# ring+conditon=field

• Jan 2nd 2009, 06:18 PM
petter
ring+conditon=field
Let A be a ring with $\displaystyle p^n$ elements with characteristic p , $\displaystyle p$ prime, $\displaystyle n\geq 1$, and $\displaystyle f\in\mathbb{Z}[X]$ a polynom of degree $\displaystyle n$ ireductibile over $\displaystyle \mathbb{Z}_p$.If $\displaystyle f$ have a solution $\displaystyle a \in A$ then A is field .
• Jan 2nd 2009, 11:52 PM
NonCommAlg
Quote:

Originally Posted by petter

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

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

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

because $\displaystyle |\mathbb{F}_{p^n}|=|A|=p^n. \ \ \Box$
• Jan 3rd 2009, 03:29 AM
petter
Can u give me a elementary solution please ?
• Jan 3rd 2009, 06:58 PM
NonCommAlg
Quote:

Originally Posted by petter

Can u give me a elementary solution please ?