Question

Let p be prime and let ${\mathbf{Z}}_{\mathbf{p}}\left(x\right)$be the field of quotients of the polynomial ring ${\mathbf{Z}}_{\mathbf{p}}\left[x\right]$. Show that ${\mathbf{Z}}_{\mathbf{p}}\left(x\right)$is an infinite field of characteristic $\mathbf{p}$.

Short answer

$Z_p\left(x\right)$ is an infinite field of characteristic p.

Step-by-step solution

Definition of field.

A field is a commutative ring with unity in which every nonzero element has a multiplication inverse.

Showing that Zp(x) is an infinite field of characteristic p

Let p be a prime and $Z_p\left(x\right)$be the field of quotient of the polynomial ring $Z_p\left[x\right]$. Since $1,\mathrm{x},{\mathrm{x}}^2.....{\mathrm{Z}}_{\mathrm{p}}\left[\mathrm{x}\right]$and $1,\mathrm{x},{\mathrm{x}}^2,.....$are linearly independent. Therefore $1,\mathrm{x},{\mathrm{x}}^2,.....$is an infinite field. Now to show $char{Z}_p\left(x\right)=p$ .

Any arbitrary element of $Z_p\left(x\right)$is of the form $\frac{q\left(x\right)}{g\left(x\right)}$where $q\left(x\right),g\left(x\right)\in Z_p\left[x\right]$. Since coefficient of $q\left(x\right),g\left(x\right)\in Z_p\left[x\right]$. Therefore $p.q\left(x\right)=0,p.g\left(x\right)=0$

This gives:

$p.\frac{q\left(x\right)}{g\left(x\right)}=0$for all $\frac{q\left(x\right)}{g\left(x\right)}\in Z_p\left(x\right)$

Therefore$char{Z}_p\left(x\right)\le p\left(1\right)$

Also each $a\in Z_p$belong to $Z_p\left(x\right)$. So, if $n.a=0$for any $n<p$. Then this will contradict the fact that $char{Z}_p\left(x\right)=p$.

This implies$char{Z}_p\left(x\right)>p\left(2\right)$

From 1 and 2:

$char{Z}_p\left(x\right)=p$

Therefore $Z_p\left(x\right)$is an infinite field of characteristic p.