The Fundamental Theorem of Algebra An (Almost) Algebraic Proof The field of complex numbers, , is algebraically closed. Robert Dr. Bob Gardner Based on Hungerfords Appendix to Section V.3 in Algebra, Springer-Verlag (1974)

Results of the Appendix to Section V.3 Lemma V.3.17. If is a finite dimensional separable extension of an infinite field , then for some . Lemma V.3.18. There are no extension fields of dimension 2 over the field of complex numbers. Theorem V.3.19. The Fundamental Theorem of Algebra. The field of complex numbers, , is algebraically closed. Corollary V.3.20. Every proper algebraic extension field of the field of real numbers is isomorphic to the field of

complex numbers. Some Results from Real Analysis Note. Every known proof of the Fundamental Theorem of Algebra depends on some result(s) from analysis. We shall give a proof which is algebraic, except for the following two results from analysis: (A) Every positive real number has a real positive square root. (B) Every polynomial in of odd degree has a root in (that is, every irreducible polynomial in of degree greater than one has even

degree). Both results actually follow from the Axiom of Completeness of the real numbers. Lemma V.3.17. If is a finite dimensional separable extension of an infinite field , then for some . Proof. Since is a separable extension of , then it is an algebraic extension and so by Theorem V.3.16(iii) there is a Galois extension of that contains (here, ). Since we hypothesize is finite then by Theorem V.3.16(iv) we have

that is finite. By the Fundamental Theorem of Galois Theory (Theorem V.2.5(i)) ) is finite (since ) and, since there is a oneto-one correspondence between the set of intermediate fields of the extension and the set of all subgroups of (by the Fundamental Theorem) with for each intermediate field , then there are only finitely many intermediate fields between and Proof (continued). Therefore, there can be only a finite number of intermediate fields in the extension of by . Since is finite, we can choose such that is maximal.

ASSUME . Then there exists . Consider all (simple extension) intermediate fields of the form with . Since is an infinite field then there are infinitely many elements of of the form where , , and . However, there are only finitely many intermediate fields between and . So for some with we must have (or else we have infinitely many simple extensions of intermediate to and ). Proof (continued). So for this and , and . Since and ,

then and so . Whence and . So and (by the choice of ), so . Whence . But this CONTRADICTS the choice of such that is maximal (for all simple extensions of ). So the assumption that is false and hence . Lemma V.3.18. There are no extension fields of dimension 2 over the field of complex numbers. Proof. ASSUME is an extension field of of dimension 2 (that is, ). Then a basis for over is of the form where by Theorem V.1.6(iv) and . In fact, for any we have (if then

for some and so and is a basis for ). By Theorem V.1.6(ii) must be a root of an irreducible monic polynomial of degree 2. We next show that no such can exist. For each , we know that has a real positive square root be Assumption (A), denoted . Proof (continued). Also by Assumption (A) the positive real numbers and have real positive square roots, say and

respectively. Now (1) Proof (continued). (2) Hence every element has a square root in (of course, and are also square roots when and , respectively). Proof (continued). Consequently, if , then has roots in

(by the quadratic equation - THANKS CLASSICAL ALGEBRA!), and so splits over . So there are no irreducible monic polynomials of degree 2 in and as explained above this CONTRADICTS the assumption of the existence of where . So there is no dimension 2 extension of . Theorem V.3.19. The Fundamental Theorem of Algebra. The field of complex numbers, , is algebraically closed. Proof. We need to show that every nonconstant

polynomial splits over . By Kroneckers Theorem (Theorem V.1.10) we know that for any algebraic over , there exists extension field where . So if we prove that has no finite dimensional extension except itself, then the result will follow. Since then every finite dimensional extension field of is a finite dimensional extension of because, by Theorem V.1.2, . Proof (continued). Now every algebraic extension field of a field of characteristic 0 is separable (see the Remark

on page 261 and Lemma before Theorem V.3.11 in the notes for Section V.3) and , so a separable extension of . By Theorem V.3.16(iii), there exists extension field of such that contains and is Galois over (here, ). By Theorem V.3.16(iv) is a finite dimensional extension of . That is, is finite. We need only show that to conclude . Proof (continued). The Fundamental Theorem of Galois Theory (Theorem V.2.5(i)) shows that is finite. So is a finite group of even order (since divides ). By the First

Sylow Theorem (Theorem II.5.7) has a Sylow 2-subgroup of order where does not divide (that is, the Sylow 2subgroup has odd index ). By the Fundamental Theorem (Theorem V.2.5(i)) for the fixed field of we have that has odd dimension over since . Similar to above, since then is separable over and so by Lemma V.3.17 (notice that the fact that is infinite is used here). Of course is algebraic over . Proof (continued). Thus the irreducible polynomial of has odd degree by Theorem V.1.6(iii). By Assumption (B),

every irreducible polynomial in of degree greater than one has even degree, so the degree of the irreducible polynomial in must be 1. Therefore and . Whence and . Consequently the subgroup of has order for some where . Proof (continued). ASSUME . Then by the First Sylow Theorem (Theorem II.5.7), has a subgroup of index 2 (that is, , or ). Let be the fixed field of . By the Fundamental Theorem of Galois Theory (Theorem V.2.5(i)) is an extension of with dimension . But this CONTRADICTS

Lemma V.3.18. This contradiction to the assumption that implies that . Proof (continued). So and by the Fundamental Theorem of Galois Theory (Theorem V.2.5(i)) we have that = ( is the number of cosets of in and so equals ). Whence and (since ) . That is, every finite dimensional algebraic extension equals and is algebraically closed.

Joseph Liouville (1809-1882) There must be an easier way Liouvilles Corollary. If then there is a root of in . Proof. ASSUME not. Then is an entire bounded function. So by my theorem, is a constant function.

CONTRADICTION. Corollary V.3.20. Every proper algebraic extension field of the field of real numbers is isomorphic to the field of complex numbers. Proof. If is an algebraic extension of and has irreducible polynomial of degree greater than one, then by the Fundamental Theorem of Algebra (Theorem V.3.19) splits over . If is a root of then by Corollary V.1.9 the identity map on extends to an isomorphism . Since and , we must

have by Theorem V.1.2, and so it must be that and . So . Corollary V.3.20. Every proper algebraic extension field of the field of real numbers is isomorphic to the field of complex numbers. Proof (continued). Therefore is an algebraic extension of (which, in turn, is an algebraic extension of ). But and is algebraically closed by the Fundamental Theorem of Algebra (Theorem V.3.19) and by Theorem V.3.3(iv) (or the definition of algebraically closed on page 258) an

algebraically closed field has no algebraic extensions (except itself). Thus it must be that .