Exercise 111.31.5. In this exercise we try to see what happens with regular functions over non-algebraically closed fields. Let $k$ be a field. Let $Z \subset k^ n$ be a Zariski locally closed subset, i.e., there exist ideals $I \subset J \subset k[x_1, \ldots , x_ n]$ such that

A function $\varphi : Z \to k$ is said to be *regular* if for every $z \in Z$ there exists a Zariski open neighbourhood $z \in U \subset Z$ and polynomials $f, g \in k[x_1, \ldots , x_ n]$ such that $g(u) \not= 0$ for all $u \in U$ and such that $\varphi (u) = f(u)/g(u)$ for all $u \in U$.

If $k = \bar k$ and $Z = k^ n$ show that regular functions are given by polynomials. (Only do this if you haven't seen this argument before.)

If $k$ is finite show that (a) every function $\varphi $ is regular, (b) the ring of regular functions is finite dimensional over $k$. (If you like you can take $Z = k^ n$ and even $n = 1$.)

If $k = \mathbf{R}$ give an example of a regular function on $Z = \mathbf{R}$ which is not given by a polynomial.

If $k = \mathbf{Q}_ p$ give an example of a regular function on $Z = \mathbf{Q}_ p$ which is not given by a polynomial.

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