305 -7. Extension fields revisited

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}.}

Using this result, we can conclude that every ideal of {{\mathbb F}[x]} is principal. To see this, consider an ideal {I\subseteq {\mathbb F}[x].} Again, the argument is very similar to the proof of the same result for {{\mathbb Z}.} There are two cases:

Either {I={0},} in which case {I=(0)} is principal.

Or else {I\ne{0},} so we can find some {a\in I} with {a\ne0.} Thus {{\mathcal B}=\{{\rm deg}(p):p\in I} and {p\ne0\}} is nonempty, and by the well-ordering principle it has a least element {k.} Let {p\in I} have degree {k.} We want to show that {I=(p).}

For this, consider an arbitrary {q\in I.} Let {r} be a greatest common divisor of {p,q.} By the above, there are polynomials {a,b\in{\mathbb F}[x]} with {r=ap+bq.} Since {I} is an ideal and {p,q\in I,} we have that {r\in I.} Since {r\mid p,} we have {{\rm deg}(r)\le {\rm deg}(p).} Since {k={\rm deg}(p)} is least in {{\mathcal B},} it follows that {{\rm deg}(r)={\rm deg}(p),} since greatest common divisors are nonzero by definition.

Since {r\mid p} and {{\rm deg}(r)={\rm deg}(p),} then {p=ri} for some unit {i.} Then {r=pi^{-1},} so {p\mid r.} Since {r\mid q,} then {p\mid q,} as we wanted.

It follows that {I=(p).}

Definition 1 A nonconstant polynomial {p\in{\mathbb F}[x]} is irreducible in {{\mathbb F}[x]} iff the equation

\displaystyle  p=rq

with {r,q\in{\mathbb F}[x]} and {{\rm deg}(r)<{\rm deg}(p),} {{\rm deg}(q)<{\rm deg}(p),} has no solutions.


Note that this definition depends on {{\mathbb F}[x].} For example, {p(x)=x^2-2} is irreducible in {{\mathbb Q}[x]} but not in {{\mathbb R}[x].}

The only irreducible polynomials in {{\mathbb C}[x]} are the linear polynomials. The only irreducible polynomials in {{\mathbb R}[x]} are the linear polynomials, and those quadratic polynomials in {{\mathbb R}[x]} that have no real roots. Both these facts are consequences of Gauß’ fundamental theorem of algebra.

Later we will learn a useful criterion for checking irreducibility of polynomials in {{\mathbb Q}[x],} Ferdinand Eisenstein’s theorem. Irreducible polynomials play the role of prime factors. 

2. Quotients.

We can now reexamine the quotient field construction.

Recall that the construction began with a commutative ring {R} with identity, and an ideal {I\subseteq R.} We then formed {R/I,} the quotient ring. Its elements are the equivalence classes {[a]} of elements {a\in R} under the equivalence relation {a\sim b} iff {a-b\in I.}

(An intuitive way of thinking about this is that elements of an ideal are in a sense “small,” and so {a\sim b} means that {a} and {b} are “close,” since their difference is small, so what we are doing is identifying two points {a} and {b} whenever they are close.)

{R/I} is itself a commutative ring with identity. Its zero element {0} is {[0]=I.} Its one element {1} is {[1].} We add elements of the quotient by setting {[a]+[b]=[a+b].} We multiply them by setting {[a][b]=[ab].}

We are interested in this construction when {R={\mathbb F}[x]} for some field {{\mathbb F}.}

As shown in the previous section, any ideal {I} of {{\mathbb F}[x]} is principal.

Suppose first that {I=(0).} Then {{\mathbb F}[x]/(0)\cong{\mathbb F}.} This is because each equivalence class {[a]} is just a singleton {{a},} and the map {h:{\mathbb F}\rightarrow{\mathbb F}[x]/(0)} given by {h(a)=[a]} is easily seen to be an isomorphism in this case.

Suppose now that {I=(a)} for some nonzero constant polynomial {a.} One can then check that {{\mathbb F}[x]/(a)\cong{0}.}

Suppose that {p} is not constant and not irreducible. Then {{\mathbb F}[x]/(a)} has zero divisors. You need to verify these two facts as part of your homework.

More interesting is to see what happens if {p(x)\in{\mathbb F}[x]} is irreducible. Here is the punchline: Suppose that {{\mathbb H}} is a field extension of {{\mathbb F},} and that in {{\mathbb H}} the polynomial {p} has a root {\alpha.} Then {{\mathbb F}(\alpha)} makes sense. Recall that this is the smallest field that contains {{\mathbb F}} and has {\alpha} as an element.

Then {{\mathbb F}[x]/(p)\cong {\mathbb F}(\alpha).}

In other words, the abstract quotient construction provides us with a way of finding field extensions in which we can find roots of polynomials that were originally irreducible.

We will examine this claim and several examples next lecture.

(The textbook does not seem to discuss the quotient ring construction and how it gives rise to field extensions. Let me know if you would like references for books that discuss this topic.)

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2 Responses to 305 -7. Extension fields revisited

  1. […] shown last lecture, is principal. Let be such […]

  2. […] shown last lecture, is principal. Let be such […]

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