M is a maximal ideal of R if and only if R/M is a field. Let f1(X), f2(X),..., fr(X), be monic polynomials belonging to F[X] such that deg f(X) > 1 (1 < i < r). Then there exists an extension E of F such that in E[X], each polynomial f1(X) splits into a product of linear factors. Let R be a ring. An ideal of R is a subring I of R such that if a I, r R, then a · r I and r · a I. An ideal A of R is said to be proper if AR. 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. If 1,...,n are algebraic over F, then F(1,...,n) is of finite degree over F and is therefore algebraic over F. Let F be a field. Then there exists an algebraically closed extension E of F. Let and be algebraic over F, then ± , · , / ( 0) are algebraic over F. Let E/F be finite. Then E is algebraic over F. Let E be algebraic over F. the zero of a monic irreducible polynomial p F[X]. The polynomial p is called the irreducible polynomial of over F denoted IrrF(,X) Let E F G be three fields. Assume that deg(F/E) and deg(G/F) are finite. Then deg(G/E) is finite and deg(G/E) = deg(G/F) · deg(F/E). Further, if {1,...,n} is a basis of G over F and {1,...,m} is a basis of F over E, then {ij} (1 < i < n, 1 < j < m) is a basis for G over E. Let be algebraic over F, n = deg(IrrF(,X)). Then - (1) deg(F()/F) = n. and (2) {1,,2,...,n-1} is a basis for F() over F. A field F is said to be algebraically closed if every nonconstant polynomial in F[X] splits into a product of linear factors in F[X]. 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. Let F be a field and let E be an extension of F. We say that E is an algebraic extension of F if every E is algebraic over F. If f has a leading coefficient 1, then we say that f is monic. The set of rational numbers. The complex numbers.

### Algebraically Closed Fields

In the beginning of the nineteenth century, Gauss proved the following remarkable property of the field C of complex numbers:

Theorem 1: Let f(X) C[X] be nonconstant. Then, in C[X], f(X) splits into a product of linear polynomials.

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 F[X] splits into a product of linear factors in F[X].

Example 1: According to Gauss's theorem C is algebraically closed.

Example 2: Let

Q = { C | is algebraic over Q}.

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 E. Then is algebraic over F. Moreover, if f = IrrF(,X) is a nonconstant, monic polynomial in F[X]. Therefore, since F is algebraically closed,

f = (X - i),   i F.

But E is a zero of f, so that X - divides f (in E[X]). Therefore, X - = X - i for some i, and thus = i for some i, so that F. Therefore, E F. But by assumption, E F, so we have proved E = F.

(2) (1). Let f be a nonconstant polynomial in F[X]. We must show that f splits into a product of linear factors in F[X]. Without loss of generality, we may restrict ourselves to the case of f monic and irreducible in F[X]. In this special case, we will prove that deg(f ) = 1 so that f is linear. Let n = deg(f ). By adjoining a root of f to F, we get an extension F() of F of degree n (Theorem 3 of the section on algebraic numbers). But by Proposition 1 of the section on Algebraic extensions, F() is an algebraic extension of F. Therefore, by our hypothesis (2), F() = F, so that n = deg(F()/F) = 1. Thus, f is linear.

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 F[X] splits into linear factors. At this point we would like to strengthen this result by proving that F is contained in an algebraically closed field. This result will prove to be of great utility to us in studying the properties of F, since we can fix, once and for all, an extension of F, which is so large as to accommodate the roots of all polynomial equations over F within it. Our main result is

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 F[X], deg(f ) > 1. In order to show that F is algebraically closed, we must show that f splits into a product of linear factors in F[X]. Without loss of generality, assume that f is monic. Since E is algebraically closed,

(1)
f = (X - i),   i E.

Let f = Xn + a1Xn-1 + ... + an, ai F. Then i is algebraic over F(a1,...,an) which implies that

deg(F(i,a1,...,an)/F(a1,...,an)) <

by Theorem 3 of algebraic extensions. But aj F (1 < j < n) aj is algebraic over F F(a1,...,an) is of finite degree over F (Corollary 5 of algebraic extensions). Therefore, by Theorem 3 of algebraic extensions

deg(F(i,a1,...,an)/F) <
i is algebraic over F (Proposition 1 of algebraic extensions)
i F.

Therefore, by (1), f splits into linear factors in F[X].

Example 3: C is an algebraic closure of R.

Example 4: Let

Q = { C | is algebraic over Q}.

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, C - Q by Lindemann's theorem.

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 PK of K such that every nonconstant polynomial f K[X] has a zero in PK.

Proof: Let

A = {f K[X] | f is nonconstant and irreducible}.

For each f A, let Yf denote an indeterminate. Let us consider the ring R = K[Yf]f A, gotten by adjoining to K all the indeterminates Yf. A typical element of R is a polynomial in finitely many indeterminates Yf, having coefficients in K. Let I denote the ideal of R generated by the polynomials

f(Yf )    (f A).

We assert that I R. For if I = R, then 1 I, so that

(2)
1 = gi f(Yf ),    gi R,

where we may assume, without loss of generality, that

gi = gi(Yf1,...,Yfm).

By Corollary 6 of the section on splitting fields, there exists an extension E of K in which each of the polynomials fi has a zero i (1 < i < m). The relation (2) only involves polynomials in K[Yf1,...,Yfm]. Let us replace Yfi in (2) by i. Then, since i is a zero of fi, equation (2) yields 1 = 0, a contradiction. Therefore, IR. Thus, I is contained in a maximal ideal M of R. Now R/M is a field by Theorem 8 if the section on maximal ideals. Moreover, the mapping

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 Pk = R/M. Let f K[X] be nonconstant and irreducible. Then

f(Yf + M) = f(Yf ) + M = 0 + M

since f(Yf ) M. Therefore, Yf + M is a zero of f in R/M. We have therefore shown that every irreducible, nonconstant polynomial in K[X] has a zero in PK, which completes the proof of Lemma 7.

Proof of Theorem 4: Let us construct a sequence of fields

E0 E1 E2 ...

as follows: Set E0 = F and for i > 0, set

Ei+1 = PEi,

where PEi is the field described in the lemma. Let

E = Ei.

If , E then for all sufficiently large i, , Ei. Therefore, ± , · / (0) are defined as elements of Ei for all i sufficiently large and are independent of the choice of i. Thus we can define the field operations on E and E becomes a field. We contend that E is algebraically closed. For let f E[X], n = deg(f ) > 1. We must show that f splits into a product of linear factors in E[X]. This assertion is trivial for n = 1. Let us proceed by induction on n. Assume that n > 1 and assume the assertion for polynomials of degree < n. Since f E[X], f Ei[X] for some i. Therefore, by construction of Ei, f has a zero in Ei+1, so that f = (X - )g in Ei+1[X], where deg(g) > 1. Therefore, f = (X - )g in E[X], where 1 < deg(g) < n. Therefore, by induction, g is a product if linear factors in E[X], so that f is a product of linear factors in E[X].