Problem Set 2

Find the irreducible components of \begin{align*} X = V(x - yz, xz - y^2) \subset {\mathbb{A}}^3/{\mathbb{C}} .\end{align*}

Since \(x=yz\) for all points in \(X\), we have \begin{align*} X &= V(x-yz, yz^2 - y^2) \\ &= V\qty{x-yz, y(z^2 - y) } \\ &= V(x-yz, y) \cup V(x-yz, z^2-y) \\ &\coloneqq X_1 \cup X_2 .\end{align*}

These two subvarieties are irreducible.

It suffices to show that the \(A(X_i)\) are integral domains. We have \begin{align*} A(X_1) \coloneqq{\mathbb{C}}[x,y,z] / \left\langle{x-yz, y}\right\rangle \cong {\mathbb{C}}[y,z]/\left\langle{y}\right\rangle \cong {\mathbb{C}}[z] ,\end{align*} which is an integral domain since \({\mathbb{C}}\) is a field and thus an integral domain, and \begin{align*} A(X_2) \coloneqq{\mathbb{C}}[x,y,z]/\left\langle{x-yz, z^2 - y}\right\rangle \cong {\mathbb{C}}[y,z]/\left\langle{z^2-y}\right\rangle \cong {\mathbb{C}}[y] ,\end{align*} which is an integral domain for the same reason.

Let \(X\subset {\mathbb{A}}^n\) be an arbitrary subset and show that \begin{align*} V(I(X)) = \mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu .\end{align*}


\(\mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu \subseteq V(I(X))\):
We have \(X\subseteq V(I(X))\) and since \(V(J)\) is closed in the Zariski topology for any ideal \(J {~\trianglelefteq~}k[x_1, \cdots, x_{n}]\) by definition, \(V(I(X))\) is closed. Thus \begin{align*} X\subseteq V(I(X)) \text{ and } V(I(X))\text{ closed } \implies \mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu \subseteq V(I(X)) ,\end{align*} since \(\mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu\) is the intersection of all closed sets containing \(X\).\

\(V(I(X)) \subseteq \mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu\):
Noting that \(V({-}), I({-})\) are individually order-reversing, we find that \(V(I({-}))\) is order-preserving and thus \begin{align*} X\subseteq \mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu \implies V(I(X)) \subseteq V(I(\mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu)) = \mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu ,\end{align*} where in the last equality we’ve used part (i) of the Nullstellensatz: if \(X\) is an affine variety, then \(V(I(X)) = X\). This applies here because \(\mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu\) is always closed, and the closed sets in the Zariski topology are precisely the affine varieties.

Let \(\left\{{U_i}\right\}_{i\in I} \rightrightarrows X\) be an open cover of a topological space with \(U_i \cap U_j \neq \emptyset\) for every \(i, j\).

  • Show that if \(U_i\) is connected for every \(i\) then \(X\) is connected.

  • Show that if \(U_i\) is irreducible for every \(i\) then \(X\) is irreducible.

Suppose toward a contradiction that \(X = X_1 {\textstyle\coprod}X_2\) with \(X_i\) proper, disjoint, and open. Since \(\left\{{U_i}\right\} \rightrightarrows X\), for each \(j\in I\) this would force one of \(U_j \subseteq X_1\) or \(U_j \subseteq X_2\), since otherwise \(U_j \cap X_1 \cap X_2\) would be nonempty.
So without loss of generality (relabeling if necessary), assume \(U_j \in X_1\) for some fixed \(j\). But then for every \(i\neq j\), we have \(U_i \cap U_j\) nonempty by assumption, and so in fact \(U_i \subseteq X_1\) for every \(i\in I\). But then \(\cup_{i\in I}U_i \subseteq X_1\), and since \(\left\{{U_i}\right\}\) was a cover, this forces \(X\subseteq X_1\) and thus \(X_2 = \emptyset\).

\(X\) is irreducible \(\iff\) any two open subsets intersect.

This follows because otherwise, if \(U, V \subset X\) are open and disjoint then \(X\setminus U,\, X\setminus V\) are proper and closed. But then we can write \(X = \qty{X\setminus U} {\textstyle\coprod}\qty{X\setminus V}\) as a union of proper closed subsets, forcing \(X\) to not be irreducible.\

So it suffices to show that if \(U, V\subset X\) then \(U\cap V\) is nonempty. Since \(\left\{{U_i}\right\} \rightrightarrows X\), we can find a pair \(i, j\) such that there is at least one point in \(U\cap U_i\) and one point in \(V \cap U_j\).\

But by assumption \(U_i\cap U_j\) is nonempty, so both \(U\cap U_i\) and \(U_j \cap U_i\) are open nonempty subsets of \(U_i\). Since \(U_i\) was assumed irreducible, they must intersect, so there exists a point \begin{align*} x_0 \in \qty{U\cap U_i} \cap\qty{U_j \cap U_i} = U\cap\qty{U_i \cap U_j} \coloneqq\tilde U .\end{align*}

We can now similarly note that \(\tilde U \cap V\) and \(U_j \cap V\) are nonempty open subsets of \(V\), and thus intersect. So there is a point \begin{align*} \tilde x_0 \in \qty{\tilde U \cap V} \cap\qty{U_j \cap V} = \tilde U\cap V = U\cap V \cap\qty{U_i \cap U_j} ,\end{align*} and in particular \(\tilde x_0 \in U\cap V\) as desired.

Let \(f:X\to Y\) be a continuous map of topological spaces.

  • Show that if \(X\) is connected then \(f(X)\) is connected.

  • Show that if \(X\) is irreducible then \(f(X)\) is irreducible.

Toward a contradiction, if \(f(X) = Y_1 {\textstyle\coprod}Y_2\) with \(Y_1, Y_2\) nonempty and open in \(Y\), then \begin{align*}f^{-1}(f(X)) \subseteq X\end{align*} on one hand, and \begin{align*}f^{-1}(f(X)) = f^{-1}(Y_1) {\textstyle\coprod}f^{-1}(Y_2)\end{align*} on the other. If \(f\) is continuous, the preimages \(f^{-1}(Y_i)\) are open (and nonempty), so \(X\) contains a disconnected subset. However, every subset of a connected set must be connected, so this contradicts the connectedness of \(X\).

Suppose \(f(X) = Y_1 \cup Y_2\) with \(Y_i\) proper closed subsets of \(Y\). Then \(f^{-1}(Y_1) \cup f^{-1}(Y^2) = (f^{-1} \circ f)(X) \subseteq X\) are closed in \(X\), since \(f\) is continuous. Since \(X\) is irreducible, without loss of generality (by relabeling), this forces \(X_1 = \emptyset\). But then \(f(X_1) = \emptyset\), forcing \(f(X) = Y_2\).

For two ideals \(J_1, J_2{~\trianglelefteq~}R\), the ideal quotient is defined by \begin{align*} J_1 : J_2 \coloneqq\left\{{f\in R {~\mathrel{\Big\vert}~}fJ_2 \subset J_1}\right\} .\end{align*}

Let \(X\) be an affine variety.

  • Show that if \(Y_1, Y_2 \subset X\) are subvarieties then \begin{align*} I(\mkern 1.5mu\overline{\mkern-1.5muY_1\setminus Y_2\mkern-1.5mu}\mkern 1.5mu) = I(Y_1): I(Y_2) .\end{align*}

  • If \(J_1, J_2 {~\trianglelefteq~}A(X)\) are radical, then \begin{align*} \mkern 1.5mu\overline{\mkern-1.5muV(J_1) \setminus V(J_2)\mkern-1.5mu}\mkern 1.5mu = V(J_1: J_2) .\end{align*}


Let \(X \subset {\mathbb{A}}^n,\, Y\subset {\mathbb{A}}^m\) be irreducible affine varieties, and show that \(X\times Y\subset {\mathbb{A}}^{n+m}\) is irreducible.

That \(X\times Y\) is again an affine variety follows from writing \(X=V(I),\, Y=V(J)\), then \(X\times Y = V(I+J)\) where \(I+J{~\trianglelefteq~}k[x_1, \cdots, x_n, y_1, \cdots, y_m]\). So let \begin{align*}X\times Y = U \cup V\end{align*} with \(U, V\) proper and closed, and let \(\pi_X, \pi_Y\) be the projections onto the factors.

For each \(x\in X\), \(\pi^{-1}(x) \cong Y\) is contained in only one of \(U\) or \(V\).

Note that if this is true, we can write \(X = G_U \cup G_V\) where \begin{align*} G_U\coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}\pi_X^{-1}(x) \subseteq U}\right\} \end{align*} are the points for which the entire fiber lies in \(U\), and similarly \(G_V\) are those for which the fiber lies in \(V\). If we can then show that \(G_U, G_V\) are closed, by irreducibility of \(X\) this will force (wlog) \(G_V = \emptyset\) and \(X = G_U\). But then \begin{align*} \pi_X^{-1}(X) = X\times Y \text{ and }\pi_X^{-1}(G_U) = U \implies X\times Y = U .\end{align*} which shows that \(X\times Y\) is irreducible.

For any fixed \(x\), we can write \begin{align*} \pi_X^{-1}(x) = \qty{\pi_X^{-1}(x) \cap U } \cup\qty{\pi_X^{-1}(x) \cap V} .\end{align*}

Since points are closed in the Zariski topology and \(\pi_X\) is continuous, each \(\pi_X^{-1}(x)\) is closed. and thus \(\pi_X^{-1}(x)\cap U\) is closed (and similarly for \(V\)). Noting that \(\pi_X^{-1}(x) \cong \left\{{x}\right\}\times Y \cong Y\), where we’ve assumed \(Y\) to be irreducible, we can conclude wlog that \(\pi_X^{-1}(x) \cap V = \emptyset\).

Wlog consider \(G_U \subseteq X\). Fixing any point \(y_0 \in Y\), we have \begin{align*}X\cong X_{y_0} \coloneqq X\times\left\{{y_0}\right\} \subseteq X\times Y,\end{align*} so we can identify \(G_U \subset X\) with \(G_U\subset X_{y_0}\) inside a \(Y{\hbox{-}}\)fiber the product. But then \begin{align*}G_U = X_{y_0} \cap U \subseteq X\times Y,\end{align*} where \(U\) is closed in \(X\times Y\) and thus closed in \(X_{y_0}\), and \(X_{y_0}\) is trivially closed in itself. This exhibits \(G_U\) as the intersection of two sets that are closed in \(X_{y_0} \cong X\).