Problem Set 3

Define \begin{align*} X \coloneqq\left\{{M \in \operatorname{Mat}(2\times 3, k) {~\mathrel{\Big\vert}~}{\operatorname{rank}}M \leq 1}\right\} \subseteq {\mathbb{A}}^6/k .\end{align*}

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

We’ll use the following fact from linear algebra:

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.

If $A\in \operatorname{Mat}(m \times n, k)$ is a matrix, then the rank of $A$ is equal to the order of largest nonzero minor.

Thus \begin{align*} M_{ij} = 0 \text{ for all $\ell\times \ell$ minors } M_{ij} \iff {\operatorname{rank}}(M) < \ell ,\end{align*} 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 $\operatorname{det}M_{ij}$ is a polynomial $f_{ij}$ in its entries, which means that $X$ can be written as \begin{align*} X = V\qty{\left\{{f_{ij}}\right\}} ,\end{align*} which exhibits $X$ as a variety. Thus \begin{align*} M = \begin{bmatrix} x & y & z \\ a & b & c \end{bmatrix} \implies X = V\qty{\left\langle{xb-ya, yc-zb, xc-za}\right\rangle} \subset {\mathbb{A}}^6 .\end{align*}

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

$\dim (X) = 4$.

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!

Let $X$ be a topological space, and show

If $\left\{{U_i}\right\}_{i\in I} \rightrightarrows X$, then $\dim X = \sup_{i\in I} \dim U_i$.
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?

Strictly for notational convenience, we’ll treat $\left\{{U_i}\right\}$ 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 $\left\{{I_j}\right\}$ of closed irreducible subsets of $V$ of length $n>\dim U$. But then $I'_j \coloneqq I_j \cap U$ 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 $s \coloneqq\sup_i \dim U_i$ and $n\coloneqq\dim X$, we want to show that $s\geq n$. Let $\left\{{I_j}\right\}_{j\leq n}$ be a maximal chain of length $n$ of closed irreducible subsets of $X$, so we have \begin{align*} \emptyset \subsetneq I_0 \subsetneq I_1 \subsetneq \cdots \subsetneq I_n \subseteq X .\end{align*}

Since $I_0\subset X$ and $\left\{{U_i}\right\}$ covers $X$, we can find some $U_{0}\in \left\{{U_i}\right\}$ such that $I_0\cap U_0$ 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 $A_j \coloneqq I_j \cap U_0$.\

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 \begin{align*} x\in A_{i+1}\setminus A_i \iff x\in \qty{I_{i+1} \cap U_0} \setminus \qty{I_{i} \cap U_0} = \qty{I_1 \setminus I_2}\cap U_0 = \emptyset .\end{align*} But this exhibits a length $n$ chain in $U_0$, so $\dim U_0 \geq n$. Taking suprema, we have \begin{align*} n \leq \dim U_0 \leq \sup_{i\in J} \dim U_i = s .\end{align*}

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 \begin{align*} X \coloneqq\left\{{a, b}\right\} \qquad \tau \coloneqq\left\{{\emptyset, X, \left\{{1}\right\}}\right\} ,\end{align*} Here $\left\{{b}\right\}$ 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 $1$, \begin{align*} \emptyset \subsetneq \left\{{b}\right\} \subsetneq X ,\end{align*} and so $\dim X = 1$. However, for $U\coloneqq\left\{{a}\right\}$, there is only one possible maximal chain: \begin{align*} \emptyset \subsetneq \left\{{a}\right\} = X ,\end{align*} so $\dim U = 0$.

Prove the following:

Every noetherian topological space is compact. In particular, every open subset of an affine variety is compact in the Zariski topology.
A complex affine variety of dimension at least 1 is never compact in the classical topology.

Let \begin{align*} R = k[x_1, x_2, x_3, x_4] / \left\langle{x_1 x_4 - x_2 x_3}\right\rangle \end{align*} and show the following:

$R$ is an integral domain of dimension 3.
$x_1, \cdots, x_4$ are irreducible but not prime in $R$, and thus $R$ is not a UFD.

$x_1 x_4$ and $x_2 x_3$ are two decompositions of the same element in $R$ which are nonassociate.
$\left\langle{x_1, x_2}\right\rangle$ is a prime ideal of codimension 1 in $R$ that is not principal.

Consider a set $U$ in the complement of $(0, 0) \in {\mathbb{A}}^2$. Prove that any regular function on $U$ extends to a regular function on all of ${\mathbb{A}}^2$.