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### Linear Independence of Square Roots of Primes

```
Date: 11/07/96 at 07:17:49
From: Mats Oldin
Subject: Linear independence of square roots of primes

I need to find a way of proving that the square roots of a finite set
of different primes are linearly independent over the field of
rationals. I've tried to solve the problem using elementary algebra
and also using the theory of field extensions, without success. To
prove linear independence of two primes is easy but then my problems
arise. I would be very thankful for an answer to this question.

Best regards,

Mats Oldin
Sweden
```

```
Date: 11/11/96 at 14:37:15
From: Doctor Rob
Subject: Re: Linear independence of square roots of primes

Actually, something more general is true.  You need only assume that
the integers you are taking the square roots of are >1, squarefree and
pairwise relatively prime.  One proof goes as follows.

We will prove that as a vector space, the dimension of the field
extension gotten from Q, the field of rational numbers, by adjoining
the square roots of s integers > 1 which are squarefree and pairwise
relatively prime, is 2^s.  Proceed by induction on the number of such

We know it works for s = 1 and s = 2, as you state in your question.
Assume it works for 1, 2, ..., s roots.  Let n(1), n(2), ... n(s),
n(s+1) be such a set of s+1 integers.  Let us use the letters E and F
to represent the following fields:

E = Q(sqrt[n(1)], ..., sqrt[n(s-1)])
F = E(sqrt[n(s)])

By induction, dim(E) = 2^(s-1) and dim(F) = 2^s.  We would be done
with the proof if we could show that sqrt[n(s+1)] is not an element
of F.

Assume otherwise, that sqrt[n(s+1)] is in F.  Then for some a, b in E,
we can write:

sqrt[n(s+1)] = a + b*sqrt[n(s)]

Squaring both sides gives:

n(s+1) = a^2 + 2*a*b*sqrt[n(s)] + b^2*n(s)
or
2*a*b*sqrt[n(s)] = n(s+1) - a^2 - b^2*n(s)

The righthand side lies in E.  Three cases are possible:

1) a = 0.
2) b = 0.
3) sqrt[n(s)] is in E.

Case 1:  a = 0.  Then sqrt[n(s+1)] = b*sqrt[n(s)].  This implies that
sqrt[n(s)*n(s+1)] = b*n(s) lies in E.  Then the set of s integers
n(1), n(2), ..., n(s-1), n(s)*n(s+1) satisfies the induction
hypothesis, so the dimension of this field extension must be 2^s.
On the other hand, this field extension must be exactly E, whose
dimension is 2^(s-1), a clear contradiction.  Thus this case is
impossible.

Case 2:  b = 0.  Then sqrt[n(s+1)] = a lies in E, and the set of
s integers n(1), n(2), ..., n(s-1), n(s+1) satisfies the induction
hypothesis, so the dimension of this field extension must be 2^s.
On the other hand, this field extension must be exactly E, whose
dimension is 2^(s-1), again a contradiction.  Thus this case is
impossible.

Case 3:  sqrt[n(s)] is in E.  Then F = E, which is also a
contradiction, since the dimension of F is twice that of E.
Thus this case, too, is impossible.

The conclusion is that sqrt[n(s+1)] cannot lie in F, so we are done.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra

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