# 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 choice 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
@GeoffRobinson: Also, yes, the argument makes no use of the specifics of what roots are adjoined, because it does not prove that the degree of the field extension is the largest possible, but only that, whatever it is, it can also be obtained by adjoining that single element, the sum. Quite funny! – paul garrett Dec 23 '11 at 18:35

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
@paul Can you please check my answer. It seems that we do not require to know that $E$ (the Galois gorup of the given extension) is abelian. Thank you. – caffeinemachine Feb 24 at 8:18
@caffeinemachine, yes your argument is correct... but perhaps the hard part is that the only element of the Galois group fixing the sum of square roots is the identity element. – paul garrett Feb 24 at 13:36
@paulgarrett I agree. That was the main part. I was just trying to clarify (to myself) that the knowledge of the structure of the Galois group is not required. – caffeinemachine Feb 24 at 16:41

$\newcommand{\Q}{\mathbf Q}$This is a simplification of Geoff Robinson's answer.

For each $i$ write $a_i=\sqrt{p_i}$. Also write $K=\Q(a_1, \ldots, a_n)$. Define $s=a_1+\cdots+a_n$. As Geoff has shown, the only element of $G:=\text{Aut}(K:\Q)$ which fixes $s$ is the identity.

Therefore $\text{Aut}(K:\Q(s))$ is a singleton. But since $K:\Q$ is Galois, so is $K:\Q(s)$, and therefore $[K:\Q(s)]=|\text{Aut}(K:\Q(s))|$ showing that $[K:\Q(s)]=1$ and hence $\Q(s)=K$.

Note that we do not require to know that $\text{Aut}(K:\Q)$ is abelian.

A neat proof of a much more general result has here been given by @orangeskid here.

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You could also say that since $K$ is a Galois extension of $\mathbb{Q}$, Galois correspondence gives a bijection between subfields of $K$ and subgroups of $Gal(K/\mathbb{Q})$ ( sending the subgroup $H$ to the fixed field $K^{H}$). The trivial subgroup oobviously goes to $K$. – Geoff Robinson Feb 24 at 10:39
@GeoffRobinson Right. That is perhaps an even simpler way to say it. – caffeinemachine Feb 24 at 10:40