I don't understand an example in Vakil's Lecture notes on Algebraic Geometry.
Consider $\mathbb{A}_k^2 = \operatorname{Spec}k[x,y]$, for $\operatorname{char}(k) \neq 2$. Then $f:=\frac{(x^2 + y^2)}{x(y^2-x^5)}$ is a function away from the $y$-axis and the curve $y^2-x^5$. So first I don't understand how you can even consider it as an element in $k[x,y]$. Shouldn't it be in $k(x)$ ?
And then I don't understand how this function is evaluated at the ideal $p:=[(x-2,y-4)]$. This should be done by taking the localisation $f_p$ and then modding out by $p k[x,y]_p$. Why does this give $\frac{(2^2 + 4^2)}{2(4^2-2^5)}$ ?
Using Vakil's notation, $f$ is a function on the open set $D(x) \cap D(y^2-x^5)$, not on $\mathbb{A}_2^k$. Indeed, the functions on this open set are given by the localization $k[x,y]_{x(y^2-x^5)}$ (since the functions $x$ and $y^2-x^5$ do not vanish on $D(x) \cap D(y^2-x^5)$, you can invert them). Moreover, since $p \in D(x) \cap D(y^2-x^5)$ (i.e. the denominators of $f$ do not vanish at $p$), you can consider $f$ as living in the localization $k[x,y]_\mathfrak{p}$, where $\mathfrak{p} = (x-2,y-4)$. To evaluate this function at $p = [\mathfrak{p}] = [(x-2,y-4)]$, you simply look at $f \text{ mod } \mathfrak{p}$, i.e. at the image of $f$ under the quotient map \begin{equation*} k[x,y]_\mathfrak{p} \longrightarrow k[x,y]_\mathfrak{p}/\mathfrak{p}k[x,y]_\mathfrak{p}. \end{equation*} Under this map, $x$ is identified with $2$ and $y$ is identified with $4$, so \begin{equation*} f \longmapsto \frac{2^2+4^2}{2(4^2-2^5)}. \end{equation*}