# Algebra, Problems For Fun (21)

• Jun 20th 2009, 04:11 PM
NonCommAlg
Algebra, Problems For Fun (21)
Let $\displaystyle K$ be a field, $\displaystyle \sigma \in \text{Aut}(K),$ and $\displaystyle x$ an indeterminate. Let $\displaystyle R=\{a_nx^n + a_{n-1}x^{n-1} + \cdots + a_mx^m : \ a_j \in K, \ n \in \mathbb{Z} \}.$ Define addition and multiplication in $\displaystyle R$ exactly the same as what we have

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

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

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

Problem: Find the center of $\displaystyle R.$

Warning:

Spoiler:
The center depends on whether $\displaystyle o(\sigma) = \infty$ or $\displaystyle o(\sigma) < \infty.$
• Jun 22nd 2009, 01:46 AM
Swlabr
If $\displaystyle b_1x^r+b_{r+1}x^{r+1} + \ldots + b_sx^s \in Z(R)$ then for all $\displaystyle a \in K, n \in \mathbb{Z}$ we have that $\displaystyle 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} = $$\displaystyle 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 \displaystyle <x> \cong C_{\infty} is a group we have that all the terms pair off, and so we have that \displaystyle b_t \sigma^t(a) = a\sigma^n(b_t) for all \displaystyle a \in K, n \in \mathbb{Z}. Let \displaystyle P=\{b \in K : \sigma(b) = b\}. If \displaystyle o(\sigma) = m then let \displaystyle P_m := <a*x^{im}:a \in P, i \in \mathbb{Z} >. It is clear that both \displaystyle P \leq Z(R) and \displaystyle P_m \leq Z(R) as they are invarient under automorphisms and are subfields of the ring. So, we wish to show that \displaystyle Z(R) \leq P_{o(\sigma)} or if \displaystyle o(\sigma) = \infty, \displaystyle Z(R) \leq P. That is to say, \displaystyle b_t \in P and either \displaystyle n | o(\sigma) or \displaystyle n=0. Rearranging \displaystyle a \sigma^n(b_t) = b_t \sigma^t(a) we get that \displaystyle \sigma(b_t)\sigma(a^{-1}) = \sigma^{-1}(b_t^{-1}a) = b_t^{-n}a^t. As \displaystyle a is arbitrary we have that \displaystyle \sigma^{n} (b_t) = b_t for all \displaystyle n \in K and \displaystyle \sigma(a^{-1}) = a^{t}. Thus, \displaystyle b_t is invariant under every single automorphism and \displaystyle a = \sigma^{t} (a) for all \displaystyle a \in K. Thus, \displaystyle t=0 or \displaystyle t | o(\sigma) (as \displaystyle \sigma^t=id_{Aut(R)}), and \displaystyle b_t \in P, and we are done...almost. It still needs to be shown what \displaystyle P is. My initial thought was that it was the prime subfield (thus, I denoted it \displaystyle P). However, if \displaystyle \sigma is trivial this is not the case. Thus, i suspect that the prime subfield is the intersection of all the \displaystyle P's (it is certainly contained in the intersection). What \displaystyle P actually is will depend on \displaystyle \sigma, and will either need someone with more knowledge of fields than I, or someone with a bigger brain... • Jun 22nd 2009, 03:17 AM NonCommAlg Quote: Originally Posted by Swlabr If \displaystyle b_1x^r+b_{r+1}x^{r+1} + \ldots + b_sx^s \in Z(R) then for all \displaystyle a \in K, n \in \mathbb{Z} we have that \displaystyle 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} =$$\displaystyle 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 $\displaystyle <x> \cong C_{\infty}$ is a group we have that all the terms pair off, and so we have that $\displaystyle b_t \sigma^t(a) = a\sigma^n(b_t)$ for all $\displaystyle a \in K, n \in \mathbb{Z}$.

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

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

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

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

notation. if $\displaystyle o(\sigma)=m < \infty,$ then the set $\displaystyle P_m,$ as you defined, cannot be equal to $\displaystyle Z(R)$ because it's not closed under addition. (Sleepy)
• Jun 22nd 2009, 03:25 AM
Swlabr
Quote:

Originally Posted by NonCommAlg
If $\displaystyle o(\sigma)=m < \infty,$ then the set $\displaystyle P_m,$ as you defined, cannot be equal to $\displaystyle Z(R)$ 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...
• Jun 22nd 2009, 06:57 AM
NonCommAlg
Quote:

Originally Posted by Swlabr
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: $\displaystyle Z(R)=P_m=\{a_n x^{nm} + a_{n-1}x^{(n-1)m} + \cdots + a_rx^{rm}: \ a_j \in K^{\sigma}, \ n \in \mathbb{Z} \}.$ the standard notation for $\displaystyle P_m$ is $\displaystyle K^{\sigma}[x^m,x^{-m}].$