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Let alpha element of E be algebraic over F. alpha the zero of a monic irreducible polynomial p element F[X]. The polynomial p is called the irreducible polynomial of alpha over F denoted IrrF(alpha,X) 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. Let f = a0+a1X+...+anXn element of Z[X] be primitive and let p be prime. Assume that p|ai (0 < i < n-1), pdoes not dividean, p2does not dividea0. Then f is irreducible in Z[X]. Let V be a vector space. If V has a finite set of generators, then we say that V is finite-dimensional and we define its dimension, denoted dimFV, to be the minimum possible number of elements in such a set. If V does not have a finite set of generators, then we say the V is infinite-dimensional, and we define dimFV to be infinity Let alpha element of E be algebraic over F, and let f element of F[X] be such that f(alpha) = 0. Then f is divisible by IrrF(alpha,X). M is a maximal ideal of R if and only if R/M is a field. An ideal M of R is said to be a maximal ideal if (a) M is a proper ideal and (b) if A is a proper ideal containing M, then M = A. Let V be a vector space over F. A basis of V is a subset {ei} of V (finite or infinite) with the property that every element v of V can be uniquely written in the form v = sum over ialphaiei. A function f : AmapsB is said to be surjective (or onto) if for every y element of B there exists x element of A such that f(x) = y. We say that alpha is algebraic over F if there exists a nonzero polynomial f element of F[X] such that f(alpha) = 0. If alpha is not algebraic over F, then we say that alpha is transcendental over F. A commutative ring R is an integral domain if R contains no zero divisors. In other words, R is an integral domain if the product of any two nonzero elements of R is nonzero. The set of rational numbers. The set integers. Let R be an integral domain. There exists a field FR and an embedding f :RmapsFR such that every element of FR is of the form f(a)·f(b)-1 (a,b element of R, bnot equal0). We say that R is a unique factorization domain if the following two conditions are satisfied: (1) If x element of Rx is not a unit of R, then x can be written as a product of irreducible elements of R.and (2) If x element of Rx is not a unit of R, and if x = pi1 ... pis = lambda1 ... lambdat are two expressions of x as a product of irreducible elements, then s = t and it is possible to renumber pi1 ... pis so that pi1 and lambda1 are associates (1 < i < s). The coefficient of the highest power of X appearing in the polynomial is 1 If f:GmapsH is a homomorphism. The kernel of f is ker(f) = {x element of G | f(x) = 1H}.

Algebraic and Transcendental Elements

Throughout this section let F be a field and E an extension of field F. Let us begin by classifying the elements of E into two broad categories with respect to F. Let alpha element of E. We say that alpha is algebraic over F if there exists a nonzero polynomial f element of F[X] such that f(alpha) = 0. If alpha is not algebraic over F, then we say that alpha is transcendental over F. In order to get some feel for this classification, let us study some examples.

Example 1: alpha element of F impliesalpha is algebraic over F. For alpha is a zero of X - alpha element of F[X].

Example 2: Let F = Q, E = C. Then alpha1 = square root of 2, alpha2 = cube root of 7, alpha3 = square root of 2 + cube root of 7 are all algebraic over Q. For alpha1 is a zero of X2 - 2; alpha2 is a zero of X3 - 7; alpha3 is a zero of X6 - 6X4 - 14X3 + 12X2 - 84X + 41.

Example 3: Let F = Q, E = C. Then pi = 3.1415926..., e = 2.718281... are transcendental over Q. We cannot come near to proving these two assertions here. Both are celebrated results of nineteenth-century mathematics. It was long suspected that neither pi nor e can satisfy a polynomial equation with rational coefficients. However, the transcendentality of pi was not proved until 1882, when it was settled by Lindemann. The transcendentality of e is somewhat easier and was first proved by Hermite in 1873.

Example 4: The theory of transcendental numbers is a vast realm of mathematics in which research is still going on in many parts of the mathematical community. It is an extremely difficult branch of mathematics, which relies for many of its proofs on analysis, especially on functions of a complex variable. Some apparently very simple questions on transcendence cannot be answered at the present time. Although we cannot go into an extended discussion of the theory of transcendental numbers, let us cite one of the crowning achievements of the theory, the Gelfond-Schneider theorem:

Theorem 1: Let alpha not equal 0, 1 be algebraic over Q. Let beta be algebraic over Q, beta not an element of Q. Then alphabeta is transcendental over Q.

Thus, for example, 2square root of 2 is transcendental over Q. The statement of this theorem was first guessed by Euler in the eighteenth century as was posed by Hilbert as one of his famous 23 problems at the International Congress of Mathematics in Paris in 1900. A proof of the result was discovered almost simultaneously in 1934 by Gelfond and Schneider, working independently.

We have not given any examples of transcendental elements which we could actually show to be transcendental. There is one obvious example however:

Example 5: Let F be a field, X an indeterminate over F. Let F(X) be the quotient field of F[X]. Then F(X) is an extension field of F, so we may set E = F(X). Then X element of E is transcendental over f. For if X satisfies the polynomial equation with coefficients in F, then

cnXn + cn-1Xn-1 + ... + c0    ci element of F
forward implicationc0 = c1 = ... = cn = 0

since X is an indeterminate over F. Thus, X cannot be the zero of any nonzero polynomial with coefficients in F.

In the remainder of this work we will have little to say concerning transcendental elements. This is for at least two reasons. First, the theory of transcendental numbers is far too complicated for the scope of this work. Second, we are striving toward a theory of polynomial equations. And zeros of a polynomial are algebraic over any field containing the coefficients of the polynomial.

Let alpha element of E be algebraic over F. Our first task is to single out a single polynomial having alpha as a zero. Let f element of F[X] have alpha as a zero. Since F[X] is a UFD, we may write f = p1,...,pn, where pi element of F[X] is irreducible. Then

0 = f(alpha) = p1(alpha)...pr(alpha).

But since E is an integral domain and p1(alpha) element of E (1 < i < r), we see that pi(alpha) = 0 for some i. Thus, alpha is a zero of some irreducible polynomial p element of F[X]. Without loss of generality, we may normalize p to be monic. We assert that p is the unique monic, irreducible polynomial in F[X] having alpha as a zero. Indeed if q were another such polynomial, then p and q are relatively prime, so that there exist polynomials a, b element of F[X] such that ap + bq = 1. But then

0 = a(alpha)p(alpha) + b(alpha)q(alpha) = 1,

which is a contradiction. Then polynomial p is called the irreducible polynomial of alpha over F and is denoted IrrF(alpha,X). Moreover, the manner in which we constructed p implies

Lemma 2: Let alpha element of E be algebraic over F, and let f element of F[X] be such that f(alpha) = 0. Then f is divisible by IrrF(alpha,X).

Example 6: X2 - 2 is monic and irreducible in Q[X]. Therefore, IrrQ(square root of 2, X) = X2 - 2.

Example 7: Let d element of Z not be a perfect square. Then X2 - d is monic and irreducible in Q[X], so that IrrQ(square root of d,X) = X2 - d.

Example 8: Let alpha element of F. Then IrrF(alpha,X) = X - alpha.

As the reader may have already guessed, it is often relatively easy to exhibit a polynomial f element of F[X] for which f(alpha) = 0. The difficult part of constructing IrrF(alpha,X) is to make sure that f is irreducible. There is a reasonable way of going about this, at least in the case F = Q. If we can exhibit some f element of F[X] such that f(alpha) = 0, then we can try to factor f and determine an irreducible factor of f which has alpha as a zero. The irreducibility developed in the previous chapters are useful in proving that a given factor of f is irreducible. Particular attention should be called to the Eisenstein irreducibility criterion.

If E is an extension of the field F, then E may be regarded as a vector space over F. The product alpha·v of a scalar alpha element of F and a vector v element of E is just the product of alpha and v, considered as elements of E. The dimension of E over F is called the degree of E over F, denoted deg(E/F).

For example, we showed in a previous section that

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

Therefore, {1,square root of 2} generates Q(square root of 2) over Q. Moreover, {1, square root of 2} is linearly independent over Q, since if a + bsquare root of 2 = 0 with a,b not both 0, we see that bnot equal0, so that square root of 2 = -a/b element ofQ, which contradicts the fact that square root of 2 is irrational. Thus we see that {1,square root of 2} is a basis of Q(square root of 2) over Q and that deg(Qsquare root of 2/Q) = 2. Note that IrrQ(square root of 2,X) = X2 - 2 and that deg(IrrQ(square root of 2,X)) = deg(Qsquare root of 2)/Q). This phenomenon is not accidental as the following result shows.

Theorem 3: Let alpha be algebraic over F, n = deg(IrrF(alpha,X)). Then

(1) deg(F(alpha)/F) = n.

(2) {1,alpha,alpha2,...,alphan-1} is a basis for F(alpha) over F.

Proof: Since {1,alpha,alpha2,...,alphan-1} contains n elements, it is clear that part (2) implies part (1). To prove part (2), we must show that every beta element of F(alpha) can be written uniquely in the form

(1)
beta = a0·1 + a1·alpha + ... + an-1alphan-1    (ai element of F).

The uniqueness is simple. For if beta = a0'·1 + a1'alpha + an-1'alphan-1, then

(a0-a0') + (a1 - a1')alpha + ... + (an-1 - an-1')alphan-1 = 0.

But then the polynomial f = (a0-a0') + (a1 - a1')X + ... + (an-1 - an-1')Xn-1 has alpha as a zero. Therefore, by Lemma 2, f is divisible by IrrQ(alpha,X). where the later has degree n. Thus, we are forced to conclude that f = 0, which implies that a0 = a0', a1 = a1', ..., an-1 = an-1', which gives us the uniqueness of the representation (1). Let us now show that every beta element of F(alpha) can be written in the form (1). Let F[alpha] denote the smallest subring of F(alpha) containing F and alpha. Then F[alpha] consists of all sums of the form

(2)
a0 + a1alpha + ... + amalpham   (ai element of F).

We assert that every sum of the form(2) can be rewritten in the form

(3)
b0 + b1alpha + ... + bn-1alphan-1   (bi element of F).

It clearly suffices to show that ar (r > n) can be so written. Let us proceed by induction on r. Let

IrrF(alpha,X) = Xn + cn-1Xn-1 + ... + c0.

Then alphan = -cn-1alphan-1-cn-2alphan-2-...-c0, so alphan can be written in the form (3). Thus the assertion is true for r = n. Assume that r > n and that alphar-1 can be written in the form (3). Then

alphar = alpha·alphar-1 = alpha·(b0 + b1alpha + ... + bn-1alphan-1)
= b0alpha + b1alpha2 + ... + bn-1alphan
= b0' + b1'alpha + ... + bn-1alphan-1,

since alphan can be written in the form of (3). Thus, the induction is complete. We have shown that

(4)
F[alpha] = {a0 + a1alpha + ... + an-1alphan-1 | ai element of F}.

It is clear that F[alpha] contains both F and alpha and F[alpha]subset ofF(alpha). Therefore, if we prove that F[alpha] is a field, then we can conclude that F[alpha] super set of F(alpha) and therefore F[alpha] = F(alpha). Therefore by (4) we conclude that every beta element of F(alpha) can be in the form (2). Thus we must show that F[alpha] is a field. Let

psialpha: F[X]mapsF[alpha]

be the homomorphism "evaluation at alpha." Then it is easy to see that psialpha is surjective. Moreover, ker(psialpha) = g · F[X], where g - IrrF(alpha,X). Therefore,

F[X]/g · F[X] isomorphic F[alpha].

However, since g is irreducible, g · F[X] is a maximal ideal by Example 5 in the section on maximal ideals and thus F[X]/g · F[X] is a field by Theorem 8 of the section on maximal ideals. Thus, F[alpha] is a field, since the image under a field under an isomorphism again a field.

In Theorem 3, we described the simple extension F(alpha) in case alpha is algebraic over F. Let us now carry out the correspondence task in case alpha is transcendental over F. The clue is provided by example 5.

Proposition 4: Let F(alpha) be a simple extension of F ant let alpha be transcendental over F. Then the mapping

psi: F(X) maps F(alpha),
p(X)/q(X)mapsp(alpha)/q(alpha)  (p(X),q(X) element of F[X], q(X)not equal0)

is a surjective F-isomorphism. In particular, F(X) and F(alpha) are F-isomorphic.

Proof: First note that q(alpha)not equal0 since alpha is transcendental over F and q(X)not equal0. Thus, p(alpha)/q(alpha) makes sense. Second, let us show that psi is well defined: If p1(X)/q1(X) = p2(X)/q2(X), then

p1(X)q2(X) = q1(X)p2(X),
impliesp1(alpha)q2(alpha) = q1(alpha)p2(alpha),
impliesp1(alpha)/q1(alpha) = p2(alpha)/q2(alpha),

Thus, psi is well defined. It is easy to check that psi is a ring homomorphism. Therefore, by Proposition 2 of the introduction to fields, psi is an isomorphism since psi is not identically zero. By (5) of the intro, psi is surjective.

We showed in Theorem 3 that if alpha is algebraic over F, then deg(F(alpha)/F) = deg(IrrF(alpha,X)). In particular, deg(F(alpha)/F) is finite. The situation is quite different if alpha is transcendental over F, however. Indeed, the set

{1,alpha,alpha2, ...} = S

is then an infinite, linearly independent subset of F(alpha). For if

sum over iaialphai = 0, ai element of F,   ai not all 0,

is a linear relation among the elements of S, then alpha is the zero of some nonzero polynomial with coefficients in F, a contradiction to the assumption that alpha is transcendental over F. Thus if alpha is transcendental over F, deg(F(alpha)/F) is finite. If E is an extension field of F, then we say that E is a finite extension of F (or that E/F is finite) if deg(E/F) is finite. Thus we have proved the following result:

Theorem 5: F(alpha)/F is finite if and only if alpha is algebraic over F.