If $F, K$ are fields, $F$ algebraically closed, and $F subseteq K$ then $K = F$?

Mathematics Asked on January 1, 2022

I want to show that an algebraically closed field $F$ cannot be contained in a larger field $K$. So if $F subseteq K$ then $F = K$ for all fields $K$.
Here’s my attempt at a proof:

For contradiction, Say $F subsetneq K$. Hence there is an element $k in K$, $k not in F$.

  • if $k$ is algebraic over $F$, then we adjoin $k$ into $F$. So consider $F(k) equiv F[X] / (p)$ for some polynomial $p in F[X]$. Since $F(k)$ should be a field, we need $(p)$ to be maximal [ring quotient maximal ideal is field]. Since we have that $F$ is algebraically closed, irreducible polynomials [which generate maximal ideals] are going to be linear. Hence $(p)$ is maximal and F$(k)$ is a field iff $p = (x – f_star)$ for some $f_star in F$. But $F[X]/ (x- f_star) simeq F$, since quotienting by $(x – f_star)$ keeps only degree 0 polynomials, which are the constant elements. So we have that $F(k) = F$. Hence we cannot have $k not in F$, giving us the desired contradiction.
  • If $k$ is transcendental over $F$, I feel that some argument ought to hold, but I don’t know how to proceed.

Is there a counter-example, where $F$ is algebraically closed, while still possessing an extension $K = F(k)$ for some $k$ that is transcendental over $F$? The only algebraically closed field I have experience with, $mathbb C$, does not allow such a thing to happen as far as I am aware.

One Answer

The algebraic numbers are an algebraically closed field that is properly contained in $mathbb{C}$ and many other fields. For example, if $F$ is the field of algebraic numbers, $K=mathbb{C}$, and $k=pi$, then we are precisely in the second case of your proof sketch, and we of course cannot show that $k$ is in $F$.

Given any (algebraically closed field) $F$ one can always construct $F(t)$ where $t$ is transcendental with respect to $F$ to get a bigger field. Then you can take the algebraic closure of $F(t)$ to get a bigger algebraically closed field.

By adjoining a lot of transcendentals you can get algebraically closed fields of arbitrarily large cardinality.

I will add some details that have by now been talked about in other comments. Let $F$ be a field and let $X$ be a set of variables. Let $F[X]$ be the ring of polynomials whose variables come from $X$ and coefficients from $F$. Now let $F(X)$ be the field of fractions of $F[X]$ (see: You can also view $F(X)$ as the field of rational functions in variables from $X$ with coefficients in $F$. Now $F(X)$ is a new field properly containing $F$. Moreover, $|F(X)|=max{|F|,|X|,|mathbb{N}|}$.

Answered by halrankard on January 1, 2022

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