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This problem is from Gathmann's notes problem 3.23 https://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2014/alggeom-2014.pdf

Let $X ⊂ \mathbb{A}^n$ be an affine variety, and let $a ∈ X$. Show that $\mathscr{O}_{X,a} \cong \mathscr{O}_{\mathbb{A}^n,a}/I(X)\mathscr{O}_{\mathbb{A}^n,a}$, where $I(X)\mathscr{O}_{\mathbb{A}^n,a}$ denotes the ideal in $\mathscr{O}_{\mathbb{A}^n,a}$ generated by all quotients $\frac{f}{ 1}$ for $f ∈ I(X)$.

My first idea is to use the first isomorphism theorem. So I want a homomorphism $φ:\mathscr{O}_{\mathbb{A}^n,a} \to \mathscr{O}_{X,a}$ such that $\varphi$ is surjective and $\ker(\varphi)=I(X)\mathscr{O}_{\mathbb{A}^n,a}$. But I am really having trouble understanding the definition of $\mathscr{O}_{\mathbb{A}^n,a}$ and what the elements looks like. Any help is appreciated, thank you.

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  • $\begingroup$ Really you just need to show that localization commutes with quotients. $\mathscr{O}_X$ is a quotient of a polynomial ring, so $\mathscr{O}_{X,a}$ means "quotient first, then localize at $a$", while $\mathscr{O}_{\mathbb{A}^n,a}/I(X)\mathscr{O}_{\mathbb{A}^n,a}$ means "localize first, then quotient". $\endgroup$ Commented Nov 4, 2019 at 20:48
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    $\begingroup$ @Cornman Let $k[X]$ be the coordinate ring of $X$. Then we have a short exact sequence $0 \to I(X) \to k[x_1, \ldots, x_n] \to k[X] \to 0$. Let $S = \{f \in k[x_1, \ldots, x_n] : f(a) \neq 0\}$. By the exactness of localization, then $0 \to S^{-1} I(X) \to S^{-1} k[x_1, \ldots, x_n] \to S^{-1} k[X] \to 0$ is exact. $\endgroup$ Commented Nov 26, 2020 at 21:19

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The elements of $\mathscr{O}_{\mathbb{A}^n,a}$ are just rational functions $\frac{g}{f}$, where $f, g$ are polynomials of $n$ variables (i.e. elements of $\mathscr{O}_{\mathbb{A}^n}$) such that $f(a) \neq 0$.

This of course is nothing but a copy of Lemma 3.21 in your linked notes.

Intuitively, you should think about $\mathbb{A}^n$ as the $n$-dimensional complex space $\mathbb{C}^n$.

A polynomial $f\in \mathbb{C}[X_1, \cdots, X_n]$ then defines a function on $\mathbb{C}^n$: it sends a point $a = (a_1, \cdots, a_n)$ to $f(a_1, \cdots, a_n)$.

Now if $f, g\in \mathbb{C}[X_1, \cdots, X_n]$ are two polynomials, in general you cannot say that $\frac{g}{f}$ defines a function on $\mathbb{C}^n$, simply because the value of $f$ may vanish at some points.

But for a given point $a$, if we have $f(a) \neq 0$, then it's clear that $f$ does not vanish in a small neighborhood of $a$ (with respect to the usual topology of $\mathbb{C}^n$), hence on that small neighborhood we may define $\frac{g}{f}$ without problem.

And if we gather all functions that can be defined as $\frac{g}{f}$ in a small neighborhood of the given point $a$, we get the local ring $\mathscr{O}_{\mathbb{A}^n,a}$.

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  • $\begingroup$ @Cornman This is a direct application of Lemma 3.21, with $X = \Bbb A^n$ (hence $A(X) = K[X_1, \dots, X_n]$). $\endgroup$
    – WhatsUp
    Commented Nov 26, 2020 at 20:52
  • $\begingroup$ @Cornman Which isomorphism? The question of the OP is to explain how the ring $\mathscr{O}_{\mathbb{A}^n,a}$ and its elements look like, not to show that isomorphism in the post. You may post a new question, so that people can answer your question more specifically. $\endgroup$
    – WhatsUp
    Commented Nov 26, 2020 at 21:31
  • $\begingroup$ I was actually answering the question concerning the definition of $\mathscr{O}_{\mathbb{A}^n,a}$. For the isomorphism, see the comment of @RichardD.James above. $\endgroup$
    – WhatsUp
    Commented Nov 26, 2020 at 21:38
  • $\begingroup$ @Cornman I think it's OK to leave them here, so that others would not misunderstand any more. If you find the answer of Richard too advanced, then probably you should learn some more commutative algebra, which is inavoidable in learning algebraic geometry. $\endgroup$
    – WhatsUp
    Commented Nov 26, 2020 at 21:43

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