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Algebraically Closed FieldsIn the beginning of the nineteenth century, Gauss proved the following remarkable property of the field C of complex numbers:
Theorem 1: Let This fact, while not surprising to a student who has spent some time trying to solve polynomial equations, is rather deep. One of the shortest proofs of Gauss's theorem relies on complex analysis and uses Liouville's theorem to the effect that a bounded, entire function is constant. Elsewhere, we will give a proof, due to Artin, which is almost purely algebraic in character, and which uses Galois theory. Artin's proof makes use of the fact that every polynomial of odd degree and having real coefficients has a real zero. However, for the rest of this section, let us assume Gauss's theorem. This will not result in any circular reasoning since we will use the theorem only for purposes of illustration. Gauss's theorem asserts that the field of complex numbers has a rather special property which we now formalize in a definition:
Definition 2: A field F is said to be algebraically closed if every nonconstant polynomial in Example 1: According to Gauss's theorem C is algebraically closed. Example 2: Let We will prove that Q is algebraically closed . Proposition 3: The following statements are equivalent: (1) F is algebraically closed. (2) If E is an algebraic extension of F, then E = F. Proof: (1) (2). Let E be an algebraic extension of F and let
f = (X  _{i}), _{i} F.
But (2) (1). Let f be a nonconstant polynomial in In the section on splitting fields, we showed that it is possible to construct an extension E of a given field F in which any given finite collection of nonconstant polynomials in Theorem 4: Let F be a field. Then there exists an algebraically closed extension E of F. The proof of Theorem 4 is rather difficult and will be delayed momentarily. It is probably best for the beginner to omit the proof of Theorem 4 entirely and concentrate effort on understanding the many significant ramifications it has. If F is a field and E is an algebraically closed field containing E, then E will usually be "too big." For example, E will usually contain elements which are transcendental over F. Since it is convenient to work only with algebraic elements, let us make the following definition. Definition 5: An algebraic closure of a field F is an extension F of F such that (1) F is algebraically closed, and (2) F is an algebraic extension of F. The main result of this section is Theorem 6: Let F be a field. Then F has an algebraic closure. Proof: Let E be an algebraically closed extension of F. E exists by Theorem 4. Set
F = { E  is algebraic over F}.
We assert that F is an algebraic closure of F. By Corollary 6 of the section on algebraic extensions, F is a field. It is clear that F is an algebraic extension of F. So we must show that F is algebraically closed. Let
f = (X  _{i}), _{i} E.
Let
deg(F(_{i},a_{1},...,a_{n})/F(a_{1},...,a_{n})) <
by Theorem 3 of algebraic extensions. But
deg(F(_{i},a_{1},...,a_{n})/F) <
_{i} is algebraic over F (Proposition 1 of algebraic extensions)
_{i} F.
Therefore, by (1), f splits into linear factors in Example 3: C is an algebraic closure of R. Example 4: Let Then, by the proof of Theorem 6, Q is an algebraic closure of Q. The field Q is called the field of algebraic numbers. Note: Q C. For example, We shall now turn our attention to the proof of Theorem 4. The principal difficulty in the proof is to pass from the finite collections of polynomials to (possibly) infinite ones. This is usually accomplished via Zorn's lemma. Indeed the present proof, which is due to Artin, makes use of Zorn's lemma, but in a rather disguised form. We use the fact that if I is a proper ideal of a commutative ring R with unity, then I is contained in a maximal ideal. And it is this fact which requires Zorn's lemma. The bulk of Theorem 4 resides in the following lemma.
Lemma 7: Let K be a field. Then there exists an extension P_{K} of K such that every nonconstant polynomial Proof: Let
A = {f K[X]  f is nonconstant and irreducible}.
For each
f(Y_{f} ) (f A).
We assert that
1 = g_{i} f(Y_{f} ), g_{i} R,
where we may assume, without loss of generality, that
g_{i} = g_{i}(Y_{f1},...,Y_{fm}).
By Corollary 6 of the section on splitting fields, there exists an extension E of K in which each of the polynomials f_{i} has a zero _{i}
xx + M (x K)
is an isomorphism of K into R/M. Let us identify K with its image in R/M. Then R/M becomes a field extension of K. Set
f(Y_{f} + M) = f(Y_{f} ) + M = 0 + M
since Proof of Theorem 4: Let us construct a sequence of fields
E_{0} E_{1} E_{2} ...
as follows: Set
E_{i+1} = P_{Ei},
where P_{Ei} is the field described in the lemma. Let
E = E_{i}.
If 
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