1.8. Algebraic and transcendental elements.

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 Let E be an extension field of F, and let

α ∈ E. Then we have a homomorphism

f(X) → f(α) : F[X]→ E.

There are two possibilites.

Case 1: The kernel of the map is (0), i.e.,

f(α) = 0, f(X) ∈ F[X] =⇒ f(X) = 0.

FIELDS AND GALOIS THEORY 7

In this case we say that α transcendental over F. The isomorphism F[X] → F[α] extends

to an isomorphism F(X) → F(α).

Case 2: The kernel is _= (0), i.e., g(α) = 0 for some nonzero g(X) ∈ F[X]. We then say

that α is algebraic over F. Let f(X) be the monic polynomial generating the kernel of the

map. It is irreducible (if f = gh is a proper factorization, then g(α)h(α) = f(α) = 0, but

g(α) _= 0 _= h(α)). We call f the minimum polynomial of α over F. It is characterized as an

element of F[X] by each of the following sets of conditions:

f is monic; f(α) = 0; g(α) = 0 and g ∈ F[X] =⇒ f|g;

f is the monic polynomial of least degree such f(α) = 0;

f is monic, irreducible, and f(α) = 0.

Note that g(X) → g(α) induces an isomorphism F[X]/(f) → F[α]. Since the first is a field,

so also is the second: F(α) = F[α]. Moreover, each element of F[α] has a unique expression

a0 + a1α + a2α2 + ・ ・ ・ + am−1αm−1, ai ∈ F,

where m = deg(f). In other words, 1, α, . . . , αm−1 is a basis for F[α] over F. Hence

[F(α) : F] = m. Since F[x] ≈ F[α], arithmetic in F[α] can be performed using the same

rules as in F[x].

Example 1.17. Let α ∈ C be such that α3 −3α−1 = 0. The minimum polynomial of α

over Q is X3 − 3X −1 (because this polynomial is monic, irreducible, and has α as a root).

The set {1, α, α2} is a basis for Q[α] over Q. The calculations in an example above show

that if β is the element α4 + 2α3 + 3 of Q[α], then β = 3α2 + 7α + 5, and

β−1 = 7

111α2 − 26

111α + 28

111.

Remark 1.18. Maple knows how to compute in Q[α]. For example,

factor(X^4+4); returns the factorization

(X2 − 2X + 2)(X2 + 2X + 2).

Now type: alias(c=RootOf(X^2+2*X+2);. Then

factor(X^4+4,c); returns the factorization

(X + c)(X − 2 − c)(X + 2+c)(X − c),

i.e., Maple has factored X4 + 4 in Q[c] where c has minimum polynomial X2 + 2X + 2.

An extension E/F is algebraic if all elements of E are algebraic over F;ot herwise it is

transcendental over F.

Proposition 1.19. (a) If [E : F] is finite, then E is algebraic over F.

(b) If E is algebraic over F and finitely generated (as a field), then [E : F] is finite.

Proof. (a) If α were transcendental over F, then 1, α, α2, . . . would be linearly independent

over F.

(b) Let E = F[α1, ..., αn];t hen F[α1] is finite over F (because α1 is algebraic over F);

F[α1, α2] is finite over F[α1] (because α2 is algebraic over F, and hence F[α1]). Hence

F[α1, α2] is finite over F. This argument can be continued.

Corollary 1.20. If E is algebraic over F then any subring R of E containing F is a

field.

8 J.S. MILNE

Proof. Let α ∈ R;the n F[α] is a field and F[α] ⊂ R. Therefore α has an inverse in

R.

A field F is said to be algebraically closed if E algebraic over F implies E = F. Equivalent

condition: the only irreducible polynomials in F[X] are of degree one;ev ery nonconstant

polynomial in F[X] has a root in F.

Example 1.21. The field of complex numbers C is algebraically closed. The set of all

complex numbers algebraic over Q is an algebraically closed field. Every field F has an algebraically

closed algebraic extension field (which is unique up to a nonunique isomorphism).

All these statements will be proved later.

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