# Galois theory

1. Mar 1, 2005

### hedlund

I haven't studied Galois theory, just seen some exercises on it. However I would like to know how one can solve this problem:

Show explicitly that $$\mathbb{Q}\left( \sqrt{2},\sqrt{3},\sqrt{5} \right)$$ is a simple extension over $$\mathbb{Q}$$. I don't think I will understand the solution, but I just want to see a solution.

2. Mar 1, 2005

### robert Ihnot

I guess a simple extension is a single element that that extends the field to the three elements. r=$$\sqrt(2)+\sqrt(3)+\sqrt(5)$$ should do it.

3. Mar 2, 2005

### matt grime

By explicit, you must show that every elemen in Q[sqrt(2),sqrt)3),sqrt(5)] lies in the subfield generated by Q[r], where r is as above.

To see how to prove this kind of thing in general, let's try a simpler case

Q[ a,b] where a^2=2, b^2=3, say, and show this is the same as Q[c], where c=a+b.

By squaring c, 5+2abis in the field on the right, thus so is (a+b)(5+2ab) = 5a+5b+4a+6b=9a+11b.

Thus 9(a+b) - 9a-11b = -2b is in it, and so is b. Hence a is in Q[c] is too, and we are done.

4. Mar 2, 2005

### robert Ihnot

We could also consider that the degree of the field extension $$Q(\sqrt2,\sqrt3,\sqrt5)$$ is build on three quadratic equations. And since the degree(A:C) =Degree(A:B) times degree(B:C), where B is an intermediate field, we have the degree of the field is 2x2x2=8.

Thus there exists a minimal polynominal of degree 8 that contains that field extension. We consider the 8 roots of a polynominal: $$r_n=\pm \sqrt(2) \pm{\sqrt3}\pm \sqrt5.$$

This is a polynominal of degree 8 which splits over an extension of Q containing its roots. (And because it is symmetric-actually consists of elementary symmetric functions- its coefficients lie in Q.)

Also, it is easy to see by adding and subtracting, dividing,etc,these roots that we have the three required square roots of the earlier extension. Since the vector space over the field in either case consists of 8 elements and their constant multiples, all of which can be found in: $$(1+\sqrt2)(1+\sqrt3)(1+\sqrt5)$$, both fields are the same.

P.S. Hopefully this is "explicit" enought, as in "show explicitly."

Last edited: Mar 2, 2005
5. Mar 3, 2005

### mathwonk

Re: "We could also consider that the degree of the field extension is build on three quadratic equations. And since the degree(A:C) =Degree(A:B) times degree(B:C), where B is an intermediate field, we have the degree of the field is 2x2x2=8."

don't you have to prove that sqrt(5) is not already contained in the extension obtaiuned say by adjoining sqrt(2) and sqrt(3)? i.e. your argument seems to show the degree of the extension divides 8.

6. Mar 3, 2005

### robert Ihnot

mathwonk: your argument seems to show the degree of the extension divides 8.

In the case you give the vector space is such that A + Bsqrt(2)+Csqrt(3) + sqrt(6) gives the elements, which does not included sqrt(5). (I have never seen a math book that thought that required an argument, and it seems, well, messy.)

As for how I wrote it up, I was too lazy to put in a third step, but considered degree(A:C) = degree(A:B) times degree (B:C) as just a rule that could be repeated as needed.

Last edited: Mar 3, 2005
7. Mar 3, 2005

### mathwonk

if you are, or aspire to be, a mathematician, then everything requires proof.

8. Mar 4, 2005

### robert Ihnot

Well, if we assume $$\sqrt5 = a+b\sqrt2+c\sqrt3+d\sqrt6$$ Then we square both sides and obtain: $$5=a^2+2b^2+3c^2+6d^2$$ and
$$\sqrt{2}(2ab+6cd)=0$$, $$\sqrt3(2ac+4bd)=0, \sqrt6(2ad+2bc)=0.$$ Because each term is linerally independent. So we have:

ab=-3cd; ac=-2bd;ad=-bc. Multiplying the first and the second and dividing out bc, we get $$a^2=6d^2$$, which is impossible since the sqrt6 is not rational.

Thus we assume b=0, then cd=0, ac=0, ad=0. Assuming c=0 is the first case, so assume d=0, then ac=0, gives a=0. Thus b=0, d=0, a=0 gives:
$$3c^2=5$$, which assumes 5/3 is a square.

Thus the remaing case b=0, [tex]c not 0, d not 0, which is false since ab=0=-3cd. So that should do it.

Last edited: Mar 4, 2005