## 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, ${R/I={[a]_sim:ain R},}$ where ${[a]_sim={b:asim b}}$ for ${sim}$ the equivalence relation defined by ${asim b}$ iff ${a-bin I.}$

Since ${sim}$ is an equivalence relation, we have that ${[a]_sim=[b]_sim}$ if ${asim b}$ and ${[a]_simcap[b]_sim=emptyset}$ if ${anotsim b.}$ 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 ${pin{mathbb F}[x],}$ i.e., given any polynomial ${qin{mathbb F}[x],}$ ${[q]_sim=0}$ iff ${p|q}$ and, more generally, ${[q]_sim=[r]_sim}$ (or, equivalently, ${qsim r}$ or, equivalently, ${rin[q]_sim}$) 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.

Theorem 1 Let ${{mathbb F}:{mathbb K}}$ be a field extension. Then ${{mathbb F}}$ with its usual addition is a vector space over ${{mathbb K},}$ where multiplication of elements of ${{mathbb F}}$ by elements of ${{mathbb K}}$ is just the usual product of ${{mathbb F}}$.