--- notoc: true title: Problem Set 2 --- # Problem Set 2 ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="Gathmann 2.17"} Find the irreducible components of ` \begin{align*} X = V(x - yz, xz - y^2) \subset {\mathbb{A}}^3/{\mathbb{C}} .\end{align*} `{=html} ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv} 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*} `{=html} ::: {.claim .proofenv .proofenv .proofenv .proofenv .proofenv} 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*} `{=html} 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*} `{=html} which is an integral domain for the same reason. ::: ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="Gathmann 2.18"} 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*} `{=html} ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv} ```{=tex} \hfill ``` $\mkern 1.5mu\overline{\mkern-1.5muX\mkern-1.5mu}\mkern 1.5mu \subseteq V(I(X))$:\ 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*} `{=html} 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*} `{=html} 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. ::: ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="Gathmann 2.21"} 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 $i, j$. a. Show that if $U_i$ is connected for every $i$ then $X$ is connected. b. Show that if $U_i$ is irreducible for every $i$ then $X$ is irreducible. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv title="a"} 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 $X_2 = \emptyset$. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv title="b"} ::: {.claim .proofenv .proofenv .proofenv .proofenv .proofenv} $X$ 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*} `{=html} 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*} `{=html} and in particular $\tilde x_0 \in U\cap V$ as desired. ::: ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="Gathmann 2.22"} Let $f:X\to Y$ be a continuous map of topological spaces. a. Show that if $X$ is connected then $f(X)$ is connected. b. Show that if $X$ is irreducible then $f(X)$ is irreducible. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv title="a"} 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*} `{=html} on one hand, and ` \begin{align*}f^{-1}(f(X)) = f^{-1}(Y_1) {\textstyle\coprod}f^{-1}(Y_2)\end{align*} `{=html} 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 $X$. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv title="b"} 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 $f(X) = Y_2$. ::: ::: {.definition .proofenv .proofenv .proofenv .proofenv .proofenv title="Ideal Quotient"} 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*} `{=html} ::: ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="Gathmann 2.23"} Let $X$ be an affine variety. a. 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*} `{=html} b. 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*} `{=html} ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv} ? ::: ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="Gathmann 2.24"} 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. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv} 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*} `{=html} with $U, V$ proper and closed, and let $\pi_X, \pi_Y$ be the projections onto the factors. ::: {.claim .proofenv .proofenv .proofenv .proofenv .proofenv} 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*} `{=html} 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*} `{=html} which shows that $X\times Y$ is irreducible. ::: {.proof .proofenv .proofenv .proofenv .proofenv .proofenv title="Every fiber is contained in one irreducible component"} 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*} `{=html} 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 $\pi_X^{-1}(x) \cap V = \emptyset$. ::: ::: {.proof .proofenv .proofenv .proofenv .proofenv .proofenv title="$G_U, G_V$ are closed"} 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*} `{=html} 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*} `{=html} 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 $X_{y_0} \cong X$. ::: :::