Question

Find an example of a field F and elements a and b from some extension field such that
F(a,b) ≠ F(a), F(a,b) ≠ F(b), and [F(a,b):F] < [F(a):F][F(b):F

Answer 1

So the first the I would try to do, is find a degree 4 extension over some field (generated by a single element). Can you give me some extension/field that satisfies this?

Answer 2

Well, in any case, the idea is as follows. Find fields \(\mathbb{F}\subseteq\mathbb{F}(\alpha),\mathbb{F}(\beta)\subseteq\mathbb{F}(\alpha,\beta)\) such that \([\mathbb{F}(\alpha,\beta):\mathbb{F}]=8\), and \([\mathbb{F}(\alpha):\mathbb{F}]=[\mathbb{F}(\beta):\mathbb{F}]=4.\)

So I guess it might be more beneficial to think of a degree 8 extension first.

Answer 3

I might suggest considering the extension\[\mathbb{Q}(\sqrt[4]2,i)/\mathbb{Q}.\]You will of course need to write it slightly differently, since \(\mathbb{Q}(i)\) has degree 2 over \(\mathbb{Q}\) and not 4.

Answer 4

I'm sorry I stepped away from my computer! I don't think I'm following your choice for an extension field

Answer 5

Did you understand why we might want a degree 8 extension?

Answer 6

because it will be a higher degree than the field F?

Answer 7

oh wait 8<4x4=16?

Answer 8

Right. And I'm pretty sure that 8 is the smallest degree where we could have such a construction.

Now we just have to be careful about what degree 8 extension we choose. Having the benefit of experience, the extension I gave above is an extension that will work - if you're careful about how you choose your generators.

Answer 9

so a and b would be the generators? (I have no experience like you haha, so this is completely new to me)

Answer 10

\(a,b\) would be generators. But the way I've written it, the extension doesn't quite meet our extensions. Can you think of another pair \(a,b\) such that\[\mathbb{Q}(a,b)=\mathbb{Q}(\sqrt[4]2,i)?\]

Answer 11

well you said above that only Q(i) didn't work....so could you just keep a and make b=i^2? Would that give you the right degree?

Answer 12

That would be a bad idea since \(i^2=-1\in\mathbb{Q}\). What if you tried a combination of \(\sqrt[4]2\) and \(i\)?

Answer 13

\[\mathbb{Q} \left( a,b \right)= \mathbb{Q} \left( \sqrt[4]{2},i \right)\]

Answer 14

so i times 4th root of 2?

Answer 15

Right. If we consider \(\mathbb{Q}(\sqrt[4]2,i\sqrt[4]2)\) instead, we can see that \(\mathbb{Q}(\sqrt[4]2,i\sqrt[4]2)=\mathbb{Q}(\sqrt[4]2,i)\). But also, \[[\mathbb{Q}(\sqrt[4]2):\mathbb{Q}]=[\mathbb{Q}(i\sqrt[4]2):\mathbb{Q}]=4.\]Do you see this is?

Answer 16

*Do you see why this is?

Answer 17

I see the first part yes, but how does Q(i)= Q(i(4th root of 2))?

Answer 18

That's not true. In fact, we don't want that to be true.

Answer 19

then you have lost me

Answer 20

There are a bunch of field extensions between \(\mathbb{Q}(\sqrt[4]2,i\sqrt[4]2)\) and \(\mathbb{Q}\) (exactly 8 if you're curious). One of those intermediate extensions is \(\mathbb{Q}(\sqrt[4]2)\). Another is \(\mathbb{Q}(i)\) and a third one is \(\mathbb{Q}(i\sqrt[4]2)\). All of these three extensions are not equal to one another. However, we do have the property that\[\mathbb{Q}(\sqrt[4]2,i)=\mathbb{Q}(\sqrt[4]2,i\sqrt[4]2).\]

Answer 21

so my generators are a=4th root of 2 and b=i times 4th root of two? and that should give 8<16=4*4?

Answer 22

Moreover, with regards to your problem, \[[\mathbb{Q}(\sqrt[4]2,):\mathbb{Q}]=[\mathbb{Q}(i\sqrt[4]2):\mathbb{Q}]=4\neq2=[\mathbb{Q}(i):\mathbb{Q}].\]So we see that\[[\mathbb{Q}(\sqrt[4]2,i\sqrt[4]2):\mathbb{Q}]=8\]while\[[\mathbb{Q}(\sqrt[4]2):\mathbb{Q}][\mathbb{Q}(i\sqrt[4]2):\mathbb{Q}]=4^2=16\]

Answer 23

thank you so much!

Answer 24

You're welcome.