Function field of a variety

Remarks

Since $X$ is irreducible, $U \cap V$ is nonempty and dense, and two regular functions that agree on a nonempty open subset of $U \cap V$ agree on all of $U \cap V$. This makes the relation transitive and the operations well defined. $k(X)$ is a field: if $f \ne 0$ on $U$, then $1/f$ is regular on the nonempty open set $U \setminus V(f)$.

If $X$ is an affine variety, then $k(X)$ is the field of fractions of the coordinate ring $k[X]$. If $U$ is a nonempty open subset of $X$, then $k(U) = k(X)$. $k(X)$ is a function field over $k$ in the algebraic sense: a finitely generated field extension of $k$.

Example: $k(\mathbb{A}^n) = k(x_1, \dots, x_n)$, the field of rational functions, since $k[\mathbb{A}^n]$ is the polynomial ring. Example: for the parabola $X = V(y - x^2)$, $k[X] \cong k[x]$, so $k(X) \cong k(x)$.

Non-example: the construction needs irreducibility. In $X = V(xy) \subseteq k^2$, the open sets $\set{x \ne 0}$ and $\set{y \ne 0}$ of $X$ are nonempty and disjoint. So the relation above is not transitive there, and the analogous ring built from $k[X] = k[x, y]/(xy)$, its total ring of fractions, is $k(x) \times k(y)$, which is not a field.