Why does $\# F(X)=\max(\# F,\# X)$ when $\# X$ is a transcendence basis? I recently asked a question about the transcendence degree of $\mathbb{C}$ over $\mathbb{Q}$. In a nice answer which I hope to understand better, two facts were given:
1: If $F$ is any infinite field and $K/F$ is an algebraic extension, then $\# K = \#F$.
2: For any infinite field $F$ and purely transcendental extension $F(X)$, we have $\# F(X) = \max (\#F, \# X)$.
For the first one, since $K$ is algebraic over $F$, then there is a surjection from $F[X]\to K$ mapping a polynomial $p(x)$ to one of its roots. Since $F$ is infinite, $\# F[X]=\# F$ so $\# F\geq\# K$. Since $K\supset F$, $\# K\geq \# F$, and equality follows. Is this correct?
I don't understand how the 2nd fact is true. Could some please explain it? Thank you.
 A: The extension $F(X)$ is isomorphic to the field of fractions of the polynomial ring $F[X]$. 
What is the cardinality of $F[X]$?  Since for every $f\in F[X]$ there exists a finite $X_0\subseteq X$ such that $f\in F[X_0]$, and $\#F[X_0]=\#F$, we have
$$\begin{align*}
\#F[X] &= \#\bigcup_{\substack{X_0\subseteq X\\ \#X_0\lt\infty}}F[X_0]\\
 &\leq \sum_{\substack{X_0\subseteq X\\ \#X_0\lt\infty}}\#F[X_0]\\
 &= \sum_{\substack{X_0\subseteq X\\ \#X_0\lt\infty}}\#F\\
&= \#\{X_0\subseteq X\mid \#X_0\lt\infty\}\#F\\
&=\left\{\begin{array}{ll}
\#X\#F &\text{if }\ \#X=\infty\\
2^{\#X}\#F &\text{if }\ \#X\lt\infty
\end{array}\right.\\
 &= \max\{\#X,\#F\}.
\end{align*}$$
On the other hand, $F\subseteq F[X]$ and $X\subseteq F[X]$, so $\max\{\#X,\#F\}\leq\#F[X]$. This gives equality.
The field of fractions can be viewed as a quotient of $F[X]\times(F[X]-\{0\})$ modulo the equivalence relation $(f,g)\sim (h,k)\Longleftrightarrow fk=gh$, and since $F[X]$ is infinite, this means that the cardinality is at most that of $F[X]$; since it contains a subring isomorphic to $F[X]$, we conclude that $\#F(X)=\#F[X]$, giving equality.
(A proof that the cardinality of the set of all finite subsets of an infinite set $X$ is equal to that of $X$ can be found here )

On rereading your argument for the first part, it is incomplete, I think.
If your $X$ represents a set of indeterminates, then you need to explain why it cannot have cardinality larger than $F$; if it represents a single indeterminate, then it is false that an algebraic extension must be a surjective image of the polynomial ring in one variable (for example, the algebraic closure of $\mathbb{Q}$ is certainly algebraic over $\mathbb{Q}$, but it is not simple, so it is not the surjective image of $\mathbb{Q}[x]$).
So you need to explain why $\#X\leq\#F$.
A better argument would be to note that there is a map from $K$ to $F[x]$ (single indeterminate) that sends every element of $K$ to its monic irreducible polynomial over $F$. This map is not one-to-one, but each element $f$ of $F[x]$ has at most $\deg(f)$ preimages. As this is finite, it follows that each fiber of the map $K\to F[x]$ is finite. Therefore,
$$\begin{align*}
\#K &\leq \sum_{f\in F[x]}\deg{f}\\
&\leq \sum_{f\in F[x]} \aleph_0\\
& = \#F[x]\aleph_0\\
&=\#F[x]\\
&=\#F
\end{align*}$$
(since $F$ is infinite). Then noting $F\subseteq K$ gives the other inequality.
A: Theorem I: If $F$ is an infinite ring then $F[x]$ has the same cardinality as $F$.
Proof: Since every polynomial in $F[x]$ is uniquely determined by its coefficients we have that $F[x]\simeq\{\langle x_0,\ldots,x_n\rangle\mid n\in\mathbb N, x_i\in F\}$, that is all the finite sequences of elements in $F$.
Since $|F|$ is infinite, we have that $|F|=|F\times F|=|\underbrace{F\times\cdots\times F}_{n\text{ times}}|=\ldots$, therefore the cardinality of $F[x]$ is the countable union of all those finite sequences therefore $|F[x]|=|\bigcup F^n|=\aleph_0\times|F|=\max\{\aleph_0,|F|\}=|F|$.
(Of course this requires quite some axiom of choice, but we're alright with that.)

Note that every algebraic number has a minimal polynomial, therefore there are at most $|F|$ many algebraic elements over $F$. Therefore every algebraic extension cannot have added more than $|F|$ new elements, thus if $K$ is algebraic over $F$ then $|K|=|F|$.
By induction (and a bit more) we have that for a countable number of variables $x_0,x_1,\ldots$ the polynomial ring in those variables has the same cardinality as $F$.
Once we add "enough" we are bound to create transcendental elements, simply because there can only be so many polynomials over $F$. Transcendental extensions correspond to the field of fractions of $F[X]$ for some set of variables $X$ (which can be infinite, of course).
The field of fraction has cardinality of $|F[X]\times F[X]|=|F[X]|$. Every polynomial in $F[X]$ is a finite sequence from $X$ and a finite sequence from $F$. As we did in Theorem I we have that the finite sequences from both $X$ and $F$ have the cardinalities of $X$ and $F$ respectively, therefore $|F[X]|=|F|\cdot|X|=\max\{|F|,|X|\}$.
