The Stacks project

Example 10.35.13. The trick in the proof of Theorem 10.35.11 really does not work if $k$ is a countable field and $I$ is countable too. Let $k$ be a countable field. Let $x$ be a variable, and let $k(x)$ be the field of rational functions in $x$. Consider the polynomial algebra $R = k[x, \{ x_ f\} _{f \in k[x]-\{ 0\} }]$. Let $I = (\{ fx_ f - 1\} _{f\in k[x] - \{ 0\} })$. Note that $I$ is a proper ideal in $R$. Choose a maximal ideal $I \subset \mathfrak m$. Then $k \subset R/\mathfrak m$ is isomorphic to $k(x)$, and is not algebraic over $k$.