Algebra, Problems For Fun (21)

**NonCommAlg** · June 20th 2009, 04:11 PM

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:

**Swlabr** · June 22nd 2009, 01:46 AM

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...

**NonCommAlg** · June 22nd 2009, 03:17 AM

Originally Posted by Swlabr

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.

**Swlabr** · June 22nd 2009, 03:25 AM

Originally Posted by NonCommAlg

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...

**NonCommAlg** · June 22nd 2009, 06:57 AM

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

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