## 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.

## 305 -Fields (3)

February 18, 2009

At the end of last lecture we arrived at the question of whether every finite field is a ${\mathbb Z}_p$ for some prime $p.$

In this lecture we show that this is not the case, by exhibiting a field of 4 elements. We also find some general properties of finite fields. Finite fields have many interesting applications (in cryptography, for example), but we will not deal much with them as our focus through the course is on number fields, that we will begin discussing next lecture.

We begin by proving the following result:

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