I am reading this document about the Bézout Theorem for curves. For some context on my doubt, we must keep in mind the following preliminary information:
Definition. Given a point $a \in \Bbb P^2$, the local ring of rational functions at $a$, denoted $R_a$, is defined as the set of all rational functions $s/t \in k(x, y, z)$ where $s, t \in k[x, y, z]$ are homogenous of the same degree and $t(a) \neq 0$.
Definition. Let $f_1,\dots,f_n \in k[x, y, z]$ be homogenous polynomials and let $a \in \Bbb P^2$ be a point.Then the ideal $(f_1,\dots, f_n)_a \subset R_a$ generated by $f_1, \dots, f_n$ is the set of rational functions in $R_a$ with numerators in $(f_1,\dots,f_n) \subset k[x, y, z]$. That is, it’s the set of all rational functions of the form $$ \frac{g_1f_1 + \dots + g_nf_n}{s},$$ where $g_1, \dots, g_n$ and $s$ are homogenous polynomials in $k[x,y,z]$ such that $s(a) \neq 0$ and the $g_jf_j$ are all of the same degree as $s$.
Definition. Givne nonzero coprime homogeneous polynomials $f,g \in k[x,y,z],$ the intersection multiplicity of $f$ and $g$ at $a$ is defined the k-dimension of $R_a/(f,g)_a$ and is denoted by $i(f \cap g,a)$.
The main result in which I have doubts follows below.
Theorem. The intersection multiplicty $i(f \cap g, a)$ is strictly positive if $a \in Z(f) \cap Z(g)$ and zero otherwise.
Proof presented in the document. Suppose first that $a \notin Z(f) \cap Z(g).$ Then, either $f$ doesn't vanish at $a$ or $g$ doesn't vanish at $a$. Without loss of generality, assume that $f$ doesn't vanish at $a$. $\color{red}{\text{Then } 1/f \in R_a},$ so $f$ is a unit and thus $(f,g)_a = R_a.$ Hence $R_a / (f,g)_a = 0$ (trivial ideal) and thus $i(f \cap g,a) = 0.$
Now, suppose that $a \in Z(f) \cap Z(g).$ Then, $f(a) = g(a) = 0$. But then every element of $(f,g)_a$ must vanish at $a$. The element $1 = 1/1 \in R_a$ certainly doesn't vanish at $a$, so $(f,g)_a \neq R_a.$ Therefore, $R_a / (f,g)_a$ is not the trivial ideal which means that $i(a, f \cap g) > 0.$
My doubts and thinking. For the first part, I don't understand why we must have $1/f \in R_a$. By definition of $R_a$ this means that $f$ must have degree $0$ and I really can't see where this comes from. Assuming this is true, it follows that $f$ is a unit that thus $(f,g)_a = R_a$ since an ideal is proper iff it contains no units.
On the other hand, for the second part, I understand that every element of $(f,g)_a$ must vanish at $a$ since the numerator will always be zero. Therefore, we come to the conclusion that $1 \notin (f,g)_a$ and thus $(f,g)_a$ is proper (an ideal is proper iff it does not contain $1$). So, basically, I think that my only issue is in understand why one must have $1/f \in R_a.$
Thanks for any help in advance.
Indeed, $1/f \notin R_a$ in general, since, by definition, elements of $R_a$ are homogeneous of degree zero.
A correct proof would be to note that, if $f(a) \neq 0$, then $1=\frac{f}{f} \in (f,g)_a$, so $(f,g)_a=R_a$.