2. Extensions by radicals
Last lecture we defined where is a subfield of a field all the roots of the polynomial are in and all the coefficients of are in Namely, if are the roots of then the field generated by over
The typical examples we will consider are those where and the coefficients of are rational or, in fact, integers.
For example, if then because the roots of are and To see this last equality, notice:
- contains since both and are in and is closed under addition.
From this it follows that contains since, as shown in the proof of Corollary 4 from last lecture, is contained in any field that contains both and
- contains since it contains both and and is closed under addition, and it also contains since because any field is closed under additive inverses.
From this it follows that is a field that contains and the numbers so it contains the field and we are done.
If is a field, the collection of polynomials with coefficients in is an important example of a mathematical structure that we have found before:
Definition 1 A commutative ring is a structure that satisfies all the field axioms except (perhaps) for the existence of multiplicative inverses of nonzero elements.
See lecture 4.1 for the definition of field.
For example, is a commutative ring, as are all the sets for Any field is a commutative ring, of course.
There are also noncommutative rings, and we will encounter them later.
Definition 2 Let be a field. Then is the set of polynomials with coefficients in
We think of polynomials as purely formal expressions (rather than functions), so two polynomials are equal iff their coefficients are equal.
We turn into a commutative ring by adding and multiplying polynomials as usual: If and where (we allow for the possibility that or ), then
The zero element of is the polynomial with all its coefficients equal to zero, and the one element is the polynomial whose constant coefficient is 1 and all its other coefficients are zero.
Notice that in the definition above is a “variable,” not an element of and that the definition is purely formal, i.e., a polynomial is not a function.
There are several ways of making the above rigorous, similar to the way we identified and For example, we could identify with a subset of the set of infinite sequences of elements of namely, is the set of all those sequences such that for all but finitely many values of Of course, we think of as In this approach there is not even a variable anywhere. This extra layer of rigor is not needed in what follows, so we can safely ignore it.
It is important, however, to keep in mind that two polynomials are equal iff their coefficients are equal. This is not the same as saying that if and only if, when evaluated in the elements of and take the same values. For example, let and let and Then, as functions, since and However, as polynomials, since the coefficients of are (in increasing order of degree) while the coefficients of are and these two sequences are different.
Since we also want to consider polynomials as functions, we introduce the evaluation map:
Definition 3 Let The evaluation map from polynomials over to elements of is the function given by We will usually just use the notation without mentioning the function unless the distinction is important at the moment.
I’ll leave the proof of Lemma 4 as an exercise. As a follow up exercise, it is useful to think whether the same result is true if a finite field is used instead of
Since we are interested in the question of whether there is a formula for the solutions of a polynomial equation, i.e., whether there is a formula that identifies from a polynomial the numbers that are solutions of the study of field extensions is important, since the roots of a polynomial with coefficients in a field will typically not be elements of For example:
- Let be the polynomial Its roots are which are irrational numbers.
- Let be the polynomial Its roots are
- Let This polynomial has no roots in since Recall that in lecture 4.3 we constructed a field of 4 elements. Using the same notation we used there, where (this easily determines all the other products and sums in ) Notice that and so the roots of are and (which do not belong to ).
The extensions that are most relevant to us in the context of this question are those that are obtained by radicals, since the formulas we are looking for (or trying to show that do not exist) are those obtained from the coefficients of and the elements of by means of and the extraction of -th roots of elements of
We need to be careful here, since there are possible choices for the -th roots of a number For example, if is a complex number (even if is real), the -th roots of are the roots of the polynomial i.e., the solutions of If then these solutions are given by
As this shows, it is more accurate to say that than it is to claim that although we do refer to as an -th root of We will continue with the convention that is only used when is a nonnegative real number, in which case denotes the unique nonnegative real whose -th power is
(This is necessary in order to be precise, but slightly annoying, since we cannot even write the quadratic formula the way we are used to. If we ever abuse language and write for some negative we mean )
We say that the equation (more precisely, ) is solvable by radicals over iff there are numbers in and (not necessarily distinct) positive integers such that
- For all with
We also say that is an extension by radicals, although this is perhaps more accurate of
The intention of this definition is that (as one can easily verify) all the elements of a field extension are of the form for some polynomials and if is an -th root of some element of then certainly any such expression qualifies as a “formula by radicals.” More generally, the same applies to any element of as in Definition 5.
For example, consider Let be a root of Then so and is a square root of either or It follows that is an extension by radicals, since one can form the following tower of extensions:
- where is a square root of
- where is another square root of
- where is a square root of and
- where is another square root of
Notice that since then certainly Moreover, this example satisfies Definition 5 with and and
Notice also that we needed to add at the beginning, since otherwise we could not add square roots of since these elements are not even present in our field
This example also illustrates that we can witness that a field is an extension by radicals in more ways than one, since and so so we could just as well begin by first adding the fourth roots of -4, which would put into our field, and then adding the square roots of these elements.
Another point to notice is that we did not take particular care in finding a tower of extensions of minimal length or anything like that, and we do not need at the end to check whether or not and coincide, since all we need is containment rather than equality.
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