# 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*}

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$$\mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu \subseteq V(I(X))$$:
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*}

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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 $$X_{y_0} \cong X$$.