If $F, K$ are fields, $F$ algebraically closed, and $F \subseteq K$ then $K = F$? 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.
 A: 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, $|F(X)|=\max\{|F|,|X|,|\mathbb{N}|\}$.
