Theorem: Field of Uncountable Cardinality K has Transcendence Degree K

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 $S_0 = \O$.
Let $S_1$ be a singleton containing some element of $F$ which is not algebraic over $\Q$.
This is possible from Algebraic Numbers are 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 $\Q \cup S_\alpha$.
{{explain|Determine whether "trivial" polynomial actually means the Definition:Null Polynomial over Ring or maybe one of Definition:Polynomial of Degree Zero or something -- whichever is meant, replace with what we have here, and make a note on that page to indicate this alternative terminology.}}
From Set of Polynomials over Infinite Set has Same Cardinality there are only $\card {\Q \cup S_\alpha} = \aleph_0 + \card \alpha < \kappa$ many such polynomials.
hence the above construction is possible.
From Polynomial over Field has Finitely Many Roots and the Cardinality of Infinite Union of Infinite Sets, the union of $\kappa$ many finite sets is at most size $\kappa$.
Thus there are are this many roots of such polynomials
Define $S_\beta$ for limit ordinals by $\ds S_\beta = \bigcup_{\alpha \mathop < \beta} S_\alpha$.
We can then define $S_\kappa$ to be $\ds \bigcup_{\alpha \mathop < \kappa} S_\alpha$.
This is a set of size $\kappa$ since its elements can be paired with the ordinals less than $\kappa$.  
It is now demonstrated that it is algebraically independent over $\Q$.
{{explain|At this point it is easy to lose track of what "it" is. I have indeed done so. So it is worth restating here what which particular object we are demonstrating to be algebraically independent.}}
Let $\map P {x_1, \ldots, x_n}$ be a non-trivial polynomial with coefficients in $\Q$ and elements $a_1, \ldots, a_n$ in $F$.  
{{WLOG}}, it is assumed that $a_n$ was added at an ordinal $\alpha + 1$ later than the other elements.  
Then $\map P {a_1, \ldots, a_{n - 1}, x_n}$ is a polynomial with coefficients in $\Q \cup S_\alpha$.  
$a_n$ was selected at stage $\alpha + 1$ so as not to be root of such a polynomial.
So $\map P {a_1, \ldots, a_{n - 1}, x_n} \ne 0$.
Thus it is algebraically independent over $\Q$.
{{qed}}
{{wtd|... implicit use of the axiom of choice. Determine if its necessary}}
Category:Field Theory
Category:Uncountable Sets
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