Field of Uncountable Cardinality K has Transcendence Degree K

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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 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 $\Q \cup S_\alpha$.



From Set of Polynomials over Infinite Set has Same Cardinality there are only $\left\vert{\Q \cup S_\alpha}\right\vert = \aleph_0 + \left\vert{\alpha}\right\vert < \kappa$ 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 $\displaystyle {S_\beta = \bigcup_{\alpha \mathop < \beta} S_\alpha}$.

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

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 $\Q$.



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

Without loss of generality, it is assumed that $a_n$ was added at an ordinal $\alpha + 1$ later than the other elements.

Then $P \left({a_1, \ldots, a_{n-1}, x_n}\right)$ 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 $P \left({a_1, \ldots, a_{n-1}, x_n}\right) \ne 0$.

Thus it is algebraically independent over $\Q$.

$\blacksquare$