# 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?

• 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. Dec 22, 2011 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. Dec 22, 2011 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.

• Nice solution !
– WLOG
Dec 22, 2011 at 14:54
• 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. Dec 22, 2011 at 21:30
• @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! Dec 23, 2011 at 18:35
• @GeoffRobinson I am not sure why we need to know that the Galois group $E$ is abelian. Let $K=\mathbf Q(\sqrt{p_1}, \ldots, \sqrt{p_n})$. We have shown that if $\sigma\in E$ is not identity then $\sigma$ does not fix $s$. We also know that the fixed field of $\text{Aut}(K:\mathbf Q(s))$ is $\mathbf Q(s)$ (because $K:\mathbf Q$ is a Galois extension). Therefore $|\text{Aut}(K:\mathbf Q(s))|=[K:\mathbf Q(s)]$. Since the former is a singleton, we have $\mathbf Q(s)=K$ and we are done. Am I making a mistake somewhere? Feb 23, 2016 at 16:28
• @GeoffRobinson The fact that the fixed field $\mathcal F$ of $\text{Aut}(K:\mathbf Q(s))$ is $\mathbf Q(s)$ follows from the fact that $K:\mathbf Q(s)$ is a Galois extension. For we have $[K:\mathbf Q(s)]=|\text{Aut}(K:\mathbf Q(s))|=[K:\mathcal F]$. In the last step I have used the result that "If $K:F$ is an extension and $\mathcal F$ is the fixed field of $\text{Aut}(K:F)$ then we have $|\text{Aut}(K:F)|=[K:\mathcal F]$". Feb 24, 2016 at 8:08

$\newcommand{\Q}{\mathbf Q}$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.

• 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 $K$. Feb 24, 2016 at 10:39
• @GeoffRobinson Right. That is perhaps an even simpler way to say it. Feb 24, 2016 at 10:40

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!

• Paul what does "independence of fields" mean?
– WLOG
Dec 22, 2011 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. Feb 24, 2016 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. Feb 24, 2016 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. Feb 24, 2016 at 16:41