# The Rational Function Field of an Affine Variety

Definition: Let $K$ be a field and let $V \subseteq \mathbb{A}^n(K)$ be a nonempty affine variety. The Rational Function Field of $V$ is defined as $K(V) = \left \{ \frac{f}{g} : f, g \in \Gamma(V), g \neq 0 \right \}$. Elements of $K(V)$ are called Rational Functions. |

*The rational function field of $V$ is also sometimes called the Quotient Field of $V$ or the Field of Fractions of $V$.*

Definition: Let $K$ be a field, $V \subseteq \mathbb{A}^n(K)$ a nonempty affine variety, and let $\mathbf{p} \in V$. A rational function $\displaystyle{f \in k(V)}$ is Defined at $\mathbf{p}$ if there exists polynomials $a, b \in \Gamma(V)$ with $\displaystyle{f = \frac{a}{b}}$ and such that $b(\mathbf{p}) \neq 0$. |

Definition: Let $K$ be a field, $V \subseteq \mathbb{A}^n(K)$ a nonempty affine variety, and let $\mathbf{p} \in V$. The Local Ring of $V$ at $\mathbf{p}$ is defined to be the subring $O_{\mathbf{p}}(V)$ of $K(V)$ of rational functions on $V$ that are defined at $\mathbf{p}$. |

In the following proposition we prove that $O_{\mathbf{p}}(V)$ is indeed a subring of $K(V)$.

Proposition 1: Let $K$ be a field, $V \subseteq \mathbb{A}^n(K)$ a nonempty affine variety, and let $\mathbf{p} \in V$. Then the local ring of $V$ at $\mathbf{p}$, $O_{\mathbf{p}}(V)$ is a subring of $K(V)$ that contains $\Gamma(V)$. |

*We use the subring test which can be found on the Subrings and Ring Extensions page to prove that $O_{\mathbf{p}}(V)$ is a subring of $K(V)$.*

- Clearly we have that:

(1)

\begin{align} \quad \Gamma(V) \subseteq O_{\mathbf{p}}(V) \subseteq K(V) \end{align}

- So all that remains to show is that $O_{\mathbf{p}} (V)$ is indeed a subring of $K(V)$.

- Let $f, g \in O_{\mathbf{p}}(V)$ with $\displaystyle{f = \frac{a}{b}}$, $\displaystyle{g = \frac{c}{d}}$ where $a, b, c, d \in \Gamma(V)$ and $b(\mathbf{p}) \neq 0$, $d(\mathbf{p}) \neq 0$. Then:

(2)

\begin{align} \quad f + g = \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} \end{align}

(3)

\begin{align} \quad fg = \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd} \end{align}

- Observe that since $b(\mathbf{p}), d(\mathbf{p}) \neq 0$ and $\Gamma(V)$ is an integral domain we have that $(bd)(\mathbf{p}) \neq 0$ and so $f + g$ and $fg$ are defined at $\mathbf{p}$ which shows that $O_{\mathbf{p}}(V)$ is closed under addition and closed under multiplication.

- Let $f \in O_{\mathbf{p}}(V)$. Then there exists $a, b \in \Gamma(V)$ such that $\displaystyle{f = \frac{a}{b}}$ and $b(\mathbf{p}) \neq 0$. Since $\Gamma(V)$ is a ring, $-a \in \Gamma(V)$ and so $\displaystyle{-f = \frac{-a}{b}}$. So $-f \in O_{\mathbf{p}}(V)$.

- Lastly we clearly see that $1$ is defined at $\mathbf{p}$ so $1 \in O_{\mathbf{p}}(V)$.

- Therefore $O_{\mathbf{p}}(V)$ is a subring of $K(V)$. $\blacksquare$

Theorem 2: Let $K$ be a field and let $V \subseteq \mathbb{A}^n(K)$ be a nonempty affine variety. Then $\Gamma(V) = \bigcap_{\mathbf{p} \in V} O_{\mathbf{p}}(V)$. |