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 ${s\in{\mathbb F}[x]}$ and ${s\mid p,q}$ then ${s\mid \alpha p+\beta q,}$ so ${s\mid g.}$

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 ${g\mid p.}$ Similarly, ${g\mid q.}$

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 -Rings, ideals, homomorphisms (3)

March 21, 2009

In order to understand the construction of the quotient ring from last lecture, it is convenient to examine some examples in details. We are interested in ideals ${I}$ of ${{\mathbb F}[x],}$ where ${{\mathbb F}}$ is a field. We write ${{\mathbb F}[x]/I}$ for the quotient ring, i.e., the set of equivalence classes ${[a]_\sim}$ of polynomials ${a}$ in ${F[x]}$ under the equivalence relation ${a\sim b}$ iff ${a-b\in I.}$

• If ${I=\{0\},}$ then for any ${a,}$ the equivalence class ${[a]_\sim}$ is just the singleton ${\{a\}}$ and the homomorphism map ${h:{\mathbb F}[x]\rightarrow{\mathbb F}[x]/I}$ given by ${h(a)=[a]_\sim}$ is an isomorphism.

To understand general ideals better the following notions are useful; I restrict to commutative rings with identity although they make sense in other contexts as well:

Definition 1 Let ${R}$ be a commutative ring with identity. An ideal ${I}$ is principal iff it is the ideal generated by an element ${a}$ of ${R,}$ i.e., it is the set ${(a)}$ of all products ${ab}$ for ${b\in R.}$

For example, ${\{0\}=(0)}$ is principal. In ${{\mathbb Z}}$ every subring is an ideal and is principal, since all subrings of ${{\mathbb Z}}$ are of the form ${n{\mathbb Z}=(n)}$ for some integer ${n.}$