Galois theory exercises

In summary, the problem is to show that the field extension \mathbb{Q}\left( \sqrt{2},\sqrt{3},\sqrt{5} \right) is a simple extension over \mathbb{Q}. This can be proven by showing that every element in the former lies in the subfield generated by a single element r=\sqrt{2}+\sqrt{3}+\sqrt{5}. This can be done by considering a simpler case, Q[a,b] where a^2=2, b^2=3, and showing that this is the same as Q[c], where c=a+b. By squaring c and using basic algebraic manipulations, it can be shown that a and
  • #1
hedlund
34
0
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 [tex] \mathbb{Q}\left( \sqrt{2},\sqrt{3},\sqrt{5} \right) [/tex] is a simple extension over [tex] \mathbb{Q} [/tex]. I don't think I will understand the solution, but I just want to see a solution.
 
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  • #2
I guess a simple extension is a single element that that extends the field to the three elements. r=[tex]\sqrt(2)+\sqrt(3)+\sqrt(5)[/tex] should do it.
 
  • #3
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
We could also consider that the degree of the field extension [tex]Q(\sqrt2,\sqrt3,\sqrt5)[/tex] 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: [tex] r_n=\pm \sqrt(2) \pm{\sqrt3}\pm \sqrt5.[/tex]

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: [tex](1+\sqrt2)(1+\sqrt3)(1+\sqrt5)[/tex], both fields are the same.

P.S. Hopefully this is "explicit" enought, as in "show explicitly."
 
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  • #5
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
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.
 
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  • #7
if you are, or aspire to be, a mathematician, then everything requires proof.
 
  • #8
Well, if we assume [tex]\sqrt5 = a+b\sqrt2+c\sqrt3+d\sqrt6[/tex] Then we square both sides and obtain: [tex]5=a^2+2b^2+3c^2+6d^2[/tex] and
[tex]\sqrt{2}(2ab+6cd)=0[/tex], [tex]\sqrt3(2ac+4bd)=0, \sqrt6(2ad+2bc)=0.[/tex] 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 [tex]a^2=6d^2[/tex], 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:
[tex]3c^2=5[/tex], which assumes 5/3 is a square.

Thus the remaining case b=0, [tex]c not 0, d not 0, which is false since ab=0=-3cd. So that should do it.
 
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1. What is Galois theory?

Galois theory is a branch of abstract algebra that studies the properties of field extensions, which are algebraic structures that extend the concept of a field. It provides a powerful tool for understanding the roots of polynomial equations and determining whether they can be solved using algebraic operations.

2. What are some common exercises in Galois theory?

Some common exercises in Galois theory include finding the degree and splitting fields of polynomials, proving the fundamental theorem of Galois theory, and determining the Galois group of a given field extension. Other exercises may involve applying Galois theory to solve problems in number theory, geometry, or other areas of mathematics.

3. How does Galois theory relate to other branches of mathematics?

Galois theory has connections to several other branches of mathematics, including algebraic geometry, number theory, and topology. It also has applications in cryptography, coding theory, and physics. Many important results in these fields have been discovered using concepts and techniques from Galois theory.

4. Can Galois theory be applied to real-world problems?

Yes, Galois theory has many practical applications in fields such as engineering, physics, and computer science. For example, it can be used in cryptography to analyze the security of encryption algorithms and in coding theory to construct error-correcting codes. Galois theory also has applications in control theory, signal processing, and robotics.

5. What are some resources for practicing Galois theory exercises?

There are many textbooks, online courses, and practice problems available for those looking to improve their skills in Galois theory. Some recommended resources include "Galois Theory" by Ian Stewart, "Galois Theory through Exercises" by Juliette Kennedy, and the exercises and solutions provided on the website of the Fields Institute for Research in Mathematical Sciences.

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