## 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 -Homework set 7

April 3, 2009

This set is due April 10 at the beginning of lecture. Details of the homework policy can be found on the syllabus and here.

1. Let $R$ be a commutative ring with identity. Let $I$ be an ideal of $R,$ and let $i$ be a unit of $R.$ Show that $I=R$ iff $i\in I.$ Conclude that the only ideals of a field ${\mathbb F}$ are $\{0\}$ and ${\mathbb F}.$ Also conclude that ${\mathbb F}[x]/I\cong\{0\}$ if $I$ is the ideal of ${\mathbb F}[x]$ generated by a constant nonzero polynomial.

2. Suppose $p(x)\in{\mathbb F}[x]$ is a nonconstant, not irreducible polynomial, and let $I=(p)$ be the ideal generated by $p.$ Show that ${\mathbb F}[x]/I$ has zero divisors.

3. Find an irreducible polynomial $p(x)$ in ${\mathbb Z}_2[x]$ of degree 3, and explicitly show that ${\mathbb Z}_2[x]/(p)$ coincides with (i.e., is isomorphic to) the field of 8 elements built in a previous homework set.

4. Either build a field of 9 elements using the kinds of arguments in the previous problems; or determine a subfield of ${\mathbb C}$ isomorphic to ${\mathbb Q}[x]/(x^3+3x+3).$ (You may assume that $x^3+3x+3$ is irreducible in ${\mathbb Q}[x].$)