Is the degree of an infinite algebraic extensions always countable?

Question

I guess this is right and try to prove it by using the fact that the polynomial ring $K[t]$ has a countable basis $1,x,x^2,\cdots$. But How to use this fact?

Aside, if this statement is true. Is the following one also true?

Let $\{L_i\}$ be a direct system of algebraic extension of $K$ and $L$ be the direct limit of them. Then $[L:K]=\underrightarrow{\lim}[L_i:K]$, where the degrees can be taken to be infinite.

Of course this is true for Galois extensions as in that case, one has \[G(L/K)=\underleftarrow{\lim}G(L_i/K).\] But how about algebraic extensions?

Edit: The statement is wrong. But I still want a proof for local fields, i.e fields with complete valuation and finite residue fields.

Answer 1

No!
Consider a set $I$ of indices and the field of rational functions $K=k(X_i|i\in I)$ over an arbitrary field $k$.
The extension field $K\subset L=k(\sqrt X_i|i\in I)$ is algebraic (since it is generated by algebraic elements), of degree $[L:K]\geq \operatorname {card} I$ because the elements $\sqrt X_i$ are linearly independent over $K$.
Thus by taking $I$ non countable you obtain an algebraic extension $K\subset L$ of non countable degree.

Answer 2

Another example of uncountable-degree algebraic extension: Let the small field be $k=\Bbb R(t)$, rational functions in one variable over the reals. Then the irrationalities $\sqrt{t-\alpha}$, for $\alpha\in\Bbb R$, are clearly $k$-linearly independent, and uncountable in number, so the compositum of all the quadratic extensions $\Bbb R(\sqrt{t-\alpha})$ over $k$ is uncountable in degree.

Not clear that $\{\sqrt{t-\alpha_i}\}$ are linearly independent over $k=\Bbb R(t)$? Let $\sqrt{t-\beta}=\sum_if_i(t)\sqrt{t-\alpha_i}$, with $\beta$ different from all $\alpha_i$. Now square both sides. As real functions, the (new) left-hand side is zero at $\beta$, with derivative $1$, while the right either has a pole, is nonzero, or is zero with derivative zero at $\beta$.