Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is Exercise 18.14 from Algebra, Isaacs.

$p_{1}\ ,\ p_{2}\ ,\ ... p_{n}$ are different prime numbers. How to show that $$\mathbb{Q}[\sqrt{p_{1}}, \sqrt{p_{2}}, \ldots \sqrt{p_{n}} ] = \mathbb{Q}[\sqrt{p_{1}}+ \sqrt{p_{2}}+\cdots + \sqrt{p_{n}}] \quad ?$$ First we can note that the Galois group of the first extension over $\mathbb{Q}$ ( which I call $E$ ) is elementary abelian of order $2^{n}$, so we can prove that the orbit of $\sqrt{p_{1}}+\sqrt{p_{2}}+ ... + \sqrt{p_{n}}$ under $\operatorname{Gal}(E/\mathbb{Q})$ contains $2^{n}$ elements, but how to do so?

share|improve this question
2  
Hiont: I think it suffices to proceed by induction and show tha the sum of the $n$ square roots is not zero. Then consider what an element of $E$ does to the individual square roots. –  Geoff Robinson Dec 22 '11 at 13:36
    
I see now that Bill Dubuque's answer to question 30687 gives some references to published papers of Besicovitch, Mordell, and Siegel on such topics. –  paul garrett Dec 22 '11 at 23:48
add comment

2 Answers

up vote 8 down vote accepted

OK, in view of what is written above, I will write down what I had in mind more carefully. Let us take the positive choce of $\sqrt{p_i}$ in each case, so clearly the sum (and each non-empty subsum) of the square roots is non-zero. Let $E$ be as stated in the question, the Galois group of the left-hand extension. The extension is certainly a Galois extension of $\mathbb{Q}.$ Let $\alpha$ be an element of $E$. Then $\alpha(\sqrt{p_i}) = \pm \sqrt{p_i}$ for each $i.$ Let $I$ be the subset of $\{i: 1 \leq i \leq n \}$ such that $\alpha$ fixes $\sqrt{p_i}$ for each $i \in I$, but no other $i.$ Let $s = \sum_{i=1}^{n} \sqrt{p_i}$. Then $s- \alpha(s) = 2\sum_{i \not \in I} \sqrt{p_i},$ which is strictly positive unless $\alpha$ is trivial. Since $E$ is elementary Abelian (I do not think we need to know that it has order $2^n$), we know that every intermediate extension is a Galois extension. Hence $s$ lies in no intermediate extension, and must generate the whole of the left-hand extension.

share|improve this answer
    
Nice solution ! –  WLOG Dec 22 '11 at 14:54
    
Cute argument! ............... –  paul garrett Dec 22 '11 at 17:23
    
To be honest, I am slightly dissatisfied with the argument, since it implicitly relies on the ordering of the real numbers, and does not seem "truly algebraic". That is, there is a "canonical" choice of square root, which is exploited, and this seems to go against the spirit of Galois theory. –  Geoff Robinson Dec 22 '11 at 21:30
    
@GeoffRobinson: Maybe the positivity can be avoided: prove by induction on the number of summands that no (non-empty) sum $\sum_{j\in J} \sqrt{p_j}$ is fixed by the Galois group $G_J$ of $\mathbb Q(\{\sqrt{p_j}:j\in J\})$. Certainly true for $|J|=1$. If $s=\sum_{j\in J} \sqrt{p_j}$ were fixed by $\alpha\in G_J$, then $0=s-\alpha s$ is (twice) a smaller sum of the same type, contradiction. –  paul garrett Dec 22 '11 at 21:56
    
@Paul: Yes, that's the sort of thing I had in mind when I wrote the earlier comment, but when I came to write it out, I noticed the positivity "shortcut", and didn't work all the way through, –  Geoff Robinson Dec 22 '11 at 22:03
show 5 more comments

I think it's not so trivial to show that $\sqrt{p_n}$ is not already in the field $\mathbb Q(\sqrt{p_1},\ldots,\sqrt{p_{n-1}})$. Bare-hands elementary algebra and unique factorization in $\mathbb Z$ do suffice for $n=1,2,3$, and perhaps a little beyond, but things quickly turn ugly. I do not know what the context is at that point in Isaacs' text, but I can offer two reasonably-structured approaches. The first is easy to explain in words, but is uses some algebraic number theory, which I suspect is not what is intended: in the corresponding rings of algebraic integers, adjoining $\sqrt{p_n}$ introduces ramification at $p_n$. Ignoring the prime $2$, the earlier adjoinings did not introduce ramification at $p_n$... done. The other approach first observes that $\sqrt{p_n}$ is a value of a quadratic Gauss sum, which exhibits that square root as lying in the cyclotomic field obtained by adjoining a $p_n$-th or $4p_n$-th root of unity. The "independence" comes from the "fact" that the degree of the $N$th cyclotomic field over $\mathbb Q$ is Euler totient-function of $N$, and that we have an explicit description of the Galois groups, if needed. Thus, there is a Galois automorphism fixing all those square roots but one, etc. The usual proofs about independence of cyclotomic fields are perhaps not so transparent... and I confess that the only proof I easily remember uses Dirichlet's theorem on primes in arithmetic progressions to obtain a prime $p$ so that all but $\mathbb Q(\zeta_{p_n})$ collapse to give trivial extensions of $\mathbb Q_p$.

EDIT: in response to comment/question... my in-quotes use of "independence" is meant to refer, albeit imprecisely, to the general problem of showing that a given polynomial in $\mathbb Q[x]$ has no root in some field extension of $\mathbb Q$. The exercise as posed asks about "independence" of square roots of primes: $\sqrt{p_n}$ ought not lie in the field extension of $\mathbb Q$ obtained by adjoining square roots of other primes. Similarly, the Gauss-sum argument still does depend on the "independence" of roots of unity of relatively-prime orders. The ramification argument from algebraic number theory is a different device to prove such things.

LATER EDIT: As in Geoff Robinson's answer, there are more elementary arguments! A day in which one learns something so basic is a good day!

share|improve this answer
    
Paul what does "independence of fields" mean? –  WLOG Dec 22 '11 at 14:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.