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Let R be a ring. An ideal of R is a subring I of R such that if a element of I, r element of R, then a · r element of I and r · a element of I. If f:GmapsH is a homomorphism. The kernel of f is ker(f) = {x element of G | f(x) = 1H}. Let G and H be groups. A function f :GmapsH which satisfies (1) for all g, g' element of G is called a homomorphism of G into H. A function f : AmapsB is said to be injective (or one to one) if whenever f(x) = f(y), we have x = y. The set of rational numbers. The set of real numbers. The coefficient of the highest power of X appearing in the polynomial is 1 The complex numbers

Introduction to the Theory of Fields

If E is a field, then a subfield of E is a subset of E which is also a field with respect to the operations of E. For example, Q is a subfield of R. If F is a subfield of E, then we say that E is an extension field of F (or simply an extension of F). Thus, R is an extension of Q. Our point of view will be roughly as follows: For a given field F, we will study the properties of extensions of F. The reason for this will become apparent as we proceed. First let us consider some nontrivial examples of extensions.

Consider the set of all real numbers of the form

(1)
Q square root of 2

where the denominator is nonzero. It is reasonably apparent that the set of all such quotients is a field containing Q and square root of 2. Let us denote this field by Q(square root of 2). If F is a subfield of R which contains Q and square root of 2, then F must contain every real number of the form (1), so that F super set of Q(square root of 2). Therefore, Q(square root of 2) is the smallest subfield of R containing Q and square root of 2. In particular, Q(square root of 2) is an extension of Q. Let us examine this extension more closely. Since square root of 22 = 2, we have square root of 23 = 2square root of 2, square root of 24 = 22,... Therefore, every quotient of the form (1) can be written in the form


(2)
(alpha+betasquare root of 2)/(gamma+deltasquare root of 2),   alpha,beta,gamma,delta element of Q.

But since (gamma + deltasquare root of 2)-1 = (gamma - deltasquare root of 2)/(gamma2 + 2delta2), we see that every element of Q(square root of 2) is of the form

(3)
a + bsquare root of 2,   a,b element ofQ.

Conversely, every element of the form (3) belongs to Q(square root of 2). Therefore,

(4)
Q(square root of 2) = {a + bsquare root of 2 | a,b element of Q}.

Let us generalize the above example into a general procedure for constructing extensions. Let F be a field, E and extension of F, S a set of elements of E. Further let collection denote the collection of all subfields of E which contain F and S. Then

intersection of subfieldsG

is a subfield of E containing F and S, and is the smallest such subfield. Let us denote this subfield by F(S). Then F(S) is called the field obtained by adjoining the elements of S to F In the above example F = Q, E = R, S ={square root of 2}. In case S = {a1, ..., an}, we will write F(a1,...,an) instead of F({a1,...,an}). A field of the form F(S) is an extension of F. Moreover, if E is an extension of F, then E = F(S) for S = E. Therefore, every extension of F can be obtained by adjoining a set of elements to F. And thus, in order to study the properties of extensions of F, we must study this process adjunction more closely.

Proposition 1: Let F be a field, E an extension of F, S a subset of E. If S = S1 union S2, then

F(S) = F(S1)F(2).

That is, F(S) can be obtained by adjoining the elements of S2 to F(S1).

Proof: F(S1)(S2) is a subfield of E containing F(S1) and S2, so that F(S1)(S2), contains F, S1, and S2. In particular, F(S1)(S2) is a subfield of E containing F and S1 union S2 = S. Therefore, F(S1)(S2) superset of F(S). Since S superset of S1, S superset of S2, we see that F(S) contains F(S1), F(S) contains S2. Therefore F(S) contains F(S1)(S2). Thus we have F(S) = F(S1)(S2).

If S = {a1,...,an}, then Proposition 1 implies that F(a1,...,an) = F(a1,...,an-1)(an). Therefore, we can reduce the process of adjoining n elements to F to n successive adjunctions of a single element: First we adjoin a1 to F; then we adjoin a2 to F(a1); then a3 to F(a1,a2); and so on. An extension of F of the form F(a) is called a simple extension of F. We have just shown that we can adjoin a set of n elements to F by forming n consecutive simple extensions. Thus, the simple extensions of a field F should be studied more closely. For now let us be content to observe that F(a) consists of all quotients of the form


(5)
F(a) quotients,   ai,bj element of F,

where the denominator is nonzero. The proof is identical to the argument we used for Q(square root of 2) above. We will examine the simple extensions to a greater extent in the next section.

We still must supply an answer to the question: Why should we study extensions of a field? Let us answer this question by constructing a very important class of extensions of Q. If f element of Q[X] is a nonconstant, monic polynomial of degree n, then f (to be more thoroughly demonstrated later) has zeros alpha1,...,alpha in C such that

f = (X - alpha1)(X - alpha2)···(X - alphan).

Let us form the extension Q(alpha1,alpha2,...,alphan) of Q. One of our major tasks is to get information about alpha1,...,alphan. We will accomplish this by studying the algebraic properties of the extension Q(alpha1,alpha2,...,alphan).

In studying field extensions, it will be necessary to study isomorphisms of one field to another. Therefore, let us clarify the notion of isomorphism we have in mind. If E1 and E2 are fields, then an isomorphism f :E1maps to E2 is a ring isomorphism of E1 into E2. Thus, an isomorphism f :E1maps to E2 is injective and preserves sums, differences, and products. But it is easy to see that if a element of E1x, we have f(a-1) = f(a)-1.

Proposition 2: Let E1 and E2 be fields, f :E1maps to E2 a ring homomorphism. Then either f(x) = 0 for all x element of E1 or f is an isomorphism.

Proof: Since E1 is a field, the only ideals of E1 are {0} and E1. But ker(f) is an ideal of E1. If ker(f) = E1, then f(x) = 0 for all x element of E1. If ker(f) = {0}, then f is an isomorphism.