Prep for Algebra Comprehensive Exam #4

[BSMSMSTMSPHD]

Determine the minimal polynomial f(x) over \mathbb{Q} of the element \sqrt3 + \sqrt7. Determine the Galois group of f(x) and all the subfields of the splitting field of f(x).

SOLUTION

This one seems pretty straightforward, I just need checks on my reasoning. I apologize in advance for any redundancies. Your comments are welcome.

The splitting field must include \sqrt3 + \sqrt7 so specifically, it must contain \sqrt3 and \sqrt7. The smallest field extension of \mathbb{Q} that contains \sqrt3 is \mathbb{Q}(\sqrt3) and likewise, the smallest field extension of \mathbb{Q} that contains \sqrt7 is \mathbb{Q}(\sqrt7). But, since neither root is contained in the other's extension, neither of these can be the splitting field.

The smallest field extension of \mathbb{Q} that contains \sqrt3 and \sqrt7 is \mathbb{Q}(\sqrt3, \sqrt7). It follows from the above argument that the degree of this extension is 4.

(Again, I realize this may be superfluous... I'm just contrasting this type of problem to, say, a similar one with the element \sqrt3 + \sqrt[6]3 which would have degree 6, and the above argument would be different)

So, we're looking for a minimal polynomial of degree 4. Since the polynomial is in \mathbb{Q} the other 3 roots must be the distinct conjugates of \sqrt3 + \sqrt7, which are \pm \sqrt3 \pm \sqrt7. Thus, the minimal polynomial is:

(x - (\sqrt3 + \sqrt7))(x - (\sqrt3 - \sqrt7))(x - (- \sqrt3 + \sqrt7))(x - (- \sqrt3 - \sqrt7)) = x^4 - 20x^2 + 16

Do I need to show irreducibility here? If so, by Gauss's Lemma, taking mod 7, and using Eisenstein (p = 2), we have it to be so.

From this, we know that the order of the Galois group is also 4, which is obvious, since any automorphism is completely determined by its action on \sqrt3 and \sqrt7, each of which can either be left alone (identity automorphism) or sent to their additive inverse (conjugate automorphism).

Let \sigma be the conjugate automorphism on \sqrt3 and let \tau be the conjugate automorphism on \sqrt7. Then, \{ 1, \sigma, \tau, \sigma\tau \} is the Galois group. Since each element has order 2, this group is isomorphic to \mathbb{Z} _2 \times \mathbb{Z} _2 which has 3 subgroups of order 2.

Therefore, the splitting field \mathbb{Q}(\sqrt3, \sqrt7) must have three subfields of index 2 (yes?) which are \mathbb{Q}(\sqrt3) , \mathbb{Q}(\sqrt7) and \mathbb{Q}(\sqrt21) which all contain the subfield \mathbb{Q}.

 

[Kummer]

As I said last time, splitting fields do not have anything to do with irreducible polynomials. If you can find an polynomial(s) for the splitting field you do not need to show they are irreducible.


Here is a nice result to know.
If p_1,p_2,...,p_n are square-free and coprime:
\mbox{Gal}(\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})/\mathbb{Q})\simeq \mathbb{Z}_2\times ... \times \mathbb{Z}_2
Where the product is taken n times.

This confirms with your result.

 

[BSMSMSTMSPHD]

Thanks Kummer - I had a feeling about that general result.

Also, just playing around a minute ago, I realized that once I knew the degree of the minimal polynomial was 4, I could just write (\sqrt3 + \sqrt7)^4 + a(\sqrt3 + \sqrt7)^3 + b(\sqrt3 + \sqrt7)^2 + c(\sqrt3 + \sqrt7) + d = 0 and then eliminate all of the radicals to find the values of a, b, c, d. This isn't necessarily easier to do, but it makes more sense to me, and I was able to do it similarly for a different problem.

 

[DeadWolfe]

The splitting field must include \sqrt3 + \sqrt7 so specifically, it must contain \sqrt3 and \sqrt7.

Isn't more reasoning needed for that?

 

[mathwonk]

yes showing that is the main point.

 

[BSMSMSTMSPHD]

Well, \mathbb{Q}( \sqrt2 + \sqrt3) is a subfield of \mathbb{Q}( \sqrt2 , \sqrt3), but I'm not sure how to start...

 

[mathwonk]

you have to perform various field operations on the sum and try to isolate one of the square roots. i would of course begin by squaring the sum and see what happens.

 

[mathwonk]

well that didnt give much so maybe then cube it.

 

[BSMSMSTMSPHD]

The square of the sum is 5 + 2 \sqrt6

This value is in \mathbb{Q}(\sqrt6)...

 

[BSMSMSTMSPHD]

well that didnt give much so maybe then cube it.

Okay, 11 \sqrt2 + 9 \sqrt3

 

[Kummer]

Theorem: For n,m\in \mathbb{Z} we have \mathbb{Q}(\sqrt{n},\sqrt{m})=\mathbb{Q}(\sqrt{n}+\sqrt{m}).

 

[mathwonk]

now subtract 9sqrt(7) + 9sqrt(3) from that.

but correct it first.

 

[BSMSMSTMSPHD]

Yeah, for some reason I used \sqrt2 instead of \sqrt7. Not sure why...

( \sqrt3 + \sqrt7 )^3 = 24 \sqrt3 + 16 \sqrt7 which, when I subtract what you have gives me 15 \sqrt3 + 7 \sqrt7. I still have no idea what I'm trying to show here.

 

[morphism]

You want to show that \sqrt3 and \sqrt7 are in the splitting field by using the field operations on \sqrt3 + \sqrt7.

Subtract 16\sqrt3 + 16\sqrt7 from what you got up there. We know that this must be in the field.

Another way would be to find \left(\sqrt3 + \sqrt7\right)^{-1} = 1/\left(\sqrt3 + \sqrt7\right). Then maybe add this or subtract it from \sqrt3 + \sqrt7.

 

[BSMSMSTMSPHD]

\frac{1}{\sqrt3 + \sqrt7} = \frac{\sqrt7 - \sqrt3}{4}

So, if I do (\sqrt3 + \sqrt7) + 4(\sqrt3 + \sqrt7)^{-1} = 2\sqrt7 that shows that \sqrt7 is in the splitting field?

 

[morphism]

It should be -\sqrt3 up there, but yes, that shows that \sqrt7 is in the splitting field. Do you understand why?

By the way, if you used mathwonk's hint, you'd have got that 8\sqrt3, and hence \sqrt3, are in the splitting field.

 

[BSMSMSTMSPHD]

Yep, I made the fix. I do understand why this shows that the splitting field for \sqrt3 + \sqrt7 includes the elements \sqrt3 and \sqrt7. Does this lead to the conclusion that the order of the extension is 4?

 

[morphism]

Yes, as long as you know that \mathbb{Q}(\sqrt3, \sqrt7) is a degree 4 extension of \mathbb{Q}.

 

[BSMSMSTMSPHD]

Which it is because \mathbb{Q}(\sqrt3, \sqrt7) is degree 2 over \mathbb{Q}(\sqrt7) which is degree 2 over \mathbb{Q}. The minimal polynomials are x^2-3 and x^2-7 respectively.