cardinality of algebraic closure

cardinality of algebraic closure

Theorem 1

.

If a field is finite, then its

algebraic closure

is

countably infinite

.

Proof.

Because a

finite field cannot be algebraically closed

, the algebraic closure
of a

finite field

must be

infinite

. Hence, it only remains to show that the
algebraic closure is

countable

. Every element of the algebraic closure is
the root of some

polynomial

. Furthermore, every polynomial
has a finite number of roots (the number is

bounded

by its degree) and there
are a countable number of polynomials whose

coefficients

belong to a given

finite set

. Since the union of a countable family of finite sets is
countable, the number of elements of the algebraic closure is countable.
∎

Theorem 2

.

If a field is infinite, then its algebraic closure has the same

cardinality

as the original field.

Proof.

Since a field is

isomorphic

to a subset of its algebraic closure, it follows
that the cardinality of the

closure

is at least the cardinality of the original field. The number of polynomials of degree

n

with coefficients in
a given set is the same as the number of

n

- tuplets of elements of

S

,
which is the cardinality of the set raised to the

n

-th power. Since an
infinite

cardinal

raised to an finite power equals itself, the number of
polynomials of a given degree equals the the cardinality the original field.
Since the cardinality of the union of a countable number of sets each of
which has the same infinite number of

elements

equals the common cardinality
of the sets, the total number of polynomials with coefficients in the field
equals the cardinality of the field. Since every element of the algebraic
closure of a field is the root of some polynomial with elements of the
field for coefficients and a polynomial has a finite number of roots, it
follows that the cardinality of the algebraic closure is bounded by the
cardinality of the original field.
∎

Theorem 3

.

For every transfinite cardinal number

N

, there exists an algebrically
closed field with exactly

N

elements.

Proof.

Let

F

be the

field of rational functions

with

integer

coefficients in

variables

x

i

, where the index

i

ranges over an

index set

I

whose cardinality is

N

. We claim that the cardinality of

F

is

N

.
The cardinality is at least

N

becasue we have the

N

rational functions

x

i

, so it only remains to show that the cardinality is not greater
than

N

. To do this, we first show that the number of polynomials in
the

x

i

with integer coefficients equals

N

. A polynomial is
determined by a finite set of coefficients and a finite set of

monomials

.
The number of possible sets of coefficients is the number of finite
tuplets of integers, which is

ℵ

0

. Since a monomial may be determined
by a

mapping

of a finite set into the set

{

x

i

∣

i

∈

I

}

, the number
of possible monomials of degree

n

is bounded by

N

n

. Since

N

is
transfinite and

n

is finite, we have

N

n

=

N

. Thus the number of
possible monomials is bounded by

N

⁢

ℵ

0

=

N

. So the number of
polynomials is bounded by the

product

of

ℵ

0

and

N

, which is

N

and the number of rational functions is bounded by

N

2

, which
equals

N

.
∎

Title

cardinality of algebraic closure

Canonical name

CardinalityOfAlgebraicClosure

Date of creation

2013-03-22 16:28:54

Last modified on

2013-03-22 16:28:54

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

22

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 12F05

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