# 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 ${\mathrm{\aleph}}_{0}$. Since a monomial may be determined
by a mapping of a finite set into the set $\{{x}_{i}\mid i\in I\}$, 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 $N{\mathrm{\aleph}}_{0}=N$. So the number of
polynomials is bounded by the product^{} of ${\mathrm{\aleph}}_{0}$ 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 |