Field of Uncountable Cardinality K has Transcendence Degree K

Theorem
Let $F$ be a field of uncountable cardinality $\kappa$.

Then $F$ has transcendence degree $\kappa$ over its prime field.

Proof
We prove the theorem for fields with characteristic $p=0$. In the case where $p$ is a prime, the proof is similar, but instead we view the fields as extensions of $\Z / \Z_p$. The main idea is to exploit the lower cardinality of sets of polynomials in order to keep finding algebraically independent elements of $F$.

Since each characteristic $0$ field contains a copy of $\Q$ as its prime field, we can view $F$ as a field extension over $\Q$.

We will show that $F$ has a subset of cardinality $\kappa$ which is algebraically independent over $\Q$.

Since $\kappa$ is the largest possible cardinality for a subset of $F$, this will establish the theorem.

We build the claimed subset of $F$ by transfinite induction and implicit use of the axiom of choice.

For each ordinal $\alpha < \kappa$ we define a set $S_\alpha$.

We will build the sets so that each $S_\alpha$ has cardinality equal to that of $\alpha$ and is algebraically independent over $\Q$.

Let $S_0 = \varnothing$.

Let $S_1$ be a singleton containing some element of $F$ which is not algebraic over $\Q$.

This is possible since the set of algebraic numbers is countable.

Define $S_\beta$ for successor ordinals $\beta=\alpha + 1 < \kappa$ to be $S_\alpha$ together with an element of $F$ which is not a root of any non-trivial polynomial with coefficients in $\Q \cup S_\alpha$.

This is possible since there are only $|\Q\cup S_\alpha| = \aleph_0 + | \alpha | < \kappa$ many such polynomials, and hence this many roots of such polynomials since each polynomial has finitely many roots and the union of $\kappa$ many finite sets is at most size $\kappa$.

Define $S_\beta$ for limit ordinals by $\displaystyle{S_\beta = \bigcup_{\alpha < \beta} S_\alpha}$.

We can then define $S_\kappa$ to be $\displaystyle{\bigcup_{\alpha < \kappa} S_\alpha}$.

This is a set of size $\kappa$ since its elements can be paired with the ordinals less than $\kappa$.

We check now that it is algebraically independent over $\Q$, which we do by contraposition.

Let $P(x_1,\dots,x_n)$ be a non-trivial polynomial with coefficients in $\Q$ and elements $a_1,\dots, a_n$ in $F$.

Without loss of generality, we can assume $a_n$ was added at an ordinal $\alpha + 1$ later than the other elements.

Then, $P(a_1,\dots,a_{n-1},x_n)$ is a polynomial with coefficients in $\Q\cup S_\alpha$.

We selected $a_n$ at stage $\alpha + 1$ so that it was not a root of such a polynomial, so $P(a_1,\dots,a_n)\ne 0$.