Results 1 to 2 of 2
Like Tree1Thanks
  • 1 Post By Deveno

Math Help - isomorphic

  1. #1
    Member
    Joined
    Jan 2011
    Posts
    83
    Thanks
    1

    isomorphic

    (i)Q[x]/<x^2+1> is    isomorphic to   Q[x]/<x^2+x+1>
    (ii)R[x]/<x^2+1> is    isomorphic to  R[x]/<x^2+x+1>

    either of the statement is correct. justify your answer.

    i think (i) is correct, but i don't know the reason, can you explain please.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,383
    Thanks
    750

    Re: isomorphic

    let's look at the second one, first.

    the coset of x in R[x]/<x2+1>, x+<x2+1> has the following property:

    writing [x] for x+<x2+1>, we get:

    [x]2 + [1] = [x2+1] = [0] (since x2+1 is clearly in <x2+1>).

    this means that [x]2 = [-1].

    identifying R with the isomorphic copy of it inside R[x]/<x2+1> (via the isomorphism a→[a] = a + <x2+1>), we have:

    [x]2 = -1.

    now all cosets of R[x]/x2+1> are of the form:

    [a+bx] (= a + bx + <x2+1>), for a,b in R. if we write [x] = i, then:

    [a+bx] = [a] + [b][x] = a + [x]b = a + ib, and these cosets obey the same algebraic rules as the complex numbers. so R[x]<x2+1> is isomorphic to C = R(i).

    in the same manner, R[x]/<x2+x+1> is isomorphic to R(ω), where ω = -1/2 + i(√3/2)

    (note that ω2+ω+1 = (-1/2 + i(√3/2))2 + (-1/2 + i(√3/2)) + 1

    = (1/4 - 3/4) + i[2(-1/2)(√3/2)] -1/2 + i(√3/2) + 1 = -1/2 - i(√3/2) - 1/2 + i(√3/2) + 1 = 1 - 1 + i0 = 0 + i0 = 0).

    now ω is a complex number, so R(ω) is clearly contained in C = R(i).

    but i = (2/√3)ω + √3, and since 2/√3 and √3 are real numbers, we see that R(i) is contained in R(ω), hence R(i) = R(ω).

    since both quotient rings are isomorphic to the same field (the complex numbers), they must be isomorphic to each other.

    now let's consider Q(i) and Q(ω). note that Q(ω) contains i√3 (= √(-3)), which is 2ω + 1.

    now Q(i) = {a + ib : a,b in Q}. can we find a,b in Q such that (a + ib)2 = -3? suppose we could:

    (a + ib)2 = -3
    (a2 - b2) + i(2ab) = -3 = -3 + i0.

    comparing sides, we must have that:

    2ab = 0
    a2 - b2 = -3

    so either a or b must be 0 (since Q is a field, and thus an integral domain). suppose b = 0.

    then a2 = -3, which is not possible, since Q is an ordered field, and a2 ≥ 0 > -3.

    thus it could only possibly be that a = 0, in which case:

    b2 = 3, implying that Q contains a square root of 3, which it does not.

    since Q(ω) has a root to x2 + 3, and Q(i) does not, they cannot be isomorphic fields

    (if x2 + 3 factors in Q(ω), then it must have a corresponding factorization in Q(i), if the 2 are isomorphic).
    Last edited by Deveno; May 13th 2012 at 10:48 AM.
    Thanks from saravananbs
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. isomorphic
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 14th 2011, 06:14 PM
  2. is U14 isomorphic to U18?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: August 21st 2010, 09:52 AM
  3. Isomorphic
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 13th 2008, 08:55 PM
  4. Isomorphic or not?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 13th 2008, 04:41 PM
  5. Isomorphic?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 15th 2007, 04:42 PM

Search Tags


/mathhelpforum @mathhelpforum