---
title: "Problem Set 3"
notoc: true
---

# Problem Set 3

:::{.exercise title="Gathmann 2.33"}
Define
\[  
X \da \ts{M \in \mat(2\times 3, k) \st \rk M \leq 1} \subseteq \AA^6/k
.\]

Show that $X$ is an irreducible variety, and find its dimension.
:::

:::{.solution}
We'll use the following fact from linear algebra:

:::{.definition title="Matrix Minor"}
For an $m\times n$ matrix, a *minor of order* $\ell$ is the determinant of a $\ell\times \ell$ submatrix obtained by deleting any $m-\ell$ rows and any $n-\ell$ columns.
:::
:::{.theorem title="Rank is a Function of Minors"}
If $A\in \Mat(m \times n, k)$ is a matrix, then the rank of $A$ is equal to the order of largest nonzero minor.
:::

Thus 
\[  
M_{ij} = 0 \text{ for all $\ell\times \ell$ minors } M_{ij} \iff \rk(M) < \ell
,\]
following from the fact that if one takes $\ell = \min(m,n)$ and all $\ell\times \ell$ minors vanish, then the largest nonzero minor must be of size $j\times j$ for $j\leq \ell -1$.
But $\det M_{ij}$ is a polynomial $f_{ij}$ in its entries, which means that $X$ can be written as
\[  
X = V\qty{\ts{f_{ij}}}
,\]
which exhibits $X$ as a variety.
Thus 
\[  
M = 
\begin{bmatrix}
x & y & z \\
a & b & c
\end{bmatrix}
\implies
X = V\qty{\gens{xb-ya, yc-zb, xc-za}} \subset \AA^6 
.\]

:::{.claim}
The ideal above is prime, and so the coordinate ring $A(X)$ is a domain and thus $X$ is irreducible.
:::

:::{.claim}
$\dim (X) = 4$.
:::
Heuristic: there are three degrees of freedom in choosing the first row $x,y,z$. 
To enforce the rank 1 condition, the second row must be a scalar multiple of the first, yielding one degree of freedom for the scalar.

> Note: I looked at this for a couple of hours, but I don't know how to prove either of these statements with the tools we have so far!

:::


:::{.exercise title="Gathmann 2.34"}
Let $X$ be a topological space, and show

a. If $\ts{U_i}_{i\in I} \covers X$, then $\dim X = \sup_{i\in I} \dim U_i$.

b. If $X$ is an irreducible affine variety and $U\subset X$ is a nonempty subset, then $\dim X = \dim U$.
  Does this hold for any irreducible topological space?
:::


:::{.solution}

> Strictly for notational convenience, we'll treat $\ts{U_i}$ is if it were a countable open cover.

**Part a:**
We first note that if $U \subseteq V$, then $\dim U \leq \dim V$.
If this were not the case, one could find a chain $\ts{I_j}$ of closed irreducible subsets of $V$ of length $n>\dim U$.
But then $I'_j \da I_j \intersect U$ would again be a closed irreducible set, yielding a chain of length $n$ in $U$.
Thus $\dim X\geq \dim U_i$, and it remains true that $\dim X \geq \sup \dim U_i$, so it suffices to show that $\dim X \leq \sup \dim U_i$.
\

Set $s \da \sup_i \dim U_i$ and $n\da \dim X$, we want to show that $s\geq n$.
Let $\ts{I_j}_{j\leq n}$ be a maximal chain of length $n$ of closed irreducible subsets of $X$, so we have
\[  
\emptyset \subsetneq I_0 \subsetneq I_1 \subsetneq \cdots \subsetneq I_n \subseteq X
.\]

Since $I_0\subset X$ and $\ts{U_i}$ covers $X$, we can find some $U_{0}\in \ts{U_i}$ such that $I_0\intersect U_0$ is nonempty, since otherwise there would be a point in $I_0 \cap \qty{X\setminus \cup_{i\in J} U_i} = \emptyset$.
We can do this for every $I_j$, so define $A_j \da I_j \intersect U_0$.
\

Each $A_j$ is now closed in $U_0$, and must remain irreducible, since any decomposition of $A_j$ would lift to a decomposition of $I_0$.
To see that $A_0 \subsetneq A_1$, i.e. that the inclusions are still proper, we can just note that 
\[  
x\in A_{i+1}\setminus A_i \iff x\in  
\qty{I_{i+1} \intersect U_0} \setminus 
\qty{I_{i} \intersect U_0} = \qty{I_1 \setminus I_2}\intersect U_0 = \emptyset
.\]
But this exhibits a length $n$ chain in $U_0$, so $\dim U_0 \geq n$.
Taking suprema, we have
\[  
n \leq \dim U_0 \leq \sup_{i\in J} \dim U_i = s
.\]

**Part b**:
The answer is **no**: we can produce a space $X$ with some $\dim X$ and a subset $U$ satisfying $\dim U < \dim X$.

Define a space and a topology by
\[  
X \da \ts{a, b} \qquad \tau \da\ts{\emptyset, X, \ts 1}
,\]
Here $\ts{b}$ is the only proper and closed subset, since its complement is open, so $X$ must be irreducible.
We can find an maximal ascending chain of length $1$,
\[  
\emptyset \subsetneq \ts{b} \subsetneq X
,\]
and so $\dim X = 1$.
However, for $U\da \ts{a}$, there is only one possible maximal chain:
\[  
\emptyset \subsetneq \ts{a} = X
,\]
so $\dim U = 0$.

:::



:::{.exercise title="Gathmann 2.36"}
Prove the following:

a. Every noetherian topological space is compact.
  In particular, every open subset of an affine variety is compact in the Zariski topology.

b. A complex affine variety of dimension at least 1 is never compact in the classical topology.

:::


:::{.exercise title="Gathmann 2.40"}
Let
\[  
R = k[x_1, x_2, x_3, x_4] / \gens{x_1 x_4 - x_2 x_3} 
\]
and show the following:

a. $R$ is an integral domain of dimension 3.

b. $x_1, \cdots, x_4$ are irreducible but not prime in $R$, and thus $R$ is not a UFD.

c. $x_1 x_4$ and $x_2 x_3$ are two decompositions of the same element in $R$ which are nonassociate.

d. $\gens{x_1, x_2}$ is a prime ideal of codimension 1 in $R$ that is not principal.


:::


:::{.exercise title="Problem 5"}
Consider a set $U$ in the complement of $(0, 0) \in \AA^2$.
Prove that any regular function on $U$ extends to a regular function on all of $\AA^2$.

:::