Varieties and ideals I'm doing the exercises from Fulton of Algebraic Geometry and I'm stuck in the problem 2.44
Let $V$ be a variety in $\mathbb{A}^{n}$, $I=I(V)\subset k[x_{1},\ldots,x_{n}]$, $P\in V$ and let $J$ be and ideal $k[x_1,\ldots,x_n]$ which contains $I$. Let $J'$ be the image of $J$ in $\Gamma(V)$. Prove that exists a natural homomorphism $\varphi$ from $\mathcal{O}_{P}(\mathbb{A}^{n})/J\mathcal{O}(\mathbb{A}^{n})$ to $\mathcal{O}_{p}(\mathbb{V})/J'\mathcal{O}(\mathbb{V})$, and prove that $\varphi$ is an isomorphism. In particular, $\mathcal{O}_{P}(\mathbb{A}^{n})/I\mathcal{O}(\mathbb{A}^{n})$ is isomorphic to $\mathcal{O}_{P}(V)$
If anyone can help I'll really appreciate it :)
 A: Let $R=k[x_1,\ldots,x_n]$ so I don't have to keep writing it.
I'm going to construct this explicitly, although I'm going to do it in two stages.
First we need a map $\psi$ from $O_P(\Bbb{A}^n)\to O_P(V)$.
Define $\psi(f/g)=\frac{\overline{f}}{\overline{g}}$, where $\overline{f}$ is the image of $f$ under the canonical projection from $R$ to $R/I$ and similarly for $g$. Since $g(P)\ne 0$, and $P\in V$, $g\not \in I$, so $\overline{g}\ne 0$. Thus this is a well defined map, and I will let you check that it is a surjective homomorphism.
Then let $\tau: O_P(V)\to O_P(V)/J'O_P(V)$ be the usual projection. $\tau$ is a surjective homomorphism as well.
Then let $\sigma= \tau \circ \psi : O_P(\Bbb{A}^n) \to O_P(V)/J'O_P(V)$.
Since $\tau$ and $\psi$ are surjective homomorphisms, $\sigma$ is as well. We just need to find the kernel of $\sigma$.
Suppose $f/g$ maps to 0 under $\sigma$. Then 
$$\frac{\overline{f}}{\overline{g}}\in J'O_P(V).$$
Multiplying by $\overline{g}$ we have $\overline{f}\in J'O_P(V)$.
Then $$\overline{f}=\sum_{i=1}^n\frac{\overline{j_i}}{\overline{g_i}}$$ for $\overline{j_i}\in J'$, $\frac{1}{\overline{g_i}}\in O_P(V)$ such that $\overline{g_i}$ is the representative of some $g_i\in R$ with $g_i(P)\ne 0$ (note this is possible by absorbing the numerator into the $\overline{j_i}$).
Now multiplying by all the $\overline{g_i}$ we have
$$\overline{f}\prod_i \overline{g_i} \in J'$$ since $J'$ is an ideal and the denominators all cancel,
so $$f\prod_i g_i=\alpha$$ for some $\alpha$ in $J$ since $I\subset J$.
Finally dividing by the product of the $g_i$ since the $g_i$ are invertible in $O_P(\Bbb{A}^n)$ gives that $f\in JO_P(\Bbb{A}^n)$. Then $f/g \in JO_P(\Bbb{A}^n)$, so $\ker \sigma \subset JO_P(\Bbb{A}^n)$.
We are almost done. To show $JO_P(\Bbb{A}^n)\subset \ker \sigma$, we note that if $a=\sum_{i} j_ia_i$ for $j_i\in J$, $a_i\in O_P(\Bbb{A}^n)$, $\sigma(a)=0$ since $\psi(j_i)\in J'$, so $\sigma(j_i)=0$.
Therefore $\ker\sigma=JO_P(\Bbb{A}^n)$, so the desired natural isomorphism exists by the first isomorphism theorem.
