Results 1 to 5 of 5

Math Help - Algebra, Problems For Fun (21)

  1. #1
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Algebra, Problems For Fun (21)

    Let K be a field, \sigma \in \text{Aut}(K), and x an indeterminate. Let R=\{a_nx^n + a_{n-1}x^{n-1} + \cdots + a_mx^m : \ a_j \in K, \ n \in \mathbb{Z} \}. Define addition and multiplication in R exactly the same as what we have

    in the ordinary polynomials but with this extra rule that xa=\sigma(a) x, for all a \in K. So, for example (x^2 -2x + x^{-1})(3x)=\sigma^2(3)x^3-2\sigma(3)x^2+\sigma^{-1}(3).

    It's easy to see that R is a ring with identity and that R is commutative if and only if \sigma=\text{id}_K. The ring R is called the ring of Laurent polynomials with the twist \sigma. A standard notation for R is

    K[x,x^{-1}, \sigma] or K[x^{\pm 1}, \sigma]. These rings are very important in ring theory and noncommutative algebraic geometry.


    Problem: Find the center of R.


    Warning:

    Spoiler:
    The center depends on whether o(\sigma) = \infty or o(\sigma) < \infty.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    If b_1x^r+b_{r+1}x^{r+1} + \ldots + b_sx^s \in Z(R) then for all a \in K, n \in \mathbb{Z} we have that b_r \sigma^r(a)x^{n+r} + b_{r+1} \sigma^{r+1}(a)x^{n+r+1} + \ldots + b_s \sigma^s(a)x^{s+n} = a \sigma^n(b_r)x^{n+r} + a \sigma^{n}(b_{r+1})x^{n+r+1} + \ldots + a \sigma^n(b_{s})x^{s+n}. As <x> \cong C_{\infty} is a group we have that all the terms pair off, and so we have that b_t \sigma^t(a) = a\sigma^n(b_t) for all a \in K, n \in \mathbb{Z}.

    Let P=\{b \in K : \sigma(b) = b\}. If o(\sigma) = m then let P_m := <a*x^{im}:a \in P, i \in \mathbb{Z} >. It is clear that both P \leq Z(R) and P_m \leq Z(R) as they are invarient under automorphisms and are subfields of the ring.

    So, we wish to show that Z(R) \leq P_{o(\sigma)} or if o(\sigma) = \infty, Z(R) \leq P. That is to say, b_t \in P and either n | o(\sigma) or n=0.

    Rearranging a \sigma^n(b_t) = b_t \sigma^t(a) we get that  \sigma(b_t)\sigma(a^{-1}) = \sigma^{-1}(b_t^{-1}a) = b_t^{-n}a^t. As a is arbitrary we have that \sigma^{n} (b_t) = b_t for all n \in K and \sigma(a^{-1}) = a^{t}. Thus,  b_t is invariant under every single automorphism and  a = \sigma^{t} (a) for all a \in K. Thus, t=0 or t | o(\sigma) (as \sigma^t=id_{Aut(R)}), and b_t \in P, and we are done...almost.

    It still needs to be shown what P is. My initial thought was that it was the prime subfield (thus, I denoted it P). However, if \sigma is trivial this is not the case. Thus, i suspect that the prime subfield is the intersection of all the P's (it is certainly contained in the intersection). What P actually is will depend on \sigma, and will either need someone with more knowledge of fields than I, or someone with a bigger brain...
    Last edited by Swlabr; June 22nd 2009 at 03:23 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Swlabr View Post
    If b_1x^r+b_{r+1}x^{r+1} + \ldots + b_sx^s \in Z(R) then for all a \in K, n \in \mathbb{Z} we have that b_r \sigma^r(a)x^{n+r} + b_{r+1} \sigma^{r+1}(a)x^{n+r+1} + \ldots + b_s \sigma^s(a)x^{s+n} = a \sigma^n(b_r)x^{n+r} + a \sigma^{n}(b_{r+1})x^{n+r+1} + \ldots + a \sigma^n(b_{s})x^{s+n}. As <x> \cong C_{\infty} is a group we have that all the terms pair off, and so we have that b_t \sigma^t(a) = a\sigma^n(b_t) for all a \in K, n \in \mathbb{Z}.

    Let P=\{b \in K : \sigma(b) = b\}. If o(\sigma) = m then let P_m := \{a*x^{im}:a \in P, i \in \mathbb{Z} \}. It is clear that both P \leq Z(R) and P_m \leq Z(R) as they are invarient under automorphisms and are subfields of the ring.

    So, we wish to show that Z(R) \leq P_{o(\sigma)} or if o(\sigma) = \infty, Z(R) \leq P. That is to say, b_t \in P and either n | o(\sigma) or n=0.

    Rearranging a \sigma^n(b_t) = b_t \sigma^t(a) we get that  \sigma(b_t)\sigma(a^{-1}) = \sigma^{-1}(b_t^{-1}a) = b_t^{-n}a^t. As a is arbitrary we have that \sigma^{n} (b_t) = b_t for all n \in K and \sigma(a^{-1}) = a^{t}. Thus,  b_t is invariant under every single automorphism and  a = \sigma^{t} (a) for all a \in K. Thus, t=0 or t | o(\sigma) (as \sigma^t=id_{Aut(R)}), and b_t \in P, and we are done...almost.

    It still needs to be shown what P is. My initial thought was that it was the prime subfield (thus, I denoted it P). However, if \sigma is trivial this is not the case. Thus, i suspect that the prime subfield is the intersection of all the P's (it is certainly contained in the intersection). What P actually is will depend on \sigma, and will either need someone with more knowledge of fields than I, or someone with a bigger brain...
    that's a good work! to make your job easier, note that u \in Z(R) if and only if ua=au, \ ux=xu, for all a \in K. as you said, if o(\sigma)=\infty, then Z(R)=K^{\sigma}, the fixed field of \sigma, or P with your

    notation. if o(\sigma)=m < \infty, then the set P_m, as you defined, cannot be equal to Z(R) because it's not closed under addition.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by NonCommAlg View Post
    If o(\sigma)=m < \infty, then the set P_m, as you defined, cannot be equal to Z(R) because it's not closed under addition.
    Sorry-that should be the set generated by the set that I gave. I've edited it, putting in the triangular brackets...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Swlabr View Post
    Sorry-that should be the set generated by the set that I gave. I've edited it, putting in the triangular brackets...
    wel, it's probably better to write: Z(R)=P_m=\{a_n x^{nm} + a_{n-1}x^{(n-1)m} + \cdots + a_rx^{rm}: \ a_j \in K^{\sigma}, \ n \in \mathbb{Z} \}. the standard notation for P_m is K^{\sigma}[x^m,x^{-m}].
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Algebra 2 problems.
    Posted in the Algebra Forum
    Replies: 3
    Last Post: July 29th 2009, 10:50 AM
  2. Algebra, Problems For Fun (22)
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: June 20th 2009, 11:59 PM
  3. Algebra, Problems For Fun (13)
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: June 5th 2009, 02:53 PM
  4. Algebra problems
    Posted in the Algebra Forum
    Replies: 6
    Last Post: February 18th 2008, 06:29 PM
  5. Need Help With All these Algebra Problems.
    Posted in the Algebra Forum
    Replies: 10
    Last Post: November 26th 2006, 05:42 PM

Search Tags


/mathhelpforum @mathhelpforum