Rational function representation in the field of fractions of an integral domain

Question

In the book Algebraic Groups and Differential Galois Theory by Crespo and Hajto, one considers $V := V(Y^{2}-X^{3}+X) \subset \mathbb{A}_{\mathbb{C}}^{2}$ and $P(0,0) \in V$. The example claims that $X/Y$ is regular at $P$ because it can be written as $Y/({X^{2}-1})$ in $\mathbb{C}(V)$. Is this representation as a rational function on $V$ arrived at by polynomial long division and then simplification?

Answer 1

You can think about this as rationalizing the denominator. Since $Y^2 = X^3 - X$ on $V$, then $$ \frac{X}{Y} \cdot \frac{Y}{Y} = \frac{XY}{Y^2} = \frac{XY}{X^3 - X} = \frac{Y}{X^2 - 1} \, . $$

The field $\mathbb{C}(V)$ of rational functions on $V$ is $\frac{\mathbb{C}(X)[Y]}{(Y^2 - (X^3 - X))}$. This is a degree $2$ extension of $\mathbb{C}(X)$ with basis $1, Y$, and thus every rational function on $V$ can be written as $f(X) + g(X)Y$ for some $f, g \in \mathbb{C}(X)$.

Consider the analogous (and perhaps more familiar) number field case. Every element of the number field $\mathbb{Q}(\sqrt{2})$ can be written as $u + v \sqrt{2}$ for some $u,v \in \mathbb{Q}$. To write the reciprocal of a nonzero element in this form we would rationalize the denominator by multiplying the top and bottom by the conjugate: $$ \frac{1}{a + b\sqrt{2}} \cdot \frac{a - b\sqrt{2}}{a - b\sqrt{2}} = \frac{a}{a^2 - 2b^2} - \frac{b}{a^2 - 2b^2} \sqrt{2} \, . $$ We can use the same method in $\mathbb{C}(V)$: $$ \frac{1}{f + g Y} \cdot \frac{f - g Y}{f - gY} = \frac{f}{f^2 - (X^3 - X)g^2} - \frac{g}{f^2 - (X^3 - X)g^2} Y \, . $$