesis says that $\beta=\frac{q}{\alpha}$ where $q\in \mathbb{Q} $ . I have to send $\alpha$ to all the roots. $\pm\alpha,\pm\frac{q}{\alpha}$. Well, starting with the automorphism $\alpha \to -\alpha$ has order two, and it's easy to see that $\alpha \to \frac{q}{\alpha}$ has also order two. Well my problem now is with $i)$ the hypothesis implies that $\beta= \frac{\alpha^2-p}{q\alpha}$ where $p,q\in \mathbb{Q}$ , sending $\alpha \to \frac{\alpha^2-p}{q\alpha}$ has no order 4 (I used maple to computation of this) so something is wrong :/! –  Daniel Sep 30 '12 at 15:19
    
In your comment, it seems you have referred to ii) when you meant i), and vice versa. Let me think a bit about the cyclic case. –  Gerry Myerson Sep 30 '12 at 23:13

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