---
title: "Algebraic Geometry Problem Sets"
subtitle: "Problem Set 1"
author:
- name: D. Zack Garza
  affiliation: University of Georgia 
  email: dzackgarza@gmail.com 
date: Spring 2022
order: 1
---

# Problem Set 1

## Problem 1

:::{.problem title="1.1"}
Recall that:

- A topology on a set $X$ is $T_{0}$ if any two points $x, y \in X$
can be topologically distinguished (by open sets).

- A topology is an Alexandrov if an intersection of any, possibly
infinite, collection of open sets is open.

- The **order topology** on a poset $(X, \leq)$ is defined in the following way: the open sets are the *upper sets*, which satisfy the property
\[
x \in U, x \leq y \Longrightarrow y \in U
\]
The closed sets are **lower sets**, which satisfy
\[
x \in Z, x \geq y \Longrightarrow y \in U
\]


Prove that a topology on $X$ is an order topology $\Longleftrightarrow$
it is $T_{0}$ and Alexandrov. As a corollary conclude that any $T_{0}$
topology on a finite set is an order topology.
:::

:::{.proposition}
A topology $\tau$ on $X$ is an order topology $\Longleftrightarrow$
$\tau$ is $T_{0}$ and Alexandrov.
:::

:::{.proof}
$\impliedby$:
Suppose $X$ is a topological space and $\tau$ is a $T_0$ Alexandrov topology on $X$.
For $U \subseteq X$, write $\cl_X(U)$ for the closure in $X$ of $U$ with respect to $\tau$, define a poset $(P, \leq)$ where $P \da X$ with an ordering defined by 
\[
x\leq y \iff x\in \cl_X(y)
.\]
Regarding $\tau$ now as a topology on $(P, \leq)$, the claim is that this is an order topology on a poset.
That this ordering defines a poset is clear, since the ordering is:

- **Reflexive**: since $x$ is contained in its closure, $x\leq x$.
- **Antisymmetric**: if $x\leq y$ and $y\leq x$, then $x$ is a limit point of $\ts{y}$ and vice-versa.
  So every neighborhood of $y$ contains $x$ and similarly every neighborhood of $x$ contains $y$.
  Since $X$ is $T_0$ and topologically distinguishes points, this can only occur if $x=y$.
- **Transitive**: if $x\leq y$ and $y\leq z$, then $x\in \cl_X(y)$ and $y\in \cl_X(z)$.
  Since $\cl_X(z)$ is a closed set containing $y$ and $\cl_X(y)$ is the *smallest* closed set in $X$ containing $y$, we have $x\in \cl_X(y) \subseteq  \cl_X(z)$, so $x\leq z$.

It thus suffices to show that if $U \ni x$ is a neighborhood of $x$ and $x\leq y$, then $y\in U$ so that $U$ is an upper set.
By definition of the closure of a set, 
\[
x\leq y \iff x\in \cl_X(y) \iff
\text{every neighborhood of $x$ intersects } \ts{y}
,\] 
so if $U_\alpha \ni x$ is any neighborhood of $x$, then $y\in U_\alpha$.
Write $\tilde U \da \intersect_{\alpha} U_\alpha$ for the neighborhood basis at $x$, the intersection of all neighborhoods of $x$.
Note that by construction, since $y\in U_\alpha$ for all $\alpha$, $y\in \tilde U$.
Since $\tau$ is $T_0$, $\tilde U$ is an open set.
Moreover, since $U$ is a neighborhood of $x$, $\tilde U \subseteq U$, so $y\in U$.

$\implies$:
Suppose $(X, \leq)$ is a poset with an order topology $\tau$, so $U$ is open iff whenever $x\in U$ and $x\leq y$ then $y\in U$.
To see that $\tau$ defines a $T_0$ topology, let $x\neq y$ in $X$.
If $x$ and $y$ are not comparable, there is nothing to show, so suppose either $x< y$ or $y< x$ -- without loss of generality, relabeling if necessary, we can assume $x< y$.
Now every neighborhood of $x$ contains $y$ by definition, but for example 
\[
U_{\geq y} \da \ts{z\in X\st z\geq y}
\]
is neighborhood of $y$ not containing $x$, topologically distinguishing $x$ and $y$.

To see that $\tau$ is Alexandrov, it suffices to show that arbitrary intersections of open sets are open.
This follows from the fact that any intersection of upper sets is again an upper set -- if $\ts{ U_i}_{i\in I}$ is an arbitrary family of upper sets, set $U \da \intersect_{i\in I} U_i$. Then if $x\in U$ with $x\leq y$, $x\in U_i$ for every $i$ and so $y\in U_i$ for every $i$, and thus $y\in U$.
:::

:::{.corollary title="?"}
If $X$ is a finite set and $\tau$ is a $T_0$ topology on $X$, then $\tau$ is an order topology.
:::

:::{.proof title="of cor"}
By the exercise, it suffices to show that any finite space is Alexandrov.
Let $(X, \tau)$ be a $T_0$ space and let $\ts{U_i}_{i\in I} \subseteq \tau$ be an arbitrary collection of open sets -- we'll show $U\da \intersect_{i\in I} U_i \in \tau$ is again open.
This follows immediately, since finite intersections of open sets are open in any topology, and since $X$ is finite and $\tau \subseteq 2^X$ is finite, $I$ can only be a finite indexing set.
:::

## Problem 2
:::{.problem title="1.2"}
Recall that:

- A paracompact space is a topological space in which every open
cover has an open refinement that is locally finite.

- A partition of unity of a topological space $X$ is a set
$f_{\alpha}$ of continous functions $f_{\alpha}: X \rightarrow[0,1]$
such that for every point $x \in X$ there exists an open neighborhood of
$x$ where all but finitely many $f_{\alpha} \equiv 0$, and such that
$\sum f_{\alpha}=1$.

Prove that

- Any Hausdorff space is paracompact iff it admits a partition of
unity subordinate to any open cover.

- Any metric space is paracompact.

> A sketch would suffice.

:::

:::{.proposition title="?"}
$X$ is Hausdorff $\iff X$ admits a partition of unity subordinate to any open cover $\mcu \covers X$.
:::

:::{.proof title="?"}
?
:::

:::{.proposition title="?"}
Metric spaces are paracompact.
:::

:::{.proof title="?"}
Let $\mcu \covers X$ be an open cover of a metric space $X$; we'll show $\mcu$ admits a locally finite refinement.
Without loss of generality, writing $\mcu = \ts{U_j}_{j\in J}$ for some index set $J$, we can assume the $U_i$ are disjoint -- this follows by invoking the axiom of choice to well-order the index set $J$ and setting
\[
\tilde U_j \da U_j \sm \Union_{k < j} U_k
.\]
Then $\tilde \mcu \da \ts{\tilde U_j}_{j\in J}$ refines $\mcu$ since $\tilde U_j \subseteq U_j$, and still covers $X$.
Moreover, we note that for every $x\in X$, we can now produce a *minimal* index $j(x)$ such that $x\in U_{j(x)}$.

The idea is to now refine $\mcu$ to a cover $\mcv$ by filling each disjoint annulus $U_j$ with balls of small enough radius.
For ease of notation and to more clearly demonstrate the following construction, suppose $J \cong \ts{0,1,\cdots}$ is countable.
For each $n\in \ZZ_{\geq 0}$, let $\delta_n < \eps_n$ be small to-be-determined real numbers depending on $n$, and define the following subsets of $X$:

\[
X_{0, n} \da \ts{x\in U_0 \st B_{\eps(n) }(x) \subseteq U_0 } 
&&
V_{0, n} \da \Union_{x\in X_{0, n}} B_{\delta_n}(x) \subseteq X_{0, n} \subseteq U_0 \\
X_{1, n} \da \ts{x\in U_1 \st B_{\eps(n) }(x) \subseteq U_1 } \sm \Union_{\ell < n} V_{0, \ell} 
&&
V_{1, n} \da \Union_{x\in X_{1, n}} B_{\delta_n}(x) \subseteq X_{1, n} \subseteq U_1 \\
X_{2, n} \da \ts{x\in U_2 \st B_{\eps(n) }(x) \subseteq U_2 } \sm \Union_{\ell < n} V_{0, \ell} \sm \Union_{\ell < n} V_{1, n} 
&&
V_{2, n} \da \Union_{x\in X_{2, n}} B_{\delta_n}(x) \subseteq X_{2, n} \subseteq U_2 \\
\vdots && \vdots \qquad \\
X_{j, n} \da \ts{x\in U_j \st B_{\eps(n) }(x) \subseteq U_j } \sm \Union_{k < j} \Union_{\ell < n} V_{k, \ell} 
&&
V_{j, n} \da \Union_{x\in X_{j, n}} B_{\delta_n}(x) \subseteq X_{j, n} \subseteq U_j 
.\]
Note that the last line prescribes a general formula which depends only on the ordering and not on the countability of $J$.

In other words, for each fixed $j_0\in J$, we consider all of those $x\in X$ such that $j(x) = j_0$, so that for each such $x$ we have $x\in U_{j_0}$ but $x\not\in U_k$ for any $k<j_0$.
For a fixed $n$, we then consider those $x\in U_{j_0}$ that are not too close to the boundary, so that a ball of radius $\eps_n$ fits entirely in $U_{j_0}$.
We then shrink these balls to a smaller radius $\delta_n$ and take their union to form an open set $V_{j_0, n}$ in the new cover, and as $n\to \infty$ these balls get smaller and fill out all of $U_{j_0}$.
However, at each stage $V_{j, n}$ we remove redundancies by discarding sets $V_{k, \ell}$ for $k<j$ and $\ell < n$.

:::{.claim}
$\mcv \da \ts{V_{j, n}}_{j\in J, n\in \ZZ_{\geq 0}}$ is a locally finite refinement of $\mcu$.
:::

:::{.proof title="of claim"}
There are several things that are clear from the construction:

- Each $V_{j, n}$ is open, as they are arbitrary unions of open balls in a metric space.
- $\mcv$ refines $\mcu$, since in fact $V_{j, n} \subseteq U_j$ for every $n$ and every $j$.
- $\mcv$ is a cover, since for any $x$ one can pick $j(x)$ minimally so that $x\in U_{j(x)}$, and since $x$ is an interior point and open balls form a basis for a metric space, for some $n$ large enough we have $x\in B_{\delta_n}(x) \subseteq V_{j(x), n}$.

So the content of this statement is that each $x\in X$ is contained in only finitely many opens from $\mcv$.
Fix $x\in X$ and pick $j(x)$ minimally as above, so $x\in V_{j(x), n}$ for every $n$ and $j(x)$ is the first such $j$ where $x$ is added. 
Then $x\in B_{\delta_n}(x')$ for some $x'$ near $x$, so choose $n$ and $k$ so that $V_x \da B_{1\over 2^k}(x) \subset B_{\delta_n}(x) \subseteq V_{j, n}$.
The claim now is that $V_x$ intersects only finitely many elements of $\mcv$.
A proof of this follows from using the triangle inequality to show that $B_{1/2^{n+k}}(x)$ does not intersect any $V_{\beta, \ell}$ for $\ell \geq n+k$, and for $\ell < n + k$ it intersects these for at most one $\beta$, leaving only finitely many such $\ell$.
:::

:::


## Problem 3

:::{.problem title="1.3"}
Let $A=\mathbb{Z}$ be an abelian group. Let $\mathcal{F}$ be
a sheaf on $X$ such that every stalk $\mathcal{F}_{x}=A .$ Does it
follow that $\mathcal{F}$ is a constant sheaf?

- Show that the answer is no in general.

- What if $X$ is an irreducible algebraic variety with Zariski
topology?

- What if $X=[0,1]$ with classical topology?
:::

:::{.proposition title="Part 1"}
There is a sheaf $\mcf$ on a space $X$ with stalks satisfying $\mcf_x = \CC$ for every $x\in X$, but $\mcf$ is not isomorphic to the constant sheaf $\ul{\CC_X}$.
:::

:::{.proof title="Part 1"}
Let $X = S^1$ in the Euclidean topology, $U = X\sm\ts{1}, Z = \ts{1}$ and let the two inclusions be
\[
j: U &\to X, \\
i: Z &\to X
.\]
Now set
\[
\mcf \da j_! \ul{\CC_U} \oplus i_* \ul{\CC_Z}
\]
We can then compute the stalks:
\[
(j_! \ul{\CC_U} )_x 
&= \colim_{W \ni x} (j_! \ul{\CC_U} )(W) \\
&= \colim_{W \ni x} 
\begin{cases}
\CC & W \subseteq U 
\\
0 & \text{else} .
\end{cases} \\
&= \colim
\begin{cases}
\cdots \to \CC \to \CC \to \CC \to \cdots & x\in U 
\\
0 \to 0\to 0 \to 0 \to \cdots & x\not\in U
\end{cases}\\
&=
\begin{cases}
 \CC & x\in U 
\\
0 & x\not\in U,
\end{cases}
\\
&=
\begin{cases}
\CC & x \neq \ts{1} 
\\
0 & x=\ts{1}.
\end{cases}
\]
where we've used that since $U$ is open, if $x\in U$ then there is an open neighborhood $W\ni x$ completely contained in $U$, making the directed system eventually constant.
Otherwise, if $x\not\in U$, then no neighborhood of $x$ is completely contained in $U$, and the sections here are zero for every $W\ni X$.

> Note: this uses that the colimit of an eventually constant diagram is isomorphic to whatever that constant object is, i.e. it satisfies the correct universal property.

Similarly, noting that for $W \subseteq X$,
\[
i\inv(W) = 
\begin{cases}
\ts{1} & \ts{1} \in W  
\\
\emptyset & \ts{1} \not\in W,
\end{cases}
\]
we have
\[
i_* \ul{\CC_Z} 
&= \colim_{W\ni x} \ul{\CC_Z}(i\inv(W)) \\
&= \colim_{W\ni x} \ul{\CC_Z}\qty{
\begin{cases}
\ts{1} & \ts{1} \in W
\\
\emptyset & \ts{1}\not\in W
\end{cases}
} \\
&=
\colim
\begin{cases}
\CC\to \CC\to \cdots & x = \ts{1} 
\\
\cdots \to 0\to 0\to \cdots & x\neq \ts{1}
\end{cases}\\
&= 
\begin{cases}
\CC & x = \ts{1} 
\\
0 & x\neq \ts{1},
\end{cases}
\]
where we've used that if $U$ is open and $x\in U$ with $x\neq \ts{1}$, there eventually all small enough neighborhoods $W \ni x$ will not intersect $X\sm U = \ts{1}$.
Thus
\[
\mcf_x =
\begin{cases}
 0 \oplus \CC  & x = \ts{1}
\\
\CC \oplus 0 & x\neq \ts{1},
\end{cases}
\]
and all of the stalks are one copy of $\CC$, as in the constant sheaf $\ul{\CC_X}$ on $X$.
However, $\mcf$ and $\ul{\CC_X}$ do not have the same sections: take $W \subseteq S^1$ to be a connected open neighborhood of $\ts{1}$, then
\[
\ts{1} \in W \implies i_* \ul{\CC_Z}(W) = \ul{\CC_Z}(i\inv(W) ) = \ul{\CC_Z}(\ts{1}) = \CC
.\]
Note that $j\inv(W) = W \sm\ts{1} = W_1 \disjoint W_2$ breaks into two connected components, so
\[
j_! \ul{\CC_U}(W) = \ul{\CC_U}(W_1 \disjoint W_2) = \CC\sumpower{2}
,\]
so
\[
\mcf(W) = \CC\sumpower{2} \oplus \CC \neq \CC = \ul{\CC_X}(W) \implies \mcf \not\cong \ul{\CC_X}
.\]
:::

:::{.proposition title="Parts 2 and 3"}
If $X$ is an irreducible algebraic variety or $X = [0, 1]$ in the Euclidean topology, the answer is still generally no.
:::

:::{.proof title="Parts 2 and 3"}
The previous example shows this, noting that $S^1 \cong \spec {\RR[x, y] \over \gens{x^2 + y^2 - 1}}$ and $f(x, y) = x^2+y^2-1$ does not factor in $\RR[x, y]$, making $S^1$ irreducible in the Zariski topology.

For $X = [0, 1]$, a modification of the previous example yields the same conclusion: set $Z = \ts{1\over 2}$ and $U = X\sm Z$; the same argument with the same sheaf goes through.
:::

## Problem 4

:::{.problem title="1.4"}
Let $X$ be a space with a poset topology (with increasing
open sets).

- Prove that a sheaf $F$ on $X$ is the same as a collection $F_{x}$
for $x \in X$ and maps $r_{x, y}: F_{x} \rightarrow F_{y}$ for all
$x \leq y$ satisfying $r_{y, z} \circ r_{x, y}=r_{x, z}$.

- Prove that for an open $U \subset X$, the set of sections $F(U)$
is $\inverselim_{U\ni x} F_{x}$.
:::

:::{.proposition title="?"}
Let $(X, \leq)$ be a poset with the order topology, then a sheaf $\mcf \in \Sh(X)$ on $X$ is is equivalently a functorial assignment on the corresponding poset category
\[
\mcf: \Poset(X) &\to \cat{C} \\
x &\mapsto \mcf_x
,\]
where the objects of $\Poset(X)$ are elements $x\in X$ where the hom spaces are defined as
\[
\Hom_{\Poset(X)}(x, y)
\da
\begin{cases}
\ts{\pt} &  x\leq y
\\
\emptyset & \text{else}.
\end{cases}
\]
Here $\cat{C} = \Ab\Grp, \Ring, \rmod, \kalg$, etc.
:::

:::{.proof title="?"}
$\implies$:
Suppose that $\mcf$ is a sheaf on $(X, \leq)$, and define
\[
\mcg: \Poset(X) &\to \cat{C} \\
x &\mapsto \mcg_x \da \mcf(U_{\geq x}) \\
(\iota_{xy}: x \to y) &\mapsto (f_{xy}: \mcg_x \to \mcg_y) 
,\]
where we define $f_{xy}$ using that
\[
x\to y \in \Poset(X) \iff x\leq y \in X \implies U_{\geq y} \injects U_{\geq x} \in \Open(X)
,\]
and since $\mcf$ is a contravariant functor, the latter inclusion induces an morphism 
\[
f_{xy}: \mcf(U_{\geq x}) \to \mcf(U_{\geq y}) \qquad \in \cat{C}
.\]
Compatibility of the $f_{xy}$ for $\mcg$ follow immediately from the fact that $\mcf$ is a functor.

$\impliedby$:
Given a functorial assignment 
\[
\mcg: \Poset(X) &\to \cat{C} \\
x &\mapsto \mcg_x
,\]
we want to construct an associated sheaf 
\[
\mcf: \Open(X)\op \to \cat{C}
.\]
By a result from class, it suffices to specify the sheaf on a basis $\mcb$ for the order topology on $X$, so let $\mcb \da \ts{U_{\geq x}}_{x\in X}$ be the basis of up-sets.
Define a presheaf by
\[
\mcf^-(U_{\geq x}) \da \mcg_x
,\]
and take $\mcf \da (\mcf^-)^+$.
:::

:::{.proposition title="?"}
For $(X, \leq)$ a poset in the order topology, $U \subseteq X$ open, and $\mcf$ a sheaf on $X$,
\[
F(U) \cong \inverselim_{x\in U} \mcf_x
.\]
:::

:::{.proof title="?"}
It suffices to show that if $\mcb$ is a basis for a topology, 
\[
\mcf(U) = \inverselim_{V\in \mcb, V \subseteq U} \mcf(V)
,\]
which follows because this precisely describes a continuous section of the *espace étalé* over $U \subseteq X$ as a compatible collection of sections on $U$ decomposed in a basis as $U = \union B_i$. 
With this, we can then directly compute
\[
\mcf(U) 
&= \inverselim_{V\in \mcb, V \subseteq U} \mcf(V) \\
&= \inverselim_{V_{\geq x} \subseteq U} \mcf(V_{\geq x}) \qquad \text{by the definition of }\mcb \\
&= \inverselim_{V_{\geq x} \subseteq U} \mcf_x \qquad \text{since $V_{\geq x}$ is the smallest open containing $x$}\\
&= \inverselim_{x\in U} \mcf_x \qquad \text{ since } V_{\geq x} \subseteq U \implies x\in U 
.\]

:::