finite transcendence degree and algebraic closure

Question

Let $k$ be an algebraically closed field. Let $K$ be an extension field of $k$ of finite transcendence degree over $k$. Intuitively, it seems to me that $K$ can not be algebraically closed. Is there a proof of that fact or a counterexample?

Answer 1

Lets just take a simple example.

$k(t)$ is not algebraically closed. For example, it doesn't have $\sqrt{t}$. (it has trdeg 1)

The algebraic closure $\overline{k(t)}$ of $k(t)$ has finite transcendence degree (ie, 1). This follows from the definition of transcendence degree, since of course $\overline{k(t)}/k(t)$ is algebraic.