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Let \(U\) be an open subvariety of a variety \(X, Y\) a closed subvariety of \(U .\) Let \(Z\) be the closure of \(Y\) in \(X\). Show that (a) \(Z\) is a closed subvariety of \(X\). (b) \(Y\) is an open subvariety of \(Z\).

Short Answer

Expert verified
Short Answer: To prove that (a) \(Z\) is a closed subvariety of \(X\) and (b) \(Y\) is an open subvariety of \(Z\), we first show that \(Z\) is a variety and closed in the Zariski topology, as it is the closure of \(Y\) in \(X\). Next, we find an open set \(U'\) in \(Z\) such that \(Y = U' \cap Z\) and show that the intersection of an open and closed set in the Zariski topology is open.

Step by step solution

01

Show that \(Z\) is a closed subvariety of \(X\)

Recall the definition of a closed subvariety. A closed subvariety is a subset of the variety that is a variety itself and also closed in the Zariski topology. To show that \(Z\) is a closed subvariety of \(X\), we need to show two things: 1. \(Z\) is a variety. 2. \(Z\) is closed in the Zariski topology. First, note that the closure of a set in the Zariski topology is indeed a variety, so we have that \(Z\) is a variety. Second, by definition of closure, the closure of a set in the Zariski topology is the smallest closed set containing the original set. Hence, \(Z\) is closed in the Zariski topology. Thus, we have shown that \(Z\) is a closed subvariety of \(X\).
02

Show that \(Y\) is an open subvariety of \(Z\)

To prove (b), we need to show that \(Y\) is an open subvariety of \(Z\). An open subvariety is a subset of the variety that is also a variety itself, but is open with respect to the Zariski topology. First, note that \(Y\) is already a variety since it is a closed subvariety of \(U\). To show that \(Y\) is open in \(Z\), we can use the definition of topology. We need to find an open set \(U'\) in \(Z\) such that \(Y = U' \cap Z\). Using the set \(U\) as our starting point, we have: $$U' = U \cap Z.$$ Now, we need to show that \(U'\) is open in \(Z\) and \(Y = U' \cap Z\). Since \(U\) is open in \(X\) and \(Z\) is closed in \(X\), it follows that \(U'\) is open in \(Z\) because the intersection of an open set and a closed set in the Zariski topology is open. Thus, \(U'\) is open in \(Z\). Finally, notice that \(Y = U' \cap Z\) because \(Y\) is a closed subvariety of \(U\) and is contained in both \(U\) and \(Z\). Hence, \(Y = U' \cap Z\). Putting it all together, we have shown that \(Y\) is an open subvariety of \(Z\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Zariski Topology
Zariski topology is a special kind of topology that is widely used in algebraic geometry. Unlike more familiar topologies, such as those on real numbers, Zariski topology has fewer open sets. This characteristic makes it quite distinct and sometimes counterintuitive for newcomers. In Zariski topology, closed sets are algebraic sets—these are the zeros of a polynomial or several polynomials.

To get a better sense of how this works, imagine you have a variety defined by the common solutions to several polynomial equations. The closed sets in the Zariski topology for this variety would then be all of its subvarieties—essentially any subset, also defined by polynomials, that fits within the original variety.

Interestingly, because Zariski topology has relatively few open sets, most sets considered open in ordinary topologies would not be open in Zariski topology. Yet, by understanding which sets in a variety are closed, you can better grasp which ones are open, as these are simply the complements of closed sets.
Closed Subvariety
A closed subvariety is a particular type of subset in a variety that holds special algebraic significance. To qualify as a closed subvariety, a subset must satisfy two primary conditions:
  • It is a variety on its own.
  • It is closed in the Zariski topology.

What does this mean practically? Consider that a variety is essentially a geometric object or space containing solutions to sets of polynomials. A closed subvariety is a subset of this space that is itself the set of all solutions to another, potentially different, set of polynomial equations and exhibits closure in the Zariski topology.

Closure in this context means that no point on the boundary of the set can be approached by any sequence of points inside the set. For example, when taking the closure of a variety, you include any points needed to make it closed in the Zariski sense. This is how we determine that a set like \(Z\) in the solved problem is a closed subvariety of \(X\). The closure of \(Y\) within \(X\) encompasses precisely these properties, making \(Z\) a closed subvariety.
Open Subvariety
Open subvarieties, although less common compared to closed subvarieties, play a significant role in algebraic geometry. An open subvariety is defined as a subset within a variety that meets the criteria of being both a variety and open in the Zariski topology.
  • It is essentially an open set that is also a variety.
  • It fits the notion of 'openness' within the Zariski framework.

Openness in Zariski topology can seem restrictive due to fewer open sets available compared to conventional topologies. However, an open subvariety can be found by taking the intersection of an open set in the larger space with a variety.

For instance, in the textbook problem, \(Y\) is shown to be an open subvariety of \(Z\) through this process. Given that \(Y\) is a variety and open relative to the larger space \(Z\), it qualifies as an open subvariety. This distinction helps describe spaces in a way that is both useful for algebraic manipulation and insightful from a geometric perspective.

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Most popular questions from this chapter

(a) Let \(X\) be a topological space, \(X=\bigcup_{\alpha \in \mathscr{A}} U_{\alpha}, U_{\alpha}\) open in \(X\). Show that a subset \(W\) of \(X\) is closed if and only if each \(W \cap U_{\alpha}\) is closed (in the induced topology) in \(U_{\alpha}\). (b) Suppose similarly \(Y=\bigcup_{\alpha \in \mathscr{A}} V_{\alpha}, V_{\alpha}\) open in \(Y\), and suppose \(f: X \rightarrow Y\) is a mapping such that \(f\left(U_{\alpha}\right) \subset V_{\alpha}\). Show that \(f\) is continuous if and only if the restriction of \(f\) to each \(U_{\alpha}\) is a continuous mapping from \(U_{\alpha}\) to \(V_{\alpha}\).

If there is a dominating rational map from \(X\) to \(Y\), then \(\operatorname{dim}(Y) \leq \operatorname{dim}(X)\).

(a) Let \(C=V\left(Y^{2} Z-X^{3}\right)\) be a cubic with a cusp, \(C^{\circ}=C \backslash\\{[0: 0: 1]\\}\) the simple points, a group with \(O=[0: 1: 0] .\) Show that the map \(\varphi: \mathbb{G}_{a} \rightarrow C^{\circ}\) given by \(\varphi(t)=\) \(\left[t: 1: t^{3}\right]\) is an isomorphism of algebraic groups. (b) Let \(C=V\left(X^{3}+Y^{3}-X Y Z\right)\) be a cubic with a node, \(C^{\circ}=C \backslash\\{[0: 0: 1]\\}, O=[1: 1: 0] .\) Show that \(\varphi: \mathbb{G}_{m} \rightarrow C^{\circ}\) defined by \(\varphi(t)=\left[t: t^{2}: 1-t^{3}\right]\) is an isomorphism of algebraic groups.

(a) If \(Y\) is an open or closed subvariety of \(X\), the inclusion \(i: Y \rightarrow X\) is a morphism. (b) The composition of morphisms is a morphism.

Let \(d \geq 1, N=\frac{(d+1)(d+2)}{2}\), and let \(M_{1}, \ldots, M_{N}\) be the monomials of degree \(d\) in \(X, Y, Z\) (in some order). Let \(T_{1}, \ldots, T_{N}\) be homogeneous coordinates for \(\mathbb{P}^{N-1} .\) Let \(V=V\left(\sum_{i=1}^{N} M_{i}(X, Y, Z) T_{i}\right) \subset \mathbb{P}^{2} \times \mathbb{P}^{N-1}\), and let \(\pi: V \rightarrow \mathbb{P}^{N-1}\) be the restriction of the projection map. (a) Show that \(V\) is an irreducible closed subvariety of \(\mathbb{P}^{2} \times \mathbb{P}^{N-1}\), and \(\pi\) is a morphism. (b) For each \(t=\left(t_{1}, \ldots, t_{N}\right) \in \mathbb{P}^{N-1}\), let \(C_{t}\) be the corresponding curve \((\S 5.2)\). Show that \(\pi^{-1}(t)=C_{t} \times\\{t\\}\) We may thus think of \(\pi: V \rightarrow \mathbb{P}^{N-1}\) as a "universal family" of curves of degree \(d\). Every curve appears as a fibre \(\pi^{-1}(t)\) over some \(t \in \mathbb{P}^{N-1}\).

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