ring+conditon=field

Printable View

January 2nd 2009, 06:18 PM
petter

ring+conditon=field

Let A be a ring with $p^n$ elements with characteristic p , $p$ prime, $n\geq 1$ , and $f\in\mathbb{Z}[X]$ a polynom of degree $n$ ireductibile over $\mathbb{Z}_p$ .If $f$ have a solution $a \in A$ then A is field .
January 2nd 2009, 11:52 PM
NonCommAlg

Quote:

Originally Posted by petter

Let A be a ring with $p^n$ elements with characteristic p , $p$ prime, $n\geq 1$ , and $f\in\mathbb{Z}[X]$ a polynom of degree $n$ ireductibile over $\mathbb{Z}_p$ .If $f$ have a solution $a \in A$ then A is field .

i'll suppose that the ring is unitary. since $\text{char} A = p,$ the map $[k] \hookrightarrow k.1_A$ is an embedding of $\mathbb{Z}_p$ into $A.$ let $\overline{f(x)}$ be the image of $f(x)$ under the natural homomorphism $\mathbb{Z}[x] \longrightarrow \mathbb{Z}_p[x].$

now let $\overline{f(a)}=0,$ where $a \in A.$ define the map $\nu: \mathbb{F}_{p^n}= \frac{\mathbb{Z}_p[x]}{<\overline{f(x)}>} \longrightarrow A$ by: $\nu(\overline{g(x)})=g(a), \ \forall g(x) \in \mathbb{Z}_p[x].$ it's easily seen that $\nu$ is a ring monomorphism. thus $\nu$ is an isomorphism

because $|\mathbb{F}_{p^n}|=|A|=p^n. \ \ \Box$
January 3rd 2009, 03:29 AM
petter

Can u give me a elementary solution please ?
January 3rd 2009, 06:58 PM
NonCommAlg

Quote:

Originally Posted by petter

Can u give me a elementary solution please ?

well, there's nothing advanced about your problem and my solution. so ... http://www.rnel.net/tutorials/upload...14-06_2044.gif