Field Extensions

**DMT** · March 16th 2006, 08:16 AM

Say:

$\alpha = \sqrt{2 + \sqrt{2}}$

And let:

$\beta = \sqrt{2 - \sqrt{2}}$

How do I show that $\beta$ is an element of the field extension:

$Q(\alpha):Q$

(the problem is to show that this field extension is normal, which I can do if I can show the above)

Any help with this would be appreciated, thanks.

**rgep** · March 16th 2006, 12:20 PM

We have $\sqrt 2 \in Q(\alpha)$ and $\alpha\beta = \sqrt 2$ .

**DMT** · March 16th 2006, 12:49 PM

Thanks! Right in front of my face the whole time.

**DMT** · March 17th 2006, 12:46 AM

Quick additional question about another method for doing the same thing.

If I compared the fields:

$ Q(\alpha):Q $

and

$ Q(\alpha, \beta):Q $

And then showed that $\alpha$ and $\beta$ have the same minimum polynomial, that would be enough to show that they are the same field, right? As would showing that the order of both extensions is 4 (which amounts to the same thing, right?).

**ThePerfectHacker** · March 17th 2006, 08:57 AM

Originally Posted by DMT

Quick additional question about another method for doing the same thing.

If I compared the fields:

$ Q(\alpha):Q $

and

$ Q(\alpha, \beta):Q $

And then showed that $\alpha$ and $\beta$ have the same minimum polynomial, that would be enough to show that they are the same field, right? As would showing that the order of both extensions is 4 (which amounts to the same thing, right?).

Your question is a very interesting one. I once asked a more general question on the forum
http://www.mathhelpforum.com/math-he...ead.php?t=1898
In general the answer is NO.
But you can do something else.
-----------------------------------
If you can show that $\mathbb{Q}\leq \mathbb{Q}(\alpha)\leq \mathbb{Q}(\beta)$ then by the theorem of finite extension fields we have,
$[\mathbb{Q}(\beta):\mathbb{Q}(\alpha)]\cdot [\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\beta):\mathbb{Q}]$
But since,
$[\mathbb{Q}(\alpha):\mathbb{Q}]=4=[\mathbb{Q}(\beta):\mathbb{Q}]$
Thus,
$4[\mathbb{Q}(\beta):\mathbb{Q}(\alpha)]=4$
Thus,
$[\mathbb{Q}(\beta):\mathbb{Q}(\alpha)]=1$
Thus,
$\mathbb{Q}(\beta)=\mathbb{Q}(\alpha)$

But the trick and problem here is to show that the subfield relation $\mathbb{Q}\leq \mathbb{Q}(\alpha)\leq \mathbb{Q}(\beta)$ is true.

**DMT** · March 18th 2006, 08:19 AM

Thanks, that's about what I had in mind.

Here's a final question though ... what would be a basis for this field extension?

I can't seem to come up with a basis with only 4 elements. My first instinct was something like:

$a + b\sqrt{2} + c\alpha + d\alpha\sqrt{2}$
$a,b,c,d \in Q$

But this doesn't pick up multiples of $\beta$ . The only combos I can find that seem to cover all the elements of the field are too long.

**DMT** · March 18th 2006, 09:27 AM

Never mind on that one ... it just hit me that since

$\beta= \sqrt{2} / \alpha$

then it's not really a problem.

The other thing I'm getting stuck with is the galois group for the field. It's supposed to be cyclic, but I seem to get the Klein 4-group instead.

I'm assuming the galois group is generated by the following two maps:

$\sigma : \alpha \mapsto -\alpha$

$\tau : \sqrt{2} \mapsto -\sqrt{2}$

Okay, I see now that $\tau$ itself is order 4 and not order 2, and can generate a cyclic group of order 4, presumably the galois group. But that doesn't seem exactly right either since I don't really get $\sigma$ in there.

Or rather, it seems $\tau^2 = \sigma$ but $\tau$ and $\tau^3$ seem to do weird things to the coeficient assuming my above basis, and not just change the signs of the root two terms (because of how they also change $\alpha$ into $\beta$ ).

I'm wondering if my basis is wrong. Since

$\beta=\alpha\sqrt{2}-\alpha$

then maybe I can use $\beta$ in my basis instead of the term with $\sqrt{2}\alpha$ ... would that work?

This is pushing the limit of my understanding of Galois theory which is what I am trying to learn.

What are the correct 4 maps for the Galois group?

Thanks.

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