Find all the subfield of $\displaystyle Q(\sqrt[4]{3},i)$
clearly the subfields $\displaystyle Q(\sqrt[4]{3})$ and $\displaystyle Q(i)$ are there. then $\displaystyle Q(\sqrt[2]{3})$and $\displaystyle Q(\sqrt[2]{3},i)$ are also there. if you draw the picture correctly i guess you will see that there is no more space to insert any more fields .. as we have broken down all extensions in degree 2 pieces.
i guess it is the correct answer ... if i come up with anything else i will let u know.
to make it clear here that i think by "subfields" ynj means the intermediate subfields, i.e. those subfields which contain $\displaystyle \mathbb{Q}.$ including $\displaystyle \mathbb{Q}$ and $\displaystyle K=\mathbb{Q}(\sqrt[4]{3},i)$ nirax has found $\displaystyle 6$ of them but $\displaystyle K$ has $\displaystyle 10$
subfields because $\displaystyle \text{Gal}(K/\mathbb{Q})=D_8,$ the dihedral group of order $\displaystyle 8$, and we know that $\displaystyle D_8$ has $\displaystyle 10$ subgroups. so, see if you can find $\displaystyle 4$ missing ones!
actually there isn't much argument. the basic thing to identify are the degree 2 intermediate extensions between them ... since there cannot be any more subfields between such a extension, so this is the end of investigation.
a field contaning fourth root must also contain the square root. so that gives you one intermediate extension and being generated by a square root, it must be of degree two.
after you have found degree two estensions, take compositum of intermediare extensions (that is the smallest subfield containing the two given extensions) .. then you try to insert more degree two extensions wherever you can and take compositums again ... such a approach is suitable only for small textbook problems ... there are more sophisticated ways of finding subfields which are not needed in this case ...
i am sorry if all this sounds sketchy but it is really an elementary problem. I dont think i can explain it any better.
I make those subgroups of D8 the trivial subgroup and the group itself for 2 of them.
Then there are 4 subgroups of order 2 with 1 reflection
1 subgroup of order 2 with 1 half turn
1 subgroup of order 4 generated by a quarter turn
2 subgroups of order 4 generated by two reflections.
But being ignorant of Galois theory I can't quite work out how they correspond to subfields since I guess the elements of the Galois group should be permuting some of the new elements introduced by the field extension. Assuming Q corresponds to the trivial subgroup and the field itself to D8 (or maybe the reverse) we need 8 proper subfields between Q and the starting field.
nirax found 4 of them, and I'm pretty sure mine are two more, but I can't find any more at all, nor see how those we have so far relate to the subgroups of D8.
what??!! those normal extensions correspond to the normal subgroups of the Galois group. see the fundamental theorem of Galois theory again!
to alunw: let $\displaystyle K=\mathbb{Q}(\alpha,i),$ where $\displaystyle \alpha=\sqrt[4]{3}.$ we have $\displaystyle \text{Gal}(K/\mathbb{Q})=<\sigma, \tau> \cong D_8,$ where $\displaystyle \sigma, \tau $ are defined by $\displaystyle \sigma(i)=-i, \ \sigma(\alpha)=\alpha$ and $\displaystyle \tau(i)=i, \ \tau(\alpha)=\alpha i.$
the subfields of $\displaystyle K$ containing $\displaystyle \mathbb{Q}$ are exactly the fixed fields of subgroups of $\displaystyle <\sigma, \tau>.$ for example $\displaystyle \text{Fix}\{<\sigma> \}=\mathbb{Q}(\alpha), \ \text{Fix}\{<\tau>\}=\mathbb{Q}(i), \ \text{Fix}\{<\tau^2> \}=\mathbb{Q}(i,\alpha^2), \cdots $
So now we can try to do the problem systematically.
Let a be NonCommAlg's $\displaystyle \tau$
Let b be NonCommAlg's $\displaystyle \sigma$
The 10 subgroups are
{1} which fixes $\displaystyle Q(\sqrt[4]3,i)$
D8 which fixes $\displaystyle Q$
{1,a,a^2,a^3} which fixes $\displaystyle Q(i)$
{1,a^2} which fixes $\displaystyle Q(\sqrt 3,i)$
{1,b} which fixes $\displaystyle Q(\sqrt[4]3)$
and
{1,ab}
{1,a^2b}
{1,a^3b}
{1,a^2b,b,a^2}
{1,ab,a^2,a^3b} which I need to think about a little.
Note that a^2b($\displaystyle \sqrt[4]3*i$) = a^2($\displaystyle \sqrt[4]3$)*b(i) (since b fixes $\displaystyle \sqrt[4]3$ and a fixes i) = $\displaystyle -\sqrt[4]3 * -i = i*\sqrt[4]3$.
So {1,a^2b} fixes $\displaystyle Q(i*\sqrt[4]3)$
And Fix {1,a^2b,b,a^2} is the intersection of $\displaystyle Q(i*\sqrt[4]3)$,$\displaystyle Q(\sqrt[4]3)$ and $\displaystyle Q(i,\sqrt 3)$ which is $\displaystyle Q(\sqrt 3)$
ab fixes $\displaystyle \sqrt-3$ since ab($\displaystyle i*(\sqrt[4]3)^2$) = b($\displaystyle i$) * a($\displaystyle (\sqrt[4]3)^2$) = $\displaystyle -i * -(\sqrt[4]3)^2 = \sqrt-3$
And a^3b fixes this element as well.
So Fix {1,ab,a^2,a^3b} is $\displaystyle Q(\sqrt-3)$
That accounts for all the subfields we identified already. We haven't found any subfields for the subgroups {1,ab} and {1,a^3b}. Neither ab or a^3b fix any of $\displaystyle \sqrt 3,i,\sqrt[4]3$, or $\displaystyle i*\sqrt[4]3$
Am I missing something, or are there actually no subfields for those two subgroups?