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Topic: Independent sqaure roots???
Replies: 17   Last Post: May 1, 2003 9:50 AM

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 Robin Chapman Posts: 3,289 Registered: 12/6/04
Re: Independent sqaure roots???
Posted: Apr 30, 2003 10:28 AM
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David C. Ullrich wrote:

> On Wed, 30 Apr 2003 13:55:33 +0100, Robin Chapman
> <rjc@ivorynospamtower.freeserve.co.uk> wrote:
>

>>David C. Ullrich wrote:
>>

>>> Are the square roots of the positive square-free integers
>>> (including 1) linearly independent over Q?

>>
>>Yes.

>
> Thanks.
>

>>This boils down to showing that K_n = Q(sqrt{p_1},...sqrt{p_n})
>>(p_n is the n-th prime) has degree 2^n over Q. But K_n/K_{n-1}
>>has degree 2 -- p_n is ramified in K_n by not in K{n-1}.

>
> Well, I was afraid someone would say something like that...<g>.
>
> Let's see:
>

>>This boils down to showing that K_n = Q(sqrt{p_1},...sqrt{p_n})
>>(p_n is the n-th prime) has degree 2^n over Q.

>
> If I'm interpreting the notation correctly, I'd probably see why
> this boiled down to that if I knew that Q(sqrt{p_1},...sqrt{p_n})
> = Q[sqrt{p_1},...sqrt{p_n}]. Which was not immediately clear
> to me, but it's true, right? Because given a/b where a and b
> are in Q[sqrt{p_1},...sqrt{p_n}] you can rationalize the
> denominator by writing b = c + d, where c is all terms containing
> p_n and d is the terms not containing p_n; then multiplying
> by (c-d)/(c-d) gets you down to something with only
> p_1, ... p_{n-1} in the denominator, and you repeat until
> done. Yes?

Oh yes. All standard field theory: if a is algebraic over a field K
then K[a] = K(a) is a field.

>>But K_n/K_{n-1}
>>has degree 2 -- p_n is ramified in K_n by not in K{n-1}.

>
> At first glance it looked like you were saying the degree
> was 2 - p_n. I take it the "--" was a whatsis, more or
> less the same as a semicolon.

It's known as a "dash".

>Have no idea what
> "ramified" means (and the "by" was supposed to
> be "but"???) but I may be able to figure out why
> K_n/K_{n-1} has degree 2 anyway...

Ah well, a prime p is ramified in a number field K if when
you split p into prime ideals of (the ring of integers of) K
then some factor is repeated. Once you are familiar enough with
this stuff then it's really obvious that K_{n-1} is unramified at p_n
but K_n is ramified (so K_n =/= K_{n-1}).

But how about this. Let p = p_n be an odd prime. Let Q_p be the
p-adic numbers and L = Q_p(sqrt(u)) where u is some quadratic
nonresidue mod p. Then show
(i) if m is an integer prime to p then m has a square root in L,
(ii) p doesn't have a square root in L.
Then it's plain that K_{n-1} embeds in L but K_n doesn't.
Bingo! K_n =/= K_{n-1}.

--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"His mind has been corrupted by colours, sounds and shapes."
The League of Gentlemen

Date Subject Author
4/30/03 David C. Ullrich
4/30/03 David C. Ullrich
4/30/03 Robin Chapman
4/30/03 David C. Ullrich
4/30/03 Robin Chapman
4/30/03 David C. Ullrich
4/30/03 fourierr
4/30/03 Robin Chapman
4/30/03 fourierr
4/30/03 Christopher J. Henrich
4/30/03 David C. Ullrich
4/30/03 fourierr
5/1/03 David C. Ullrich
4/30/03 fourierr
5/1/03 David C. Ullrich
4/30/03 Chris Hughes
4/30/03 Paul Pollack
4/30/03 Gerry Myerson

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