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 E. We say that is algebraic over F if there exists a nonzero polynomial f F[X] such that f() = 0. If is not algebraic over F, then we say that is transcendental over F. In order to get some feel for this classification, let us study some examples.

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

Example 2: Let F = Q, E = C. Then ₁ = , ₂ = , ₃ = + are all algebraic over Q. For ₁ is a zero of X² - 2; ₂ is a zero of X³ - 7; ₃ is a zero of X⁶ - 6X⁴ - 14X³ + 12X² - 84X + 41.

Example 3: Let F = Q, E = C. Then = 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 nor e can satisfy a polynomial equation with rational coefficients. However, the transcendentality of 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 0, 1 be algebraic over Q. Let be algebraic over Q, Q. Then is transcendental over Q.

Thus, for example, 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 E is transcendental over f. For if X satisfies the polynomial equation with coefficients in F, then

c_nXⁿ + c_n-1X^n-1 + ... + c₀    c_i F

c₀ = c₁ = ... = c_n = 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 E be algebraic over F. Our first task is to single out a single polynomial having as a zero. Let f F[X] have as a zero. Since F[X] is a UFD, we may write f = p₁,...,p_n, where p_i F[X] is irreducible. Then

0 = f() = p₁()...p_r().

But since E is an integral domain and p₁() E (1 < i < r), we see that p_i() = 0 for some i. Thus, is a zero of some irreducible polynomial p 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 as a zero. Indeed if q were another such polynomial, then p and q are relatively prime, so that there exist polynomials a, b F[X] such that ap + bq = 1. But then

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

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

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

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

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

Example 8: Let F. Then Irr_F(,X) = X - .

As the reader may have already guessed, it is often relatively easy to exhibit a polynomial f F[X] for which f() = 0. The difficult part of constructing Irr_F(,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 F[X] such that f() = 0, then we can try to factor f and determine an irreducible factor of f which has 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 ·v of a scalar F and a vector v E is just the product of 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() = {a + b | a,b Q}.

Therefore, {1,} generates Q() over Q. Moreover, {1, } is linearly independent over Q, since if a + b = 0 with a,b not both 0, we see that b0, so that = -a/b Q, which contradicts the fact that is irrational. Thus we see that {1,} is a basis of Q() over Q and that deg(Q/Q) = 2. Note that Irr_Q(,X) = X² - 2 and that deg(Irr_Q(,X)) = deg(Q)/Q). This phenomenon is not accidental as the following result shows.

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

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

(2) {1,,²,...,^n-1} is a basis for F() over F.

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

= a₀·1 + a₁· + ... + a_n-1^n-1    (a_i F).

The uniqueness is simple. For if = a₀'·1 + a₁' + a_n-1'^n-1, then

(a₀-a₀') + (a₁ - a₁') + ... + (a_n-1 - a_n-1')^n-1 = 0.

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

a₀ + a₁ + ... + a_m^m   (a_i F).

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

b₀ + b₁ + ... + b_n-1^n-1   (b_i F).

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

Irr_F(,X) = Xⁿ + c_n-1X^n-1 + ... + c₀.

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

^r = ·^r-1 = ·(b₀ + b₁ + ... + b_n-1^n-1)

= b₀ + b₁² + ... + b_n-1ⁿ

= b₀' + b₁' + ... + b_n-1^n-1,

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

F[] = {a₀ + a₁ + ... + a_n-1^n-1 | a_i F}.

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

: F[X]F[]

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

F[X]/g · F[X] F[].

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[] is a field, since the image under a field under an isomorphism again a field.

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

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

: F(X) F(),

p(X)/q(X)p()/q()  (p(X),q(X) F[X], q(X)0)

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

Proof: First note that q()0 since is transcendental over F and q(X)0. Thus, p()/q() makes sense. Second, let us show that is well defined: If p₁(X)/q₁(X) = p₂(X)/q₂(X), then

p₁()q₂() = q₁()p₂(),

p₁()/q₁() = p₂()/q₂(),

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

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

{1,,², ...} = S

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

a_iⁱ = 0, a_i F,   a_i not all 0,

is a linear relation among the elements of S, then is the zero of some nonzero polynomial with coefficients in F, a contradiction to the assumption that is transcendental over F. Thus if is transcendental over F, deg(F()/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()/F is finite if and only if is algebraic over F.