Example 9.7.4 (Degree of a rational function field). If $k$ is any field, then the rational function field $k(t)$ is *not* a finite extension. For example the elements $\left\{ t^ n, n \in \mathbf{Z}\right\} $ are linearly independent over $k$.

In fact, if $k$ is uncountable, then $k(t)$ is *uncountably* dimensional as a $k$-vector space. To show this, we claim that the family of elements $\{ 1/(t- \alpha ), \alpha \in k\} \subset k(t)$ is linearly independent over $k$. A nontrivial relation between them would lead to a contradiction: for instance, if one works over $\mathbf{C}$, then this follows because $\frac{1}{t-\alpha }$, when considered as a meromorphic function on $\mathbf{C}$, has a pole at $\alpha $ and nowhere else. Consequently any sum $\sum c_ i \frac{1}{t - \alpha _ i}$ for the $c_ i \in k^*$, and $\alpha _ i \in k$ distinct, would have poles at each of the $\alpha _ i$. In particular, it could not be zero.

Amusingly, this leads to a quick proof of the Hilbert Nullstellensatz over the complex numbers. For a slightly more general result, see Algebra, Theorem 10.34.11.

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