## 305 -9. Groups

April 24, 2009

We want to define the notion of group that will be fundamental to determine which polynomials are solvable by radicals. This notion is very important and appears in every area of mathematics.

We motivate the definition through the example that most concerns us: automorphisms of fields. They are particular class of isomorphisms, so we begin with them.

Some of the arguments below have been discussed in previous lectures.

## 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 ${p\mid q}$ and, more generally, ${[q]_\sim=[r]_\sim}$ (or, equivalently, ${q\sim r}$ or, equivalently, ${r\in[q]_\sim}$) iff ${p\mid (q-r).}$

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.