## 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 -Extension fields revisited (2)

April 15, 2009

Most of our work from now on depends on the following simple, but very useful observation:

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 ${{\mathbb F}}$.

## 305 -7. Extension fields revisited

April 3, 2009

1. Greatest common divisors.

Let’s conclude the discussion from last lecture.

If ${{\mathbb F}}$ is a field and ${p(x),q(x)\in{\mathbb F}[x]}$ are nonzero, then we can find polynomials ${\alpha(x),\beta(x)\in{\mathbb F}[x]}$ such that ${\alpha p+\beta q}$ is a gcd of ${p}$ and ${q.}$

To see this, consider ${{\mathcal A}=\{{\rm deg}(a(x)):0\ne a(x)\in{\mathbb F}[x]}$ and for some polynomials ${\alpha,\beta\in{\mathbb F}[x],}$ we have ${a=\alpha p+\beta q\}.}$

We see that ${{\mathcal A}\ne\emptyset,}$ because both ${p}$ and ${q}$ are nonzero linear combinations of ${p}$ and ${q,}$ so their degrees are in ${{\mathcal A}.}$ Each element of ${{\mathcal A}}$ is a natural number because ${{\rm deg}(a)=-\infty}$ only for ${a=0.}$ By the well-ordering principle, there is a least element of ${{\mathcal A}.}$

Let ${n}$ be this least degree, and let ${g=\alpha p+\beta q}$ have degree ${n.}$

First, if ${s\in{\mathcal F}[x]}$ and ${s|p,q}$ then ${s|\alpha p+\beta q,}$ so ${s|g.}$

Second, by the division algorithm, we can write ${p=gm+r}$ for some polynomials ${m,r\in{\mathbb F}[x]}$ with ${{\rm deg}(r)<{\rm deg}(g).}$ Then ${r=p-gm=(1-\alpha m)p+(-\beta m)q}$ is a linear combination of ${p,q.}$ Since ${{\rm deg}(r)<{\rm deg}(g),}$ and ${n={\rm deg}(g)}$ is the smallest number in ${{\mathcal A},}$ it follows that ${{\rm deg r}=-\infty,}$ i.e., ${r=0.}$ This is to say that ${p=gm,}$ so ${g|p.}$ Similarly, ${g|q.}$

It follows that ${g}$ is a greatest common divisor of ${p,q.}$

Since any other greatest common divisor of ${p,q}$ is ${ig}$ for some unit ${i,}$ it follows that any gcd of ${p}$ and ${q}$ is a linear combination of ${p}$ and ${q.}$

Notice that this argument is very similar to the proof of the same result for ${{\mathbb Z}.}$

## 305 -5. Extensions by radicals

March 4, 2009

(This post was typeset using Luca Trevisan‘s LaTeX2WP program.)

Last lecture we characterized subfields and used the characterization to provide many new examples of fields. Now we start to explore systematically which subfields of the complex numbers are suitable to study the question of which polynomial equations can be solved. Read the rest of this entry »