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.

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: https://en.wikipedia.org/wiki/Field_of_fractions). 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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