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 $k$.

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: