A natural isomorphism obtained from ideals of finitely many points.

Question

Suppose $K$ is an algebraically closed field and $\mathbb A^n$ be the affine $n$-space over the field $K$.Then the coordinate ring $A(\mathbb A^n)$ is given by the polynomial ring $K[X_1,...,X_n]$.Let $I$ be an ideal of this ring defined by $I=\mathbf I(p_1,..,p_r)$ where $p_i$'s are finitely many points in the affine $n$-space and $\mathbf I$ is the ideal consisting of those polynomials which vanish on all of these points.Then my claim (actually saw in a book on algebraic geometry) is that $K[X_1,...,X_n]/I\cong\prod\limits_{i=1}^r\mathcal O_i/I\mathcal O_i$ where $\mathcal O_i:=\mathcal O_{p_i}(\mathbb A^n)$ the local ring of $\mathbb A^n$ at $p_i$ which is defined by $\mathcal O_p(X)=\{[(U,f)]: U$ is open $\ni p$ and $f\in \mathcal O(U)\}$ where $\mathcal O(U)$ is the ring of all functions regular at every point of $U$ and the equivalence class $[$ $]$ is taken under the equivalence relation $(U,f)\sim (V,g)$ if and only if $f=g$ on $U\cap V$.I think we can obtain a natural map $K[X_1,...,X_n]/I\to \mathcal O_i/I\mathcal O_i$ by the properties of localization as we note that $\mathcal O_p(X)=(A(X))_{{\scr m}_p}$ where ${\scr m}_p =\{f\in A(X): f(p)=0\}$.But I cannot do it as I am not so strong in commutative algebra.Can someone answer my question by helping me solving this problem?

Answer 1

In this answer, I'm going to characterise the various rings that you're dealing with, describing the various isomorphisms that exist between them. This is probably not the most direct answer to your question - it's more of a summary of my natural thought process.

Part 1: Reducing to the case of a single point.

It's annoying to work with products, so why don't we reduce our problem the case of a single point in $\mathbb A^n$?

So let $I(p_1, \dots, p_r)$ denote be the ideal of polynomials that vanish at all of the points $p_1, \dots, p_r$, and let $I(p_i)$ denote the ideal of polynomials that vanish at the single point $p_i$, for $i \in \{ 1, \dots, r \}$.

The ideal $I(p_1, \dots, p_r)$ is equal to the product $I(p_1)\dots I(p_r)$. The $I(p_i)$'s are pairwise comaximal, so the Chinese Remainder Theorem gives us a ring isomorphism, $$ \frac{k[x_1, \dots, x_n] }{I(p_1, \dots, p_r)} \cong \frac{k[x_1, \dots, x_n] }{I(p_1)} \times \dots \times \frac{k[x_1, \dots, x_n] }{I(p_r)}.$$

Thus, if we want to make sense of $\frac{k[x_1, \dots, x_n] }{I(p_1, \dots, p_r)}$, then all we have to do is characterise each $\frac{k[x_1, \dots, x_n] }{I(p_1, \dots, p_r)} $ individually!

Part 2: Characterising $k[x_1, \dots, x_n] / I(p_i) $.

Let $\text{ev} : k[x_1, \dots, x_n] \to k$ be the evaluation map, which evaluates polynomials in $k[x_1, \dots, x_n]$ at the point $p_i$. Clearly, $\text{ev}$ is a ring homomorphism, and clearly, it is surjective, and its kernel is $I(p_i)$. (Indeed, $I(p_i)$ is by definition is the ideal of polynomials that evaluate to zero at $p_i$.)

This gives us a simple characterisation of $k[x_1, \dots, x_n] / I(p_i)$: it is isomorphic to $k$, via the evaluation map.

Part 3: Relating $k[x_1, \dots, x_n] / I(p_i)$ to $\mathcal O_{p_i} / ( I(p_i) \mathcal O_{p_i}).$

$\mathcal O_{p_i}$ is the ring of rational functions regular at the point $p_i$; it is obtained by localising $k[x_1, \dots, x_n]$ at the maximal ideal $I(p_i)$.

Similar to what we did earlier, we can consider the ring homomorphism $\text{ev} : \mathcal O_{p_i} \to k$ which evaluates functions in $\mathcal O_{p_i}$ at the point $p_i$. $\text{ev}$ is surjective, and its kernel is $I(p_i)O_{p_i}$.

Now let's put the pieces together. We have the following commutative diagram. $\require{AMScd}$ \begin{CD} \tfrac{k[x_1, \dots, x_n]}{I(p_i)} @>{\text{ev}}>{\cong}> k \\ @V{\iota}VV @V{\text{id}}V{\cong}V \\ \tfrac{\mathcal O_{p_i}} {(I(p_i) \mathcal O_{p_i})} @>{\text{ev}}>{\cong}> k \end{CD}

The two horizontal morphisms are the evaluation maps that we've discussed, which are isomorphisms. The right-hand vertical morphism is the identity morphism on $k$, which is certainly an isomorphism. The left-hand morphism $\iota$ is the inclusion morphism that from $\tfrac{k[x_1, \dots, x_n]}{I(p_i)}$ to $\tfrac{\mathcal O_{p_i}} {(I(p_i) \mathcal O_{p_i})}$. It is clear that the four morphisms form a commutative diagram as shown.

Since the horizontal morphisms and the right-hand vertical morphism are isomorphisms, the left-hand morphism must be an isomorphism too. Thus $\tfrac{k[x_1, \dots, x_n]}{I(p_i)}$ is isomorphic to $\tfrac{\mathcal O_{p_i}} {(I(p_i) \mathcal O_{p_i})}$, via the inclusion morphism $\iota$.

Part 4: Relating $\mathcal O_{p_i} / (I(p_i)\mathcal O_{p_i})$ to $\mathcal O_{p_i} / (I(p_1, \dots, p_r)\mathcal O_{p_i})$.

Having gone to the effort of writing all of this, I spotted to my annoyance that the thing that appears in your book is not $\mathcal O_{p_i} / (I(p_i)\mathcal O_{p_i})$, but $\mathcal O_{p_i} / (I(p_1, \dots, p_r)\mathcal O_{p_i})$!

Not to worry. It turns out that $I(p_i)\mathcal O_{p_i}$ and $I(p_1, \dots, p_r)\mathcal O_{p_i}$ are the same thing.

To see this, observe that $I(p_1, \dots, p_r)\mathcal O_{p_i}$ is the product of the ideals $I(p_1)\mathcal O_{p_i}, \dots, I(p_r) \mathcal O_{p_i}$. Then observe that for any $j \neq i$, there certainly exists a regular function $f$ that vanishes at $p_j$ but does not vanish at $p_i$, i.e. there exists an element $f \in I(p_j)$ that is invertible in $\mathcal O_{p_i}$. Thus, writing $1 = f.\frac{1}{f}$, we see that $1$ is an element of $I(p_j)\mathcal O_{p_i}$, so $I(p_j)\mathcal O_{p_i}$ is in fact the whole of $\mathcal O_{p_i}$. Hence $I(p_1, \dots, p_r)\mathcal O_{p_i}$ is the same thing as $I(p_i)\mathcal O_{p_i}$.