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 closureof a finite field must be infinite. Hence, it only remains to show that thealgebraic closure is countable. Every element of the algebraic closure isthe root of some polynomial. Furthermore, every polynomialhas a finite number of roots (the number is bounded by its degree) and thereare a countable number of polynomials whose coefficients belong to a givenfinite set. Since the union of a countable family of finite sets iscountable, the number of elements of the algebraic closure is countable.∎

Theorem 2.

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

Proof.

Since a field is isomorphic to a subset of its algebraic closure, it followsthat the cardinality of the closure is at least the cardinality of the original field. The number of polynomials of degree $n$ with coefficients ina 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 aninfinite cardinal raised to an finite power equals itself, the number ofpolynomials of a given degree equals the the cardinality the original field.Since the cardinality of the union of a countable number of sets each ofwhich has the same infinite number of elements equals the common cardinalityof the sets, the total number of polynomials with coefficients in the fieldequals the cardinality of the field. Since every element of the algebraicclosure of a field is the root of some polynomial with elements of thefield for coefficients and a polynomial has a finite number of roots, itfollows that the cardinality of the algebraic closure is bounded by thecardinality of the original field.∎

Theorem 3.

For every transfinite cardinal number $N$, there exists an algebricallyclosed field with exactly $N$ elements.

Proof.

Let $F$ be the field of rational functions with integer coefficients invariables $x_i$, 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 greaterthan $N$. To do this, we first show that the number of polynomials inthe $x_i$ with integer coefficients equals $N$. A polynomial isdetermined by a finite set of coefficients and a finite set of monomials.The number of possible sets of coefficients is the number of finitetuplets of integers, which is ${\mathrm{\aleph }}_0$. Since a monomial may be determinedby a mapping of a finite set into the set $\{x_i\mid i\in I\}$, the numberof possible monomials of degree $n$ is bounded by $N^n$. Since $N$ istransfinite and $n$ is finite, we have $N^n=N$. Thus the number ofpossible monomials is bounded by $N{\mathrm{\aleph }}_0=N$. So the number ofpolynomials is bounded by the product of ${\mathrm{\aleph }}_0$ and $N$, which is$N$ and the number of rational functions is bounded by $N^2$, whichequals $N$.∎