Robert Dr Bob Gardner Based on Hungerfords Appendix to Section V3 in Algebra Springer Verlag 1974 The field of complex numbers is algebraically closed Lemma V317 ID: 690205
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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
.