Intuition behind the theorem about ideals with finite number of zeroes

Question

So I've been studying algebraic geometry and found the theorem which states that for an ideal $I$ in $K[x_{1}, ..., x_{n}]$ such that $V(I)$ is finite it holds that $K[x_{1}, ..., x_{n}]/I$ is naturally isomorphic to the product of $O_{i}/IO_{i}$, where $O_{i}$ denotes the localization of $i$-th zero of $I$ over a whole affine space $A^{n}$ (this theorem can be found in Fulton's book about algebraic curves on the page 27). I do understand every step of the proof but I cannot find any intuition behind it. More important, I cannot realise how can someone even come up with the statement of this theorem, though I do understand the concept of the coordinate ring and it's localization in a point.

Answer 1

First, by the generalized chinese remainder theorem, we have that $K[x_1,\cdots,x_n]/I$ is isomorphic to $\prod K[x_1,\cdots,x_n]/\mathfrak{m}_i^{e_i}$, where the $\mathfrak{m}_i$ are the maximal ideals corresponding to the finite collection of points $P_i\in V(I)$ and $e_i$ are the multiplicity of $I$ at these points. This corresponds to the statement that a regular function on a finite set is determined by what happens at each point.

The reason why we want to show that this is the same as the version coming from the local rings is that our original setup with $K[x_1,\cdots,x_n]$ contains some amount of "unnecessary choice": we've picked a coordinate system on $\Bbb A^n_k$, while dealing with local rings removes such choices.

To translate to the local ring side, we will show that $R_\mathfrak{m}/\mathfrak{m}^n_\mathfrak{m}\cong R/\mathfrak{m}^n$ for any ring $R$ with a maximal ideal $\mathfrak{m}$. Since localization commutes with quotients, $R_\mathfrak{m}/\mathfrak{m}^n_\mathfrak{m}\cong (R/\mathfrak{m}^n)_\mathfrak{m}$, but every element in $R\setminus\mathfrak{m}$ is already invertible in $R/\mathfrak{m}$. This shows that $K[x_1,\cdots,x_n]/I\cong \prod \mathcal{O}_i/I\mathcal{O}_i$.

Answer 2

Since the other answer gave the (rigorous) algebraic intuition behind this result, I'll try give the geometric intuition. The idea is that "functions" on an arbitrary "space" form a ring, and we can interpret arbitrary rings in this way geometrically. Setting this up rigorously isn't particularly enlightening, and its not required to see the geometric intution behind this result.

So in this setup, we have our ring of functions $R$ on a space $X$, and an ideal $I$ that is the ideal of functions that vanish on a subset $V$. The ring $R/I$ is the functions on $V$, and the natural map $R\rightarrow R/I$ is restricting functions from $X$ to $V$.

We are told that $V$ is finite, and thus intuitively, functions on a finite set are just determined by their values on each point on the set. So we expect functions on $V$ to be a tuple of values, one for each point of $V$, under pointwise multiplication and addition. That is, we expect a product ring, where each component is "functions on a point". Now we must confront the fact that our functions on "a point" may depend on the point, and this is why we see the rings $R_x/I_x$ occurring, where we use $R_x$ to mean the localisation of $R$ at the prime ideal corresponding to the geometric point $x$. If we assume that $K$ is algebraically closed and $I$ is radical for simplicity, this is just $K$, and weakening those assumptions requires enlarging your definition of "point", but the intuition is the same.

So to summarise, the geometry behind this result is that if we have $X$ a space, and a finite set of point $V$ in $X$, then the functions on $V$ are a product of functions on each point, and the restriction map from functions on $X$ to functions on $V$ is surjective, and has kernel exactly the ideal $I$ that vanishes on $V$.