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 $\Q \cup S_\alpha$. {{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 $\left\vert{\Q \cup S_\alpha}\right\vert = \aleph_0 + \left\vert{\alpha}\right\vert < \kappa$ 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 $\Q$. {{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 $P \left({x_1, \ldots, x_n}\right)$ be a non-trivial polynomial with coefficients in $\Q$ and elements $a_1, \ldots, a_n$ in $F$.   {{WLOG}}, 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 $\Q$. {{qed}} {{wtd|... implicit use of the axiom of choice. Determine if its necessary}} Category:Field Theory Category:Uncountable Sets h0pxhtc8ixdqu51pz8zi9efmgtibyuy