# Units

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

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

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

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

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

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

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

Originally Posted by chiph588@

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

i don't think you can make this conclusion!

well, if $(a_nx^n + \cdots + a_0)(b_mx^m + \cdots + b_0)=1, \ a_n \neq 0, \ b_m \neq 0,$ and n > 0 or m > 0, then we must have $a_nb_m=0,$ 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 $R,$ the set of units of $R[x]$ is exactly equal to the set of units of $R.$

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