El grupo al cual envías entradas es un grupo Usenet. Si envías mensajes a este grupo, cualquier usuario de Internet podrá ver tu dirección de correo electrónico
I would like to show that Q(sqrt(2)) is not isomorphic to Q(sqrt(3))--well, at least I believe this to be the case.
Here Q(sqrt(2)) = {a + b*sqrt(2) : a,b rational} and similarly for Q(sqrt(3)).
I've tried playing about to deduce a contradiction but nothing seems to work really. Any ideas? I'm sure it is relatively simple but it's not clicking.
On Nov 14, 6:07 am, "Solomon Welsh" <no-re...@solomon.com> wrote:
> Hello,
> I would like to show that Q(sqrt(2)) is not isomorphic to Q(sqrt(3))--well, > at least I believe this to be the case.
Indeed; in fact, Q(sqrt(d)) is isomorphic to Q(sqrt(f)) for integer squarefree d and f, both different from 1, if and only if d=f.
> Here Q(sqrt(2)) = {a + b*sqrt(2) : a,b rational} and similarly for > Q(sqrt(3)).
> I've tried playing about to deduce a contradiction but nothing seems to work > really. Any ideas? I'm sure it is relatively simple but it's not clicking.
Well, any isomorphism must fix Q pointwise. If something maps to sqrt (3), then it's square must be 3, so in fact you would have *equality*, not merely isomorphism. And...
(And then you can use this for the more general statement: if Q(sqrt (d)) is isomorphic to Q(sqrt(f)), then sqrt(f) must lie in Q(sqrt (d)).
Indeed; in fact, Q(sqrt(d)) is isomorphic to Q(sqrt(f)) for integer squarefree d and f, both different from 1, if and only if d=f.
Well, any isomorphism must fix Q pointwise. If something maps to sqrt (3), then it's square must be 3, so in fact you would have *equality*, not merely isomorphism. And...
(And then you can use this for the more general statement: if Q(sqrt (d)) is isomorphic to Q(sqrt(f)), then sqrt(f) must lie in Q(sqrt (d)).
-- Arturo Magidin
Thanks, Arturo. Your kind reminder that any isomorphism must fix Q pointwise helped me deduce that 2=3, an obvious contradiction, and thus I showed the two objects are not isomorphic.
Can I follow this up with a related question? Q(sqrt(2)) is isomorphic (again, so I believe) to Q(1-sqrt(8)). I wonder whether the following argument is sufficient:
Q(1-sqrt(8)) = Q(1-2sqrt(2)) = Q(2sqrt(2)) as 1 is rational = Q(sqrt(2)) as 2 is rational
If not, another idea I had was to write Q(1-sqrt(8)) = {a + b*(1-sqrt(8)) : a,b rational} and then set a= c+d/2 and b=-d/2 where c and d are two rational numbers but I feel forming a relationship between a and b in this way might have its consequences.
On Nov 14, 4:40 pm, "Solomon Welsh" <no-re...@solomon.com> wrote:
> Thanks, Arturo. Your kind reminder that any isomorphism must fix Q > pointwise helped me deduce that 2=3, an obvious contradiction, and thus I > showed the two objects are not isomorphic.
> Can I follow this up with a related question? Q(sqrt(2)) is isomorphic > (again, so I believe) to Q(1-sqrt(8)).
Yes. Q(sqsrt(2)) is *equal* to Q(1-sqrt(8)).
> I wonder whether the following > argument is sufficient:
> Q(1-sqrt(8)) = Q(1-2sqrt(2)) > = Q(2sqrt(2)) as 1 is rational > = Q(sqrt(2)) as 2 is rational
Yes: you are showing that any field that contains Q and 1-sqrt(8) must contain sqrt(2) and vice-versa. That gives the double inclusion.
> If not, another idea I had was to write Q(1-sqrt(8)) = {a + b*(1-sqrt(8)) : > a,b rational} and then set a= c+d/2 and b=-d/2 where c and d are two > rational numbers but I feel forming a relationship between a and b in this > way might have its consequences.
Huh?
If you want to show explicitly that
{a + b*(1-sqrt(8)) : a,b in Q}
and
{c + d*sqrt(2) : c,d in Q}
are equal, you can do that by showing that each choice of a and b corresponds to a choice of c and d, and vice versa. Given c+d*sqrt(2), if you set a = c + (d/2), b = -(d/2), then you have that
a + b*(1-sqrt(8)) = c+(d/2) -(d/2) +(d/2)2*sqrt(2) = c + d*sqrt(2), which gives you one inclusion.
If you are *given* a and b and they are fixed, then from a = c+(d/2), b=-(d/2) you can solve for c and d to get the other equality. This does not "form[] a relationship between a and b"; a and b are *given and fixed*. What you are doing is putting conditions on c and d (which may or may not be inconsistent).
On Sat, 14 Nov 2009, Arturo Magidin wrote: >> I would like to show that Q(sqrt(2)) is not isomorphic to Q(sqrt(3))--well, >> at least I believe this to be the case.
> Indeed; in fact, Q(sqrt(d)) is isomorphic to Q(sqrt(f)) for integer > squarefree d and f, both different from 1, if and only if d=f.
Assume h:Q(sqr n) -> Q(sqr m) an isomorphism, n,m in N\1. For all a in Q, h(a) = a. Some a with h(a) = sqr m. h(a^2) = h(a)^2 = m = h(m); a^2 = m
Some r,s in Q with (r + s.sqr n)^2 = m. r^2 + ns^2 + 2rs.sqr n = m If s = 0, then sqr m is rational. If r = 0, then m/n is rational. If rs /= 0, then sqr n is rational.
|
