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.

Read the rest of this entry »

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}.}

Read the rest of this entry »