# How to prove that $\mathbb{Q}[\sqrt{p_1}, \sqrt{p_2}, \ldots \sqrt{p_n} ] = \mathbb{Q}[\sqrt{p_1}+ \sqrt{p_2}+\cdots + \sqrt{p_n}]$, for $p_i$ prime?

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?

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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

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.

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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

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!

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Paul what does "independence of fields" mean? –  WLOG Dec 22 '11 at 14:23