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


COROLLARY: If is a root of , where , then in .

DEFINITION: The element is a root of of multiplicity if




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.

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.

An immediate consequence of this theorem is the

COROLLARY: If , then there is a finite extension of in which has a root. Moreover,


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).

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

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


2. for every .

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.)

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.