Kolten Conrad

2022-07-09

Suppose $f\in {O}_{p}(V)$ is a rational function on V that has a value at $p$. Then write $f=a/b={a}^{\prime}/{b}^{\prime}$ where $a,b,{a}^{\prime},{b}^{\prime}\in \mathrm{\Gamma}(V)$, the coordinate ring of V. Want to show the value of $f$ at $p$ is well-defined, i.e. $a(p)/b(p)={a}^{\prime}(p)/{b}^{\prime}(p)$.

So since $a/b={a}^{\prime}/{b}^{\prime}$ are the same equivalence class, there is some non-zero poly $x\in \mathrm{\Gamma}(V)$ such that $x(a{b}^{\prime}-{a}^{\prime}b)=0$. Then $x(p)(a(p){b}^{\prime}(p)-{a}^{\prime}(p)b(p))=0$. Then what? How do we know $x(p)\ne 0$?

So since $a/b={a}^{\prime}/{b}^{\prime}$ are the same equivalence class, there is some non-zero poly $x\in \mathrm{\Gamma}(V)$ such that $x(a{b}^{\prime}-{a}^{\prime}b)=0$. Then $x(p)(a(p){b}^{\prime}(p)-{a}^{\prime}(p)b(p))=0$. Then what? How do we know $x(p)\ne 0$?

Keegan Barry

Beginner2022-07-10Added 18 answers

So $x\ne \overline{0}\in \mathrm{\Gamma}(V)$. Since $\mathrm{\Gamma}(V)$ 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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