Roots of Polynomials In Section 5.1 we discussed elements in a given extension of which were algebraic over , that is, elements which satisfied polynomials in . We now turn the problem around; given a polynomial in we wish to find a field which is an extension of in which has a root. No longer is the field available to us; in fact it is our prime objective to construct it. Once it is constructed, we shall examine it more closely and see what consequences we can derive. DEFINITION: If then an element lying in some extension field of is called a root of if . We begin with the familiar result known as the Remainder Theorem. LEMMA 5.3.1: If and if is an extension of , then for any element where and where . Proof: Since is contained in whence we can consider to be lying in . By the division algorithm for polynomials in where and where or . Thus either or ; in either case must be an element of . But exactly what element of is it? Since Therefore, . That the degree of is one less than that of is easy to verify and is left to the reader. COROLLARY: If is a root of , where , then in . Proof: From Lemma 5.3.1, in since . Thus in . DEFINITION: The element is a root of of multiplicity if , whereas . A reasonable question to ask is: How many roots can a polynomial have in a given field? Before answering we must decide how to count a root of multiplicity . We shall always count it as roots. Even with this convention we can prove LEMMA 5.3.2: A polynomial of degree over a field can have at most roots in any extension field. Proof: We proceed by induction on , the degree of the polynomial . If is of degree 1, then it must be of the form where are in a field and where . Any such that must then imply that from which we conclude that . That is, has the unique root whence the conclusion of the lemma certainly holds in this case. Assuming the result to be true in any field for all polynomials of degree less than , let us suppose that is of degree over . Let be any extension of . If has no roots in then we are certainly done, for the number of roots in , namely zero, is definitely at most . So, suppose that has at least one root and that is a root of multiplicity . Since follows. Now where is of degree . From the fact that we get that whence, by the corollary to Lemma 5.3.1, is not a root of . If is a root, in , of then however, since and since we are in a field, we conclude that That is, any root of in other than must be a root of Since is of degree , by our induction hypothesis, has at most roots in which, together with the other root , counted times, tells us that has at most roots in . This completes the induction and proves the lemma. One should point out that commutativity is essential in Lemma 5.3.2. If we consider the ring of real quaternions, which falls short of being a field only in that it fails to be commutative, then the polynomial has at least 3 roots, (in fact, it has an infinite number of roots). In a somewhat different direction we need, even when the ring is commutative, that it be an integral domain, for if with and in the commutative ring , then the polynomial of degree 1 over has at least two distinct roots and in . The previous two lemmas, while interesting, are of subsidiary interest. We now set ourselves to our prime task, that of providing ourselves with suitable extensions of in which a given polynomial has roots. Once this is done, we shall be able to analyze such extensions to a reasonable enough degree of accuracy to get results. The most important step in the construction is accomplished for us in the next theorem. The argument used will be very reminiscent of some used in Section 5.1. THEOREM 5.3.1: If is a polynomial in of degree and is irreducible over then there is an extension of , such that in which has a root. Proof: Let be the ring of polynomials in over and let be the ideal of generated by . By Lemma 3.9.6, is a maximal ideal of , whence by Theorem 3.5.1, is a field. This will be shown to satisfy the conclusions of the theorem. First we want to show that is an extension of ; however, in fact, it is not! But let be the image of in ; that is, We assert that is a field isomorphic to ; in fact, if is the mapping from into defined by , then the restriction of to induces an isomorphism of onto . (Prove!) Using this isomorphism, we identify and in this way we can consider to be an extension of . We claim that is a finite extension of of degree , for the elements form a basis of over . (Prove!) For convenience of notation let us denote the element in the field as . Given , what is ? We claim that it is merely for since is a homomorphism, if then and using the identification indicated above of with , we see that . In particular, since ; however, . Thus the element in is a root of . The field has been shown to satisfy all the properties required in the conclusion of Theorem 5.3.1, and so this theorem is now proved. An immediate consequence of this theorem is the COROLLARY: If , then there is a finite extension of in which has a root. Moreover, . Proof: Let be an irreducible factor of ; any root of is a root of . By the theorem there is an extension of with in which , and so, has a root. Although it is, in actuality, a corollary to the above corollary, the next theorem is of such great importance that we single it out as a theorem. THEOREM 5.3.2: Let be of degree . Then there is an extension of of degree at most in which has roots (and so, a full complement of roots). Proof: In the statement of the theorem, a root of multiplicity is, of course, counted as roots. By the above corollary there is an extension of with in which has a root . Thus in factors as where is of degree . Using induction (or continuing the above process), there is an extension of of degree at most in which has roots. Since any root of is either or a root of , we obtain in all roots of . Now, All the pieces of the theorem are now established. Theorem 5.3.2 asserts the existence of a finite extension in which the given polynomial , of degree , over has roots. If and if the roots in are making use of the corollary to Lemma 5.3.1, can be factored over as . Thus splits up completely over as a product of linear (first degree) factors. Since a finite extension of exists with this property, a  finite extension of of minimal degree exists which also enjoys this property of decomposing as a product of linear factors. For such a minimal extension, no proper subfield has the property that factors over it into the product of linear factors. This prompts the DEFINITION: If , a finite extension of is said to be a splitting field over for if over (that is, in ), but not over any proper subfield of can be factored as a product of linear factors. We reiterate: Theorem 5.3.2 guarantees for us the existence of splitting fields. In fact, it says even more, for it assures us that given a polynomial of degree over there is a splitting field of this polynomial which is an extension of of degree at most over . We shall see later that this upper bound of is actually taken on; that is, given , we can find a field and a polynomial of degree in such that the splitting field of over has degree . Equivalent to the definition we gave of a splitting field for over is the statement: is a splitting field of over if is a minimal extension of in which has roots, where . An immediate question arises: given two splitting fields and of the same polynomial in , what is their relation to each other? At first glance, we have no right to assume that they are at all related. Our next objective is to show that they are indeed intimately related; in fact, that they are isomorphic by an isomorphism leaving every element of fixed. It is in this direction that we now turn. Let and be two fields and let be an isomorphism of onto . For convenience let us denote the image of any under by ; that is, . We shall maintain this notation for the next few pages. Can we make use of to set up an isomorphism between and , the respective polynomial rings over and ? Why not try the obvious? For an arbitrary polynomial we define by It is an easy and straightforward matter, which we leave to the reader, to verify LEMMA 5.3.3: defines an isomorphism of onto with the property that for every . If is in we shall write as . Lemma 5.3.3 immediately implies that factorizations of in result in like factorizations of in , and vice versa. In particular, is irreducible in if and only if is irreducible in . However, at the moment, we are not particularly interested in polynomial rings, but rather, in extensions of . Let us recall that in the proof of Theorem 5.1.2 we employed quotient rings of polynomial rings to obtain suitable extensions of . In consequence it should be natural for us to study the relationship between and , where denotes the ideal generated by in and that generated by in . The next lemma, which is relevant to this question, is actually part of a more general, purely ring-theoretic result, but we shall content ourselves with it as applied in our very special setting. LEMMA 5.3.4: There is an isomorphism onto with the property that for every Proof: Before starting with the proof proper, we should make clear what is meant by the last part of the statement of the lemma. As we have already done several times, we can consider as imbedded in by identifying the element with the coset in . Similarly, we can consider to be contained in . The isomorphism is then supposed to satisfy We seek an isomorphism of onto . What could be simpler or more natural than to try the defined by for every ? We leave it as an exercise to fill in the necessary details that the so defined is well defined and is an isomorphism of onto with the properties needed to fulfill the statement of Lemma 5.3.4. For our purpose --- that of proving the uniqueness of splitting fields --- Lemma 5.3.4 provides us with the entering wedge, for we can now prove THEOREM 5.3.3: If is irreducible in and if is a root of , then is isomorphic to where is a root of ; moreover, this isomorphism can so be chosen that 1. 2. for every . Proof: Let be a root of the irreducible polynomial lying in some extension of . Let Trivially is an ideal of , and . Since and is an irreducible polynomial, we have that . As in the proof of Theorem 5.1.2, map into by the mapping defined by for every We saw earlier (in the proof of Theorem 5.1.2) that maps onto . The kernel of is precisely , so must be . By the fundamental homomorphism theorem for rings there is an isomorphism of onto . Note further that for every . Summing up: is an isomorphism of onto leaving every element of fixed and with the property that . Since is irreducible in is irreducible in (by Lemma 5.3.3), and so there is an isomorphism of onto where is a root of such that leaves every element of fixed and such that . We now stitch the pieces together to prove Theorem 5.3.3. By Lemma 5.3.4 there is an isomorphism of onto which coincides with on and which takes onto . Consider the mapping (motivated by of onto . It is an isomorphism of onto since all the mapping , and are isomorphisms and onto. Moreover, since Also, for We have shown that is an isomorphism satisfying all the requirements of the isomorphism in the statement of the theorem. Thus Theorem 5.3.3 has been proved. A special case, but itself of interest, is the COROLLARY: If is irreducible and if are two roots of , then is isomorphic to by an isomorphism which takes onto and which leaves every element of fixed. We now come to the theorem which is, as we indicated earlier, the foundation stone on which the whole Galois theory rests. For us it is the focal point of this whole section. THEOREM 5.3.4: Any splitting fields and of the polynomials and respectively, are isomorphic by an isomorphism with the property that for every . (In particular, any two splitting fields of the same polynomial over a given field are isomorphic by an isomorphism leaving every element of fixed.) Proof: We should like to use an argument by induction; in order to do so, we need an integer-valued indicator of size which we can decrease by some technique or other. We shall use as our indicator the degree of some splitting field over the initial field. It may seem artificial (in fact, it may even be artificial), but we use it because, as we shall soon see, Theorem 5.3.3 provides us with the mechanism for decreasing it. If then , whence splits into a product of linear factors over itself. By Lemma 5.3.3 splits over into a product of linear factors, whence . But then provides us with an isomorphism of onto coinciding with on . Assume the result to be true for any field and any polynomial provided the degree of some splitting field of has degree less than over , that is, . Suppose that , where is a splitting field of over . Since has an irreducible factor of degree . Let be the corresponding irreducible factor of . Since splits , a full complement of roots of , and so, a priory, of roots of are in . Thus there is a such that ; by Theorem 5.1.3, . Similarly, there is a such that . By Theorem 5.3.4 there is an isomorphism of onto with the property that for every . Since , We claim that is a splitting field for considered as a polynomial over , for no subfield of containing and hence , can split , since is assumed to be a splitting field of over . Similarly is a splitting field for over . By our induction hypothesis there is an isomorphism of onto such that for all . But for every hence for every . This completes the induction and proves the theorem. To see the truth of the ``(in particular...)'' part, let and let be the identity map for every . Suppose that and are two splitting fields of . Considering and , and applying the theorem just proved, yields that and are isomorphic by an isomorphism leaving every element of fixed. In view of the fact that any two splitting fields of the same polynomial over are isomorphic and by an isomorphism leaving every element of fixed, we are justified in speaking about the splitting field, rather than a splitting field, for it is essentially unique. EXAMPLES: 1. Let be any field and let be in . If is any extension of in which has a root, , then the element also in is also a root of . If it is easy to check that must then be , and so both roots of are in . If then again both roots of are in . Consequently, can be split by an extension of degree 2 of . We could also get this result directly by invoking Theorem 5.3.2. 2. Let be the field of rational numbers and let In the field of complex numbers the three roots of are where and where is a real cube root of 2. Now cannot split , for, as a subfield of the real field, it cannot contain the complex, but not real, number . Without explicitly determining it, what can we say about , the splitting field of over ? By Theorem 5.3.2, ; by the above remark, since is irreducible over and since , by the corollary to Theorem 5.1.1, Finally, . The only way out is . We could, of course, get this result by making two extensions and and showing that satisfies an irreducible quadratic equation over . 3. Let be the field of rational numbers and let We claim that where is a splitting field of . Thus , far short of the maximum possible .