I am studying algebraic geometry using the Fulton. They say that the local ring at p of a variety V mod the maximal ideal consisting of $m_p = \{ f\in O_p(V) : f(p) = 0\}$ is isomorphic to the field k where the coefficients of the polynomials come from. This is due to the fact, that the evaluation function is onto k with kernel $m_p$, which is clear.
But if I take $k = \mathbb{C}$, $V = \mathbb{A}(\mathbb{C})$, then $O_p(V) = \{f = \frac{a}{b} : a,b \in \mathbb{C}[X], b(p) \neq 0\}$ and $m_p(V) = \{f\in O_p(V) : a(p) = 0\}$. Thus $f:= 3\cdot X \neq g:=2\cdot X + 1$ in $O_1(V)/m_1(V)$, but clearly $f(1) = g(1) = 3$. How does this make sense?
Best Answer
As you say, for a variety $V$ over a field $k$ we have the isomorphism $ \mathcal{O}_{V, p} / \mathfrak{m}_p \mathcal{O}_{V, p} \cong k $ given by $$ \frac{g}{h} \mapsto \frac{g(p)}{h(p)} $$ where $\mathfrak{m}_p \mathcal{O}_{V, p}$ is the unique maximal ideal corresponding to the point $p \in V$.
In your case, let $f=3x$ and $g=2x+1$, both elements of $\mathcal{O}_{V, p}$. They are indeed not equal in $\mathcal{O}_{V, p}$. Now, $\mathfrak{m}_p \mathcal{O}_{V, p} = \{ q \in \mathcal{O}_{V, p} \mid q(p)=0 \}$ and in your example you are letting $p=1$. Recall that $[x]=[y] \in R/I \iff x-y \in I$. Here, $f(1)-g(1) = 0$ so $f-g \in \mathfrak{m}_1 \mathcal{O}_{V, 1}$, so they are equivalent in the quotient ring as required.