0

This is from the Stacks project.

Let $A_k$ be the subring of $(\mathbb{C}[x])^{2^k + 1}$ such that $(f_0, ... f_{2^k}) \in A_k$ if $f_i(1) = f_{i+1}(0)$ for all $i = 0 ,... 2^k -1$. Define a map $A_k \rightarrow A_{k+1}$ as $(f_0, ... f_{2^k}) \rightarrow (f_0, f_0(1), f_1, f_1(1), ..., f_{2^k})$. Let $A$ be the direct limit of the $A_k$.

The claim is that $\operatorname{Spec}(A)$ is connected, not integral, and the stalk at every point is an integral domain. I don't understand the proof that the stalk at every point is an integral domain.

I don't know how to visualize this scheme, so I don't understand the proof. Is there a purely algebraic way to show that the stalks are all integral domains?

Edit : The parts I find unclear is that the proof doesn't give a nice description of the points of $\operatorname{Spec}(A)$ and $\operatorname{Spec}(A_k)$. Is there an easy description in the form of "the points are ..., the irreducible components are ... and they intersect at ..."?

At the smooth points the stalk is an integral domain by definition since regular local rings are integral domains. On the other points it says "two components of $X_k$ pass through $y$". But nowhere does the proof explain how many irreducible components of $X_k$ there are and how they intersect.

The next sentence is "However, on one of these two components (the one with odd index), both $f$ and $g$ are constant". So there should be two components containing $y$? Why is this? Also, pullbacks of functions on $X_{k-1}$ aren't constant everywhere, only half of the entries are constants, so I don't understand the next line either.

Also, based on the description it seems like I'm supposed to visualize this as lines meeting each other at a point. But then it seems like the stalks wouldn't be integral domains since it isn't on, for example, $\operatorname{Spec}(k[x,y]/(xy))$.

KReiser
  • 65,137
David Lui
  • 6,295

1 Answers1

1

Yes, there's an easy description of $X_k=\operatorname{Spec} A_k$ and the map $X_k\to X_{k-1}$. I'm going to slightly change the indexing conventions to try and make it a little easier to understand how this chain is being constructed and how the final scheme is put together. Instead of adding new links at the end of the chain, you replace each point where two links come together with a smaller link joining the two. Here's a description:

  • $X_0$ is isomorphic to the cross in the plane, $\operatorname{Spec} k[x_0,x_1]/(x_0x_1)$. Call the irreducible components $Y_{0,0}$ and $Y_{0,1}$, each isomorphic to $\Bbb A^1$, and the components intersect transversally at one point.
  • $X_1$ is isomorphic to three affine lines, each meeting its neighbors. Call the three irreducible components $Y_{1,0}, Y_{1,\frac12}, Y_{1,1}$. Each irreducible component meets its neighbors (the components with the closest indicies) transversally in a single point. The map $X_1\to X_0$ sends $Y_{1,0}$ isomorphically on to $Y_{0,0}$, $Y_{1,1}$ isomorphically on to $Y_{0,1}$, and contracts $Y_{1,\frac12}$ to the point of intersection of $Y_{0,0}$ and $Y_{0,1}$.
  • $X_2$ is isomorphic to five affine lines, each meeting its neighbors transversally in a single point. Call the irreducible components $Y_{2,a}$ for $a\in C_2=\{0,\frac14,\frac12,\frac34,1\}$; the map $X_2\to X_1$ sends $Y_{2,a}$ to $Y_{1,a}$ if $a\in\{0,\frac12,1\}$ and to the point of intersection of $Y_{1,a-\frac14}$ and $Y_{1,a+\frac14}$ otherwise.
  • $X_k$ is isomorphic to $2^k+1$ affine lines, each meeting its neighbors transversally in a single point. Writing $Y_{k,a}$ for the irreducible components with $a\in C_k = \{\frac{n}{2^k}\mid 0\leq n \leq 2^k\}$, the map $X_k\to X_{k-1}$ is given by sending $Y_{k,a}$ to $Y_{k-1,a}$ if $a\in C_{k-1}$ and to the point of intersection of $Y_{k-1,a-\frac{1}{2^k}}$ and $Y_{k-1,a+\frac{1}{2^k}}$ otherwise.

I'm not so sure that there's an easy description of $X=\operatorname{Spec} A$, though. Intuitively, what you have is a scheme where the irreducible components (all isomorphic to $\Bbb A^1$) are in bijection with the dyadic rationals in $[0:1]$, and the scheme is connected, but it's tough to think of where an intersection point of irreducible components is because we've blown them all up.

The good news is that the proof of integrality of stalks can be thought of just in terms of the finite steps in this chain. Let $x\in X$ be a point with image $x_i \in X_i$ and let $f,g\in A_{k-1}$. Now we have two cases:

  1. $x_{k-1}$ is on only one irreducible component, say $Y_{k-1,a}$. Then the local ring of $x_{k-1}$ in $X_{k-1}$ is $k[t]_{(t)}$, so one of $f$ or $g$ vanishes in this local ring, which implies that it vanishes on an open neighborhood of $x_{k-1}$, which implies it vanishes on an open neighborhood of $x$ by taking the preimage along the natural map $X\to X_{k-1}$.

  2. $x_{k-1}$ is on two irreducible components, say $Y_{k-1,a}$ and $Y_{k-1,a+\frac{1}{2^k}}$. If $x_k$ is only on one irreducible component (the one mapping down to $x_{k-1}$), then we can reuse the proof from step 1. So assume $x_k$ is on both $Y_{k,a}$ and $Y_{k,a+\frac{1}{2^{k+1}}}$. For $fg=0$ in the local ring of $x_{k-1}$, we must have $f$ vanishing on $Y_{k-1,a}$ and $g$ vanishing on $Y_{k-1,a+\frac{1}{2^{k}}}$ (or vice-versa). Now the pullbacks of $f$ and $g$ to $X_k$ are zero on $Y_{k,a+\frac{1}{2^{k+1}}}$, since $f(x_{k-1})=g(x_{k-1})=0$. Therefore on a neighborhood of $x_k$ which is entirely contained in $Y_{k,a}$ and $Y_{k,a+\frac{1}{2^{k+1}}}$, we have $f=0$, and we're done by the last line of 1.

KReiser
  • 65,137