1. Greatest common divisors.
Let’s conclude the discussion from last lecture.
If is a field and are nonzero, then we can find polynomials such that is a gcd of and
To see this, consider and for some polynomials we have
We see that because both and are nonzero linear combinations of and so their degrees are in Each element of is a natural number because only for By the well-ordering principle, there is a least element of
Let be this least degree, and let have degree
First, if and then so
Second, by the division algorithm, we can write for some polynomials with Then is a linear combination of Since and is the smallest number in it follows that i.e., This is to say that so Similarly,
It follows that is a greatest common divisor of
Since any other greatest common divisor of is for some unit it follows that any gcd of and is a linear combination of and
Notice that this argument is very similar to the proof of the same result for
Using this result, we can conclude that every ideal of is principal. To see this, consider an ideal Again, the argument is very similar to the proof of the same result for There are two cases:
Either in which case is principal.
Or else so we can find some with Thus and is nonempty, and by the well-ordering principle it has a least element Let have degree We want to show that
For this, consider an arbitrary Let be a greatest common divisor of By the above, there are polynomials with Since is an ideal and we have that Since we have Since is least in it follows that since greatest common divisors are nonzero by definition.
Since and then for some unit Then so Since then as we wanted.
It follows that
Definition 1 A nonconstant polynomial is irreducible in iff the equation
with and has no solutions.
Note that this definition depends on For example, is irreducible in but not in
The only irreducible polynomials in are the linear polynomials. The only irreducible polynomials in are the linear polynomials, and those quadratic polynomials in 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 Ferdinand Eisenstein’s theorem. Irreducible polynomials play the role of prime factors.
We can now reexamine the quotient field construction.
Recall that the construction began with a commutative ring with identity, and an ideal We then formed the quotient ring. Its elements are the equivalence classes of elements under the equivalence relation iff
(An intuitive way of thinking about this is that elements of an ideal are in a sense “small,” and so means that and are “close,” since their difference is small, so what we are doing is identifying two points and whenever they are close.)
is itself a commutative ring with identity. Its zero element is Its one element is We add elements of the quotient by setting We multiply them by setting
We are interested in this construction when for some field
As shown in the previous section, any ideal of is principal.
Suppose first that Then This is because each equivalence class is just a singleton and the map given by is easily seen to be an isomorphism in this case.
Suppose now that for some nonzero constant polynomial One can then check that
Suppose that is not constant and not irreducible. Then has zero divisors. You need to verify these two facts as part of your homework.
More interesting is to see what happens if is irreducible. Here is the punchline: Suppose that is a field extension of and that in the polynomial has a root Then makes sense. Recall that this is the smallest field that contains and has as an element.
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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