The field of rational functions of $k$ contains an algebraic closure of $k$?

Question

If $k$ is a field and $k(\mathbf{x}) = k(x_1,\dots,x_m)$ is the field of rational functions in the indeterminates $x_1,\dots,x_m$, then if $f(y) \in k(\mathbf{x})[y]$ is irreducible, take the extension $K$ of $k(\mathbf{x})$ given by $$K = k(\mathbf{x})[y]/(f).$$ The book I'm reading (Groups as Galois Groups) then makes the statement "let $\hat k$ be the algebraic closure of $k$ in $K$". I have no idea why $K$ contains an algebraic closure of $k$; in particular, does $k(\mathbf{x})$ contain an algebraic closure of $k$ (this I'm pretty sure is false)?

Answer 1

If $K/k$ is an extension of fields, the algebraic closure of $k$ in $K$ is the subfield of $K$ of those elements which are algebraic over $k$.