# Prep for Algebra Comprehensive Exam #4

1. Aug 13, 2007

### 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 $$\mathbb{Q}$$.

2. Aug 13, 2007

### 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.

Last edited: Aug 13, 2007
3. Aug 13, 2007

### BSMSMSTMSPHD

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.

4. Aug 13, 2007

Isn't more reasoning needed for that?

5. Aug 13, 2007

### mathwonk

yes showing that is the main point.

6. Aug 13, 2007

### BSMSMSTMSPHD

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

7. Aug 13, 2007

### 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.

8. Aug 13, 2007

### mathwonk

well that didnt give much so maybe then cube it.

9. Aug 13, 2007

### BSMSMSTMSPHD

The square of the sum is $$5 + 2 \sqrt6$$

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

10. Aug 13, 2007

### BSMSMSTMSPHD

Okay, $$11 \sqrt2 + 9 \sqrt3$$

11. Aug 13, 2007

### Kummer

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

Last edited: Aug 13, 2007
12. Aug 13, 2007

### mathwonk

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

but correct it first.

13. Aug 14, 2007

### 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.

14. Aug 14, 2007

### 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 $\sqrt3 + \sqrt7$.

Last edited: Aug 14, 2007
15. Aug 14, 2007

### 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?

Last edited: Aug 14, 2007
16. Aug 14, 2007

### 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.

17. Aug 14, 2007

### 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?

18. Aug 14, 2007

### morphism

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

19. Aug 14, 2007

### 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.