The Fundamental Theorem of Algebra – An (Almost) Algebraic Proof - PowerPoint Presentation

Slide1

The Fundamental Theorem of Algebra – An (Almost) Algebraic Proof

Robert “Dr. Bob” Gardner

Based on Hungerford’s

Appendix to Section V.3 in Algebra, Springer-Verlag (1974)

The field of complex numbers, , is algebraically closed.

 Slide2

Lemma

V.3.17.

If is a finite dimensional separable extension of an infinite field

, then for some .  Theorem V.3.19. The Fundamental Theorem of

Algebra.The field of complex numbers,

, is algebraically closed.

Lemma V.3.18.

There are no extension fields of dimension 2 over the field of complex numbers.

Corollary V.3.20.

Every proper algebraic extension field of the field of real numbers is isomorphic to the field of complex numbers.

Results of the Appendix to Section V.3Slide3

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.

Some Results from Real AnalysisSlide4

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 one-to-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

Lemma

V.3.17

.

If is a finite dimensional separable extension of an infinite field , then for some .

 Slide5

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 ).

 Slide6

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 .

 Slide7

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. Lemma V.3.18. There are no extension fields of dimension 2 over the field of complex numbers.For each

, we know

that

has a real positive square root be

Assumption (A), denoted

.

 Slide8

Proof (continued).

Also

by Assumption (A) the positive real numbers

and

have

real positive square roots,

say

and

respectively.

Now

(1)

 Slide9

Proof (continued).

(2

)

Hence

every

element

has

a square root

in

(

of course,

and

are also square roots

when

and

, respectively).

 Slide10

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

.

 Slide11

Proof.

We need to show that every

nonconstant

polynomial splits over . By Kronecker’s 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

,

.

Theorem V.3.19. The Fundamental Theorem of

Algebra.

The field of complex numbers, , is algebraically closed. Slide12

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 .  Slide13

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

2-subgroup 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 .  Slide14

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

.

 Slide15

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

.

 Slide16

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

.

 Slide17

There must be an easier way…

Joseph Liouville (

1809-1882

)Liouville’s 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.

 Slide18

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

.

 Slide19

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

.