Isomorphism in coordinate ring Let $x_{1},x_{2},...,x_{m}$ be elements of $\mathbb{A}^{n}$, where $\mathbb{A}^{n}$ is the n-affine space over an algebraically closed field $k$. Now define $X=\{x_{1},x_{2},...,x_{m}\}$. Why is the coordinate ring $A(X)$, isomorphic to $\oplus_{j=1}^{m} k = k^{m}$?
 A: The coordinate ring $A(X)$ is defined to be the quotient of $k[z_1,\ldots,z_n]$ (I'm using $z_i$ because you are using $x_i$ to denote the points of affine space) by the ideal $I(X)$. The ideal $I(X)$ is the ideal of all elements of $k[z_1,\ldots,z_n]$ that are zero at $X$.
By the Nullstellensatz, the ideal corresponding to a single point $x_i = (x_{i1},\ldots,x_{in})$ is given by $(z_1-x_{i1}, z_2-x_{i2},\ldots,z_n-x_{in})$. 
The ideal of a union is the intersection of the ideals; so you are trying to mod out by
$$\bigcap_{i=1}^m (z_1-x_{i1},\ldots,z_n-x_{in}).$$
But the ideals $(z_1-x_{i1},\ldots,z_n-x_{in})$ are maximal ideals, since $k[z_1,\ldots,z_n]/(z_1-x_{i1},\ldots,z_n-x_{in}) \cong k$. If they are all distinct, then they are pairwise comaximal, hence pairwise coprime. By the Chinese Remainder Theorem, we know that if $\mathfrak{p}_1,\ldots,\mathfrak{p}_k$ are pairwise coprime ideals in $R$, then
$$\frac{R}{\mathfrak{p}_1\cap\cdots\cap\mathfrak{p}_k} \cong \frac{R}{\mathfrak{p}_1}\oplus\cdots\oplus\frac{R}{\mathfrak{p}_k}.$$
So we have that
$$A(X) = \frac{k[z_1,\ldots,z_n]}{\cap_{i=1}^m(z_1-x_{i1},\ldots,z_n-x_{in})} \cong \mathop{\bigoplus}_{i=1}^m \frac{k[z_1,\ldots,z_n]}{(z_1-x_{i1},\ldots,z_m-x_{im})} \cong \mathop{\bigoplus}_{i=1}^m k.$$
A: For each $i$, $A(\{x_i\}) = k[x]/I(x_i) \cong k$, so each $I(x_i)$ is a maximal ideal of
$k[x]$.  I assume the points $x_1,\ldots,x_n$ are distinct, from which it follows easily
that the ideals $I(x_1),\ldots,I(x_n)$ are distinct maximal ideals.  Thus they are pairwise comaximal and the Chinese Remainder Theorem -- see e.g. $\S 4.3$ of these notes -- applies.  I leave it to you to check that it gives the conclusion you want.
