# Units

• Apr 21st 2009, 11:03 AM
mathman88
Units
Show that $\displaystyle \left( \mathbb{Z}_{[x]} \right)^* = \{-1,1\}$
• Apr 21st 2009, 04:20 PM
chiph588@
Suppose $\displaystyle a_n x^n + \cdots + a_1 x + a_0 \in \mathbb{Z} [x]$ is a unit.

Then there exists $\displaystyle p(x)$ s.t. $\displaystyle (a_n x^n + \cdots + a_1 x + a_0)p(x) = 1$. It is evident $\displaystyle p(x) \neq 0$.

$\displaystyle (a_n p(x)) x^n + \cdots + (a_1 p(x)) x + a_0 p(x) = 1$

Therefore $\displaystyle a_i p(x) = 0 \; 1 \leq i \leq n$ and $\displaystyle a_0 p(x) = 1$.

Since $\displaystyle \mathbb{Z} [x]$ is an integral domain and $\displaystyle p(x) \neq 0$, $\displaystyle a_i = 0 \; 1 \leq i \leq n$.

The only way for $\displaystyle a_0 p(x) = 1$ is if $\displaystyle p(x) = c$, where $\displaystyle c \in \mathbb{Z}$. So $\displaystyle a_0 c = 1 \Longleftrightarrow a_0 = 1 , \; c = 1$ or $\displaystyle a_0 = -1 , \; c = -1$

Hence $\displaystyle a_n x^n + \cdots + a_1 x + a_0 = -1$ or $\displaystyle 1$. Which means $\displaystyle \left( \mathbb{Z} [x] \right)^* = \{-1,1\}$.
• Apr 21st 2009, 09:42 PM
NonCommAlg
Quote:

Originally Posted by chiph588@

Therefore $\displaystyle a_i p(x) = 0 \; 1 \leq i \leq n$ and $\displaystyle a_0 p(x) = 1$.

i don't think you can make this conclusion!

well, if $\displaystyle (a_nx^n + \cdots + a_0)(b_mx^m + \cdots + b_0)=1, \ a_n \neq 0, \ b_m \neq 0,$ and n > 0 or m > 0, then we must have $\displaystyle a_nb_m=0,$ which is not possible in an integral domain. so n = m = 0, and the rest is

obvious. this also shows that, in general, for any integral domain $\displaystyle R,$ the set of units of $\displaystyle R[x]$ is exactly equal to the set of units of $\displaystyle R.$

Remark: there's a well-known result that in any commutative ring with unity $\displaystyle R,$ a polynomial $\displaystyle \sum_{k=0}^n a_kx^k \in R[x]$ is a unit iff $\displaystyle a_0$ is a unit of $\displaystyle R$ and all $\displaystyle a_k, \ k \neq 0,$ are nilpotent.