In our discussion of rings we have already singled out a special class which we called fields. A field, let us recall, is a commutative ring with unit element in which every nonzero element has a multiplicative inverse. Put another way, a field is a commutative ring in which we can divide by any nonzero element.
Fields play a central role in algebra. For one thing, results about them find important applications in the theory of numbers. For another, their theory encompasses the subject matter of the theory of equations which treats questions about the roots of polynomials.
In our development we shall touch only lightly on the field of algebraic numbers. Instead, our greatest emphasis will be on aspects of field theory which impinge on the theory of equations. Although we shall not treat the material in its fullest or most general form, we shall go far enough to introduce some of the beautiful ideas, due to the brilliant French mathematician Evariste Galois, which have served as a guiding inspiration for algebra as it is today.
Section 5.1 Extension Fields
In this section we shall be concerned with the relation of one field to another. Let be a field; a field is said to be an extension of if contains . Equivalently, is an extension of if is a subfield of . Throughout this chapter will denote a given field and an extension of .
As was pointed out earlier, in the chapter on vector spaces, if is an extension of , then, under the ordinary field operations in is a vector space over . As a vector space we may talk about linear dependence, dimension, bases, etc., in relative to .
DEFINITION: The degree of over is the dimension of as a vector space over .
We shall always denote the degree of over by . Of particular interest to us is the case in which is finite, that is, when is finite-dimensional as a vector space over . This situation is described by saying that is a finite extension of .
We start off with a relatively simple but, at the same time, highly effective result about finite extensions, namely,
THEOREM 5.1.1: If is a finite extension of and if is a finite extension of then is a finite extension of F. Moreover,
Proof: The strategy we employ in the proof is to write down explicitly a basis of over In this way not only do we show that
is a finite extension of , but we actually prove the sharper result and the one which is really the heart of the theorem, namely that
.
Suppose, then, that and that . Let be a basis of over and let be a basis of over . What could possibly be nicer or more natural than to have the elements where , serve as a basis of over ? Whatever else, they do at least provide us with the right number of elements. We now proceed to show that they do in fact form a basis of over . What do we need to establish this? First we must show that every element in is a linear combination of them with coefficients in , and then we must demonstrate that these elements are linearly independent over .
Let be any element in . Since every element in is a linear combination of with coefficients in , in particular, must be of this form. Thus
where the elements are all in . However, every element in is a linear combination of with coefficients in . Thus
where every is in .
Substituting these expressions for into
we obtain
Multiplying this out, using the distributive and associative laws, we finally arrive at
Since the are in , we have realized as a linear combination over of the elements . Therefore, the elements do indeed span all of over , and so they fulfill the first requisite property of a basis.
We still must show that the elements are linearly independent over . Suppose that
where the are in . Our objective is to prove that each . Regrouping the above expression yields
Since the are in , and since all the elements
are in . Now
with . But, by assumption, form a basis of over , so, in particular they must be linearly independent over . The net result of this is that . Using the explicit values of the , we get
But now we invoke the fact that the are linearly independent over ; this yields that each . In other words, we have proved that the are linearly independent over . In this way they satisfy the other requisite property for a basis.
We have now succeeded in proving that the elements form a basis of over . Thus since and we have obtained the desired result
.
Suppose that are three fields in the relation and, suppose further that is finite. Clearly, any elements in linearly independent over are, all the more so, linearly independent over . Thus the assumption that is finite forces the conclusion that is finite. Also, since is a subspace of is finite. By the theorem, , whence . We have proved the
COROLLARY: If is a finite extension of and is a subfield of which contains , then
Thus, for instance, if is a prime number, then there can be no fields properly between and . A little later, in Section 5.4, when we discuss the construction of certain geometric figures by straightedge and compass, this corollary will be of great significance.
DEFINITION: An element is said to be algebraic over if there exist elements in , not all , such that
.
If the polynomial the ring of polynomials in over , and if
,
then for any element , by we shall mean the element in . In the expression commonly used, is the value of the polynomial obtained by substituting for . The element is said to satisfy if . In these terms is algebraic over if there is a nonzero polynomial which satisfies, that is, for which .
Let be an extension of and let be in . Let be the collection of all subfields of which contain both and is not empty, for itself is an element of . Now, as is easily proved, the intersection of any number of subfields of is again a subfield of . Thus the intersection of all those subfields of which are members of is a subfield of . We denote this subfield by . What are its properties? Certainly it contains both and , since this is true for every subfield of which is a member of . Moreover, by the very definition of intersection, every subfield of in
contains , yet itself is in . Thus is the smallest subfield of containing both and . We call the subfield obtained by adjoining to .
Our description of , so far, has been purely an external one. We now give an alternate and more constructive description of . Consider all
these elements in which can be expressed in the form here the 's can range freely over and can be any nonnegative integer. As elements in , one such element can be divided by another, provided the latter is not . Let be the set of all such quotients. We leave it as an exercise to prove that is a subfield of .
On one hand, certainly contains and , whence . On the other hand, any subfield of which contains both and , by virtue of closure under addition and multiplication, must contain all the elements
where each . Thus must contain all these elements; being a subfield of must also contain all quotients of such elements. Therefore, . The two relations of course imply that . In this way we have obtained an internal construction of , namely as .
We now intertwine the property that is algebraic over with macroscopic properties of the field itself. This is
THEOREM 5.1.2: The element is algebraic over if and only if is a finite extension of .
Proof: As is so very common with so many such ``if and only if'' propositions, one-half of the proof will be quite straightforward and easy, whereas the other half will be deeper and more complicated.
Suppose that is a finite extension of and that . Consider the elements they are all in and are in number. By Lemma 4.2.4, these elements are linearly dependent over . Therefore, there are elements in , not all , such that
Hence is algebraic over and satisfies the nonzero polynomial in of degree at most . This proves the ``if'' part of the theorem.
Now to the ``only if'' part.
Suppose that in is algebraic over . By assumption, satisfies some nonzero polynomial in ; let be a polynomial in of smallest positive degree such that . We claim that is irreducible over . For, suppose that where then
(see Problem 1) and, since and are elements of the field , the fact that their product is forces or . Since is of lowest positive degree with , we must conclude that one of or must hold. But this proves the irreducibility of .
We define the mapping from into as follows. For any . We leave it to the reader to verify that is a ring homomorphism of the ring into the field (see Problem 1). What is the kernel of ? By the very definition of
Also, is an element of lowest degree in the ideal of . By the results of Section 3.9,
every element in is a multiple of , and since
is irreducible, by Lemma 3.9.6, is a maximal ideal of . By Theorem 3.5.1 is a field. Now by the general homomorphism theorem for rings (Theorem 3.4.1), is isomorphic to the image of under . Summarizing, we have shown that the image of under is a subfield of . This image contains and, for every . Thus the image of under is a subfield of which contains both and ; by the very definition of we are forced to conclude that the image of under is all of . Put more succinctly, is isomorphic to .
Now, , the ideal generated by ; from this we claim that the dimension of , as a vector space over , is precisely equal to (see Problem 2). In view of the isomorphism between and we obtain the fact that
.
Therefore, is certainly finite; this is the contention of the ``only if'' part of the theorem. Note that we have actually proved more, namely that is equal to the degree of the polynomial of least degree satisfied by over .
The proof we have just given in Version 1 has been somewhat long-winded, but deliberately so. The route followed contains important ideas and ties in results and concepts developed earlier with the current exposition. No part of mathematics is an island unto itself.
We now redo the ``only if'' part, working more on the inside of . This reworking is, in fact, really identical with the proof already given; the constituent pieces are merely somewhat differently garbed.
Again let be a polynomial over of lowest positive degree satisfied by . Such a polynomial is called a minimal polynomial for over . We may assume that its coefficient of the highest power of is 1, that is, it is monic; in that case we can speak of the minimal polynomial for over for any two minimal, monic polynomials for over are equal. (Prove!) Suppose that is of degree ; thus where the are in . By assumption, , whence
What about ? From the above,
if we substitute the expression for into the right-hand side of this relation, we realize as a linear combination of the elements 1, over . Continuing this way, we get that , for , is a linear combination over of 1, .
Now consider
Clearly, is closed under addition; in view of the remarks made in the paragraph above, it is also closed under multiplication. Whatever further it may be, has at least been shown to be a ring. Moreover, contains both and . We now wish to show that is more than just a ring, that it is, in fact, a field.
Let be in and let
.
Since , and , we have that
, whence . By the irreducibility of and must therefore be relatively prime. Hence we can find polynomials and in such that
But then
since ; putting into this that , we obtain . The inverse of is thus ; in all powers of higher than can be replaced by linear combinations of 1, over , whence . We have shown that every nonzero element of has its inverse in ; consequently, is a field. However, , yet and are both contained in , which results in . We have identified as the set of all expressions .
Now is spanned over by the elements 1, in consequence of which . However, the elements 1, are linearly independent over , for any relation of the form , with the elements , leads to the conclusion that satisfies the polynomial over of degree less than . This contradiction proves the linear independence of 1, , and so these elements actually form a basis of over , whence, in fact, we now know that . Since the result follows.
DEFINITION: The element is said to be algebraic of degree over if it satisfies a nonzero polynomial over of degree but no nonzero polynomial of lower degree.
In the course of proving Theorem 5.1.2 (in each proof we gave), we proved a somewhat sharper result than that stated in that theorem, namely,
THEOREM 5.1.3: If is algebraic of degree over then
.
This result adapts itself to many uses. We give now, as an immediate consequence thereof, the very interesting
THEOREM 5.1.4: If in are algebraic over then , , and (if ) are all algebraic over . In other words, the elements in which are algebraic over form a subfield of .
Proof: Suppose that is algebraic of degree over while is algebraic of degree over . By Theorem 5.1.3 the subfield of is of degree over . Now is algebraic of degree over , a fortiori it is algebraic of degree at most over which contains . Thus the subfield of , again by Theorem 5.1.3 is of degree at most over . But
by Theorem 5.1.1; therefore,
and so is a finite extension of . However, and are both in , whence all of , , and are in . By Theorem 5.1.2, since is finite, these elements must be algebraic over , thereby proving the theorem.
Here, too, we have proved somewhat more. Since , every element in satisfies a polynomial of degree at most over , whence the
COROLLARY: If and in are algebraic over of degrees and respectively, then , , and (if ) are algebraic over of degree at most .
In the proof of the last theorem we made two extensions of the field . The first we called ; it was merely the field . The second we called and it was . Thus
;
it is customary to write it as . Similarly, we could speak about ; it is not too difficult to prove that . Continuing this pattern, we can define for elements in .
DEFINITION: The extension of is called an algebraic extension of if every element in is algebraic over .
We prove one more result along the lines of the theorems we have proved so far.
THEOREM 5.1.5: If is an algebraic extension of and if is an algebraic extension of then is an algebraic extension of .
Proof: Let be any arbitrary element of ; our objective is to show that satisfies some nontrivial polynomial with coefficients in . What information do we have at present? We certainly do know that satisfies some polynomial
where are in . But is algebraic over ; therefore, by several uses of Theorem 5.1.3, is a finite extension of . Since satisfies the polynomial whose coefficients are in is algebraic over . Invoking Theorem 5.1.2 yields that is a finite extension of . However, by Theorem 5.1.1,
whence is a finite extension of . But this implies that is algebraic over , completing the proof of the theorem.
A quick description of Theorem 5.1.5: algebraic over algebraic is algebraic.
The preceding results are of special interest in the particular case in which is the field of rational numbers and the field of complex numbers.
DEFINITION: A complex number is said to be an algebraic number if it is algebraic over the field of rational numbers.
A complex number which is not algebraic is called transcendental. At the present stage we have no reason to suppose that there are any
transcendental numbers. In the next section we shall prove that the familiar real number is transcendental. This will, of course, establish the existence of transcendental numbers. In actual fact, they exist in great abundance; in a very well-defined way there are more of them than there are algebraic numbers.
Theorem 5.1.4 applied to algebraic numbers proves the interesting fact that {\it the algebraic numbers form a field}; that is, the sum, products, and quotients of algebraic numbers are again algebraic numbers.
Theorem 5.1.5 when used in conjunction with the so-called ``fundamental theorem of algebra,'' has the implication that the roots of a polynomial whose coefficients are algebraic numbers are themselves algebraic number.