0
$\begingroup$

suppose $E=Q(\sqrt{2},\sqrt3,u)$ ,where $u^2=(9-5\sqrt{3})(2-\sqrt{2})$ The question is to prove it is Galois extension,and compute its Galois group.

Notice that the characteristic of $Q$ is infinity, so very irreducible polynomial is separable. Then just to prove $E/Q$ is the splitting field of the minimal polynomial of $ \sqrt{2},\sqrt3,u$.
My problem is compute the minimal polynomial of $u$,is there any easy way to compute its minimal polynomial?I think only know its minimal polynamial,then I can compute its roots ,which are possible image that $Q-$homophorsims send $u$ to

$\endgroup$
2
$\begingroup$

For clarity, consider first the general problem of embedding a finite galois extension $K/k$, with group $G$, into a finite overextension $E/F/k$, and give a necessary and sufficient for $E/k$ to be galois. Almost by the definition of normality, such a NSC is that all the extensions of the $k$-automorphisms $s$ of $F$ (i.e. of the elements of $G$) to $k$-embeddings $\hat s$ of $E$ in a separable closure of $k$, should stabilize $E$. If $E$ is given by $E=F(\alpha)$, this means that all the conjugates $\hat s(\alpha)$ belong to $E$.

In your particular case, $k=\mathbf Q$, $F=\mathbf Q (\sqrt 2, \sqrt 3)$ is a biquadratic field, and $E=F(u)$ with $u^2=v=(2-\sqrt 2)(9-5\sqrt 3) \in F$. The group $G$ is obviously $\cong C_2\times C_2$, generated by $s$ s.t. $s(\sqrt 2)=-\sqrt 2, s(\sqrt 3)=\sqrt 3$, and $t$ defined likewise by permuting $2$ and 3. Because the quadratic extension $F(\sqrt v)$ depends only on the class of $v$ mod ${F^*}^2$, the stabilization condition above readily amounts to $s(v)/v$ and $t(v)/v \in {F^*}^2$. This is easily checked, so that $E/\mathbf Q$ is galois of degree $8$.

There are only $3$ possibilities for a group of order $8$ admitting a quotient $\cong C_2\times C_2$, which are $C_2\times C_2\times C_2$, the dihedral $D_8$, or the quaternionic $H_8$. The first case can be immediately discarded on looking at $u$ (why ?). To distinguish between the $2$ latter cases, a pedestrian way is to extend $s$ and $t$ explicitly to $E$ and compute the (non)commutation relations between $\hat s$ and $\hat t$. For a (possible) shortcut, see e.g. https://math.stackexchange.com/a/2046653/300700 .

$\endgroup$
  • $\begingroup$ What's meaning of finite overextension? $\endgroup$ – Bruce Jan 1 '18 at 12:15
  • $\begingroup$ A tower of finite extensions as indicated in the notations $\endgroup$ – nguyen quang do Jan 3 '18 at 18:15

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.