## 305 -Extension fields revisited (3)

April 15, 2009

1. Isomorphisms

We return here to the quotient ring construction. Recall that if ${R}$ is a commutative ring with identity and ${I}$ is an ideal of ${R,}$ then ${R/I}$ is also a commutative ring with identity. Here, ${R/I=\{[a]_\sim:a\in R\},}$ where ${[a]_\sim=\{b:a\sim b\}}$ for ${\sim}$ the equivalence relation defined by ${a\sim b}$ iff ${a-b\in I.}$

Since ${\sim}$ is an equivalence relation, we have that ${[a]_\sim=[b]_\sim}$ if ${a\sim b}$ and ${[a]_\sim\cap[b]_\sim=\emptyset}$ if ${a\not\sim b.}$ In particular, any two classes are either the same or else they are disjoint.

In case ${R={\mathbb F}[x]}$ for some field ${{\mathbb F},}$ then ${I}$ is principal, so ${I=(p)}$ for some ${p\in{\mathbb F}[x],}$ i.e., given any polynomial ${q\in{\mathbb F}[x],}$ ${[q]_\sim=0}$ iff ${p|q}$ and, more generally, ${[q]_\sim=[r]_\sim}$ (or, equivalently, ${q\sim r}$ or, equivalently, ${r\in[q]_\sim}$) iff ${p|(q-r).}$

In this case, ${{\mathbb F}[x]/(p)}$ contains zero divisors if ${p}$ is nonconstant but not irreducible.

If ${p}$ is 0, ${{\mathbb F}[x]/(p)\cong{\mathbb F}.}$

If ${p}$ is constant but nonzero, then ${{\mathbb F}[x]/(p)\cong\{0\}.}$

Finally, we want to examine what happens when ${p}$ is irreducible. From now on suppose that this is the case.

At the end of last lecture we arrived at the question of whether every finite field is a ${\mathbb Z}_p$ for some prime $p.$
Lemma 13. Suppose that ${\mathbb F}$ is a finite field. Then there is some natural number $n>0$ such that the sum of $n$ ones vanishes, $1+\dots+1=0.$ The least such $n$ is a prime that divides the size of the field