Re: adjoining elements to rings
From: Arturo Magidin (magidin_at_math.berkeley.edu)
Date: 11/28/04
- Next message: Mensanator: "Re: Delphi or VB Column generation ?"
- Previous message: Richard Henry: "Re: JSH: Being me"
- In reply to: Jing M Lim: "adjoining elements to rings"
- Next in thread: Jing M Lim: "Re: adjoining elements to rings"
- Reply: Jing M Lim: "Re: adjoining elements to rings"
- Reply: Jing M Lim: "Re: adjoining elements to rings"
- Messages sorted by: [ date ] [ thread ]
Date: Sun, 28 Nov 2004 22:45:21 +0000 (UTC)
In article <1101672501.691290.38400@f14g2000cwb.googlegroups.com>,
Jing M Lim <jingobingo@gmail.com> wrote:
>What happens when we adjoin the inverse of 2 to the ring Z_12 i.e. what
>is the field Z_12[x]/(2x-1) isomorphic to? I have reason to believe
>that it is simply the field of 3 elements, because of the following:
>
>In Z_12[x]/(2x-1) we have that 12=0 and 2x-1=0. Therefore 12x-6=0 ==>
>6=0. Similarly we have 6x-3=0 ==> 3=0. This makes me believe that we
>are talking about F_3 here.
>
>However, I cannot prove this. I tried setting up the following maps:
>
>j:Z ---> Z_12
>
>pi:Z_12 ---> Z_12[x]/(2x-1)
>
>i:Z ---> Z_3
>
>If ker i = ker (pi o j) then we have that Z_12[x]/(2x-1) is isomorphic
>to Z_3 by the first isomorphism theorem. I know that
>
>ker j = 12Z
>ker pi = (2x-1)
>ker i = 3Z
>
>But for some reason I cannot compute the kernel of pi o j. Is this even
>the right track to go on?
Well, you know that the kernel contains (3). Since (3) is a maximal
ideal of Z, the only possibilities are for the kernel to be all of Z,
or else to be just equal to (3). So the question then becomes: is 1 in
the kernel?
This will happen if and only if 1 = 0 (mod (2x-1)) in Z_{12}[x]. That
is, if and only if there exists a polynomial f(x) in Z_{12}[x] such
that f(x)(2x-1) = 1.
Write f(x) = a_n*x^n + ... + a_1*x + a_0 with a_i in Z. Then
(2x-1)f(x) = 2a_n*x^{n+1} + (2a_{n-1}-a_n)x^n + ... (2a_0-a_1)x - a_0.
So you must have a_0 = - 1 (mod 12)
2a_{i-1} - a_i = 0 (mod 12) for i=1,...,n
2a_n = 0 (mod 12).
So, from a_0 = -1, we have -2 - a_1 = 0 (mod 12), or a_1 = -2.
Then 2a_1 - a_2 = 0 (mod 12), so -4 = a_2 (mod 12).
2a_2 - a_3 = 0 (mod 12) so a_3 = -8 (mod 12).
Continuing in this way, a_{i} = -2^i (mod 12).
Since we also need 2a_n = 0 (mod 12), that means that -2^{n+1}=0 (mod
12). But this never happens. So no such polynomial exists. Therefore,
1 is not in the kernel of the map. Since the kernel already contains
3, and is not everything, the kernel of pi o j must be (3).
HOWEVER: you have not proven that Z_{12}[x]/(2x-1) is isomorphic to
F_3; you've only shown that it CONTAINS F_3; you would also have to
prove that the induced map from Z is surjective in order to prove
that. (I mean, what if it is some other field of characteristic 3?)
Slightly easier: you have already shown that 3 is in (2x-1), so the
quotient Z_{12}[x]/(2x-1) factors through Z_{3}[x]/(2x-1). And since
Z_3[x]/(2x-1) is isomorphic to Z_3, you are done.
>Similarly I am trying to describe Z[i]/(2+i).
This is a bit simpler since Z[i] is not only a domain, but a UFD.
> Since 2+i=0 we must have
>5=0, which leads me to believe that Z[i]/(2+i) is simply Z_5. I know
>that 2+i is prime in Z[i] and therefore the quotient should be a field.
It's maximal, and therefore the quotient should be a field (prime and
maximal are equivalent in Z[i], but not in general).
>I tried a similar approach to the above but did not get far.
Yes, it is isomorphic to Z/5Z. Certainly, it is a field a
characteristic 5, since a similar approach to the above will establish
that the kernel of the map Z -> Z[i] -> Z[i]/(2+i) is just (5), by
noting that 1 is not in (2+i). But you would still have to show that
this map is surjective, which is not too hard: you can easily show
that every element of Z[i] can be written as (a+bi)(2+i) + r, with r =
0, 1, 2, 3, or 4.
This because 5 = (2+i)(2-i). So, given x+yi, with x and y in Z, then
we have x-2y = (x+yi) - y(2+i). Now let x-2y = 5t + r, with t an
integer, r=0,1,2,3, or 4; Then
x - 2y = (2t-ti)(2+i) + r
So
(2t-ti)(2+i) + r = (x+yi)-y(2+i)
hence
(x+yi) = (2t+y - ti)(2+i) + r.
Thus, every element in Z[i]/(2+i) is congruent modulo 2+i to 0, 1, 2,
3, 4, or 5, so the map Z->Z[i]->Z[i]/(2+i) is surjective, kernel (5),
and you are done.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
- Next message: Mensanator: "Re: Delphi or VB Column generation ?"
- Previous message: Richard Henry: "Re: JSH: Being me"
- In reply to: Jing M Lim: "adjoining elements to rings"
- Next in thread: Jing M Lim: "Re: adjoining elements to rings"
- Reply: Jing M Lim: "Re: adjoining elements to rings"
- Reply: Jing M Lim: "Re: adjoining elements to rings"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
