More about Roots
We return to the general exposition. Let be any field and, as usual, let be the ring of polynomials in over .
in , then the derivative of written as , is the polynomial
To make this definition or to prove the basic formal properties of the derivative, as applied to polynomials, does not require the concept of a limit. However, since the field is arbitrary, we might expect some strange things to happen.
At the end of Section 3.2, we defined what is meant by characteristic of a field. Let us recall it now. A field is said to be of characteristic 0 if for in and an integer. If for some and some then is said to be of finite characteristic. In this second case, the characteristic of is defined to be the smallest positive integer such that for all It turned out that if is of finite characteristic then its characteristic is a prime number.
We return to the question of the derivative. Let be a field of characteristic . In this case, the derivative of the polynomial is . Thus the usual result from the calculus that a polynomial whose derivative is must be a constant no longer need hold true. However, if the characteristic of is and if for it is indeed true that (see Problem 1). Even when the characteristic of is we can still describe the polynomials with zero derivative; if then is a polynomial in (see Problem 2).
We now prove the analogs of the formal rules of differentiation that we know so well.
LEMMA 5.5.1: For any and any
Proof: The proofs of parts 1 and 2 are extremely easy and are left as exercises. To prove part 3, note that from parts 1 and 2 it is enough to prove it in the highly special case and where both and are positive. But then whence
Recall that in elementary calculus the equivalence is shown between the existence of a multiple root of a function and the simultaneous vanishing of the function and its derivative at a given point. Even in our setting, where is an arbitrary field, such an interrelation exists.
LEMMA 5.5.2: The polynomial has a multiple root if and only if and have a nontrivial (that is, of positive degree) common factor.
Proof: Before proving the lemma proper, a related remark is in order, namely, if and in have a nontrivial common factor in , for an extension of , then they have a nontrivial common factor in . For, were they relatively prime as elements in then we would be able to find two polynomials and in such that
Since this relation also holds for those elements viewed as elements of , in they would have to be relatively prime.
Now to the lemma itself. From the remark just made, we may assume, without loss of generality, that the roots of all lie in (otherwise extend to , the splitting field of . If has a multiple root then
where . However, as is easily computed, whence, by Lemma 5.5.1,
since . But this says that and have the common factor , thereby proving the lemma in one direction.
On the other hand, if has no multiple root then
where the 's are all distinct (we are supposing to be monic). But then
where the denotes the term is omitted. We claim no root of is a root of , for
since the roots are all distinct. However, if and have a nontrivial common factor, they have a common root, namely, any root of this common factor. The net result is that and have no nontrivial common factor, and so the lemma has been proved in the other direction.
COROLLARY 1: If is irreducible, then
1. If the characteristic of is has no multiple roots.
2. If the characteristic of is has a multiple root only if it is
of the form .
Proof: Since is irreducible, its only factors in are 1 and . If has a multiple root, then and have a nontrivial common factor by the lemma, hence . However, since the degree of is less than that of , the only possible way that this can happen is for to be . In characteristic this implies that is a constant, which has no roots; in characteristic , this forces .
We shall return in a moment to discuss the implications of Corollary 1 more fully. But first, for later use in Chapter 7 in our treatment of finite fields, we prove the rather special
COROLLARY 2: If is a field of characteristic then the polynomial , for , has distinct roots.
Proof: The derivative of is
since is of characteristic . Therefore, and its derivative are certainly relatively prime, which, by the lemma, implies that has no multiple roots.
Corollary 1 does not rule out the possibility that in characteristic an irreducible polynomial might have multiple roots. To clinch matters, we exhibit an example where this actually happens. Let be a field of characteristic 2 and let be the field of rational functions in over . We claim that the polynomial in is irreducible over and that its roots are equal. To prove irreducibility we must show that there is no rational function in whose square is ; this is the content of Problem 4. To see that has a multiple root, notice that its derivative (the derivative is with respect to ; for being in is considered as a constant) is . Of course, the analogous example works for any prime characteristic.
Now that the possibility has been seen to be an actuality, it points out a sharp difference between the case of characteristic and that of characteristic . The presence of irreducible polynomials with multiple roots in the latter case leads to many interesting, but at the same time complicating, subtleties. These require a more elaborate and sophisticated treatment which we prefer to avoid at this stage of the game. Therefore, we make the flat assumption for the rest of this chapter that all fields occurring in the text material proper are fields of characteristic .
DEFINITION: The extension of is a simple extension of if for some in .
In characteristic (or in properly conditioned extensions in characteristic ; see Problem 14) all finite extensions are realizable as simple extensions. This result is
THEOREM 5.5.1: If is of characteristic and if , are algebraic over then there exists an element such that .
Proof: Let and , of degrees and , be the irreducible polynomials over satisfied by and respectively. Let be an extension of in which both and split completely. Since the characteristic of is all the roots of are distinct, as are all those of . Let the roots of be and those of .
If , then , hence the equation
has only one solution in , namely,
Since is of characteristic it has an infinite number of elements, so we can find an element such that for all and for all . Let
our contention is that . Since , we certainly do have that . We will now show that both and are in from which it will follow that .
Now satisfies the polynomial over , hence satisfies considered as a polynomial over . Moreover, if
since . Thus in some extension of and have as a common factor. We assert that is in fact their greatest common divisor. For, if is another root of , then since by our choice of for avoids all roots of . Also, since cannot divide the greatest common divisor of and . Thus is the greatest common divisor of and over some extension of . But then they have a nontrivial greatest common divisor over , which must be a divisor of . Since the degree of is 1, we see that the greatest common divisor of and in is exactly . Thus , whence ; remembering that , we obtain that . Since , and since , we get that , whence . The two opposite containing relations combine to yield .
A simple induction argument extends the result from 2 elements to any finite number, that is, if are algebraic over , then there is an element such that . Thus the
COROLLARY: Any finite extension of a field of characteristic is a simple extension.