Not isomorphic, quadratic fields
Solomon Welsh
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.
Thanks very much.
SW
Arturo Magidin
> 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)).
--
Arturo Magidin
Solomon Welsh
"Arturo Magidin" <mag...@member.ams.org> wrote in message
news:71cec9d4-7d8d-4702...@k19g2000yqc.googlegroups.com...
On Nov 14, 6:07 am, "Solomon Welsh" <no-re...@solomon.com> wrote:
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.
Any pointers appreciated. Thanks again.
SW
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)).
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).
--
Arturo Magidin
Solomon Welsh
On Nov 14, 4:40 pm, "Solomon Welsh" <no-re...@solomon.com> wrote:
--
Arturo Magidin
Thanks for taking the time to clarify that for me. Much appreciated indeed.
SW.
William Elliot
>> 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.