 # Suppose f &#x2208;<!-- ∈ --> O p </msub> ( V ) is a rational func Kolten Conrad 2022-07-09 Answered
Suppose $f\in {O}_{p}\left(V\right)$ is a rational function on V that has a value at $p$. Then write $f=a/b={a}^{\prime }/{b}^{\prime }$ where $a,b,{a}^{\prime },{b}^{\prime }\in \mathrm{\Gamma }\left(V\right)$, the coordinate ring of V. Want to show the value of $f$ at $p$ is well-defined, i.e. $a\left(p\right)/b\left(p\right)={a}^{\prime }\left(p\right)/{b}^{\prime }\left(p\right)$.
So since $a/b={a}^{\prime }/{b}^{\prime }$ are the same equivalence class, there is some non-zero poly $x\in \mathrm{\Gamma }\left(V\right)$ such that $x\left(a{b}^{\prime }-{a}^{\prime }b\right)=0$. Then $x\left(p\right)\left(a\left(p\right){b}^{\prime }\left(p\right)-{a}^{\prime }\left(p\right)b\left(p\right)\right)=0$. Then what? How do we know $x\left(p\right)\ne 0$?
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So $x\ne \overline{0}\in \mathrm{\Gamma }\left(V\right)$. Since $\mathrm{\Gamma }\left(V\right)$ is an integral domain, ${a}^{\prime }b-a{b}^{\prime }=\overline{0}$. Actually one can remove $x$ from the definition of fraction field altogether.

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