Exercise 111.6.1. Compute \mathop{\mathrm{Spec}}(\mathbf{Z}) as a set and describe its topology.
111.6 The Spectrum of a ring
Exercise 111.6.2. Let A be any ring. For f\in A we define D(f):= \{ \mathfrak p \subset A \mid f \not\in \mathfrak p\} . Prove that the open subsets D(f) form a basis of the topology of \mathop{\mathrm{Spec}}(A).
Exercise 111.6.3. Prove that the map I\mapsto V(I) defines a natural bijection
\{ I\subset A\text{ with }I = \sqrt{I}\} \longrightarrow \{ T\subset \mathop{\mathrm{Spec}}(A)\text{ closed}\}
Definition 111.6.4. A topological space X is called quasi-compact if for any open covering X = \bigcup _{i\in I} U_ i there is a finite subset \{ i_1, \ldots , i_ n\} \subset I such that X = U_{i_1}\cup \ldots U_{i_ n}.
Exercise 111.6.5. Prove that \mathop{\mathrm{Spec}}(A) is quasi-compact for any ring A.
Definition 111.6.6. A topological space X is said to verify the separation axiom T_0 if for any pair of points x, y\in X, x\not= y there is an open subset of X containing one but not the other. We say that X is Hausdorff if for any pair x, y\in X, x\not= y there are disjoint open subsets U, V such that x\in U and y\in V.
Exercise 111.6.7. Show that \mathop{\mathrm{Spec}}(A) is not Hausdorff in general. Prove that \mathop{\mathrm{Spec}}(A) is T_0. Give an example of a topological space X that is not T_0.
Remark 111.6.8. Usually the word compact is reserved for quasi-compact and Hausdorff spaces.
Definition 111.6.9. A topological space X is called irreducible if X is not empty and if X = Z_1\cup Z_2 with Z_1, Z_2\subset X closed, then either Z_1 = X or Z_2 = X. A subset T\subset X of a topological space is called irreducible if it is an irreducible topological space with the topology induced from X. This definition implies T is irreducible if and only if the closure \bar T of T in X is irreducible.
Exercise 111.6.10. Prove that \mathop{\mathrm{Spec}}(A) is irreducible if and only if Nil(A) is a prime ideal and that in this case it is the unique minimal prime ideal of A.
Exercise 111.6.11. Prove that a closed subset T\subset \mathop{\mathrm{Spec}}(A) is irreducible if and only if it is of the form T = V({\mathfrak p}) for some prime ideal {\mathfrak p}\subset A.
Definition 111.6.12. A point x of an irreducible topological space X is called a generic point of X if X is equal to the closure of the subset \{ x\} .
Exercise 111.6.13. Show that in a T_0 space X every irreducible closed subset has at most one generic point.
Exercise 111.6.14. Prove that in \mathop{\mathrm{Spec}}(A) every irreducible closed subset does have a generic point. In fact show that the map {\mathfrak p} \mapsto \overline{\{ {\mathfrak p}\} } is a bijection of \mathop{\mathrm{Spec}}(A) with the set of irreducible closed subsets of X.
Exercise 111.6.15. Give an example to show that an irreducible subset of \mathop{\mathrm{Spec}}(\mathbf{Z}) does not necessarily have a generic point.
Definition 111.6.16. A topological space X is called Noetherian if any decreasing sequence Z_1\supset Z_2 \supset Z_3\supset \ldots of closed subsets of X stabilizes. (It is called Artinian if any increasing sequence of closed subsets stabilizes.)
Exercise 111.6.17. Show that if the ring A is Noetherian then the topological space \mathop{\mathrm{Spec}}(A) is Noetherian. Give an example to show that the converse is false. (The same for Artinian if you like.)
Definition 111.6.18. A maximal irreducible subset T\subset X is called an irreducible component of the space X. Such an irreducible component of X is automatically a closed subset of X.
Exercise 111.6.19. Prove that any irreducible subset of X is contained in an irreducible component of X.
Exercise 111.6.20. Prove that a Noetherian topological space X has only finitely many irreducible components, say X_1, \ldots , X_ n, and that X = X_1\cup X_2\cup \ldots \cup X_ n. (Note that any X is always the union of its irreducible components, but that if X = {\mathbf R} with its usual topology for instance then the irreducible components of X are the one point subsets. This is not terribly interesting.)
Exercise 111.6.21. Show that irreducible components of \mathop{\mathrm{Spec}}(A) correspond to minimal primes of A.
Definition 111.6.22. A point x\in X is called closed if \overline{\{ x\} } = \{ x\} . Let x, y be points of X. We say that x is a specialization of y, or that y is a generalization of x if x\in \overline{\{ y\} }.
Exercise 111.6.23. Show that closed points of \mathop{\mathrm{Spec}}(A) correspond to maximal ideals of A.
Exercise 111.6.24. Show that {\mathfrak p} is a generalization of {\mathfrak q} in \mathop{\mathrm{Spec}}(A) if and only if {\mathfrak p}\subset {\mathfrak q}. Characterize closed points, maximal ideals, generic points and minimal prime ideals in terms of generalization and specialization. (Here we use the terminology that a point of a possibly reducible topological space X is called a generic point if it is a generic points of one of the irreducible components of X.)
Exercise 111.6.25. Let I and J be ideals of A. What is the condition for V(I) and V(J) to be disjoint?
Definition 111.6.26. A topological space X is called connected if it is nonempty and not the union of two nonempty disjoint open subsets. A connected component of X is a maximal connected subset. Any point of X is contained in a connected component of X and any connected component of X is closed in X. (But in general a connected component need not be open in X.)
Exercise 111.6.27. Let A be a nonzero ring. Show that \mathop{\mathrm{Spec}}(A) is disconnected iff A\cong B \times C for certain nonzero rings B, C.
Exercise 111.6.28. Let T be a connected component of \mathop{\mathrm{Spec}}(A). Prove that T is stable under generalization. Prove that T is an open subset of \mathop{\mathrm{Spec}}(A) if A is Noetherian. (Remark: This is wrong when A is an infinite product of copies of {\mathbf F}_2 for example. The spectrum of this ring consists of infinitely many closed points.)
Exercise 111.6.29. Compute \mathop{\mathrm{Spec}}(k[x]), i.e., describe the prime ideals in this ring, describe the possible specializations, and describe the topology. (Work this out when k is algebraically closed but also when k is not.)
Exercise 111.6.30. Compute \mathop{\mathrm{Spec}}(k[x, y]), where k is algebraically closed. [Hint: use the morphism \varphi : \mathop{\mathrm{Spec}}(k[x, y]) \to \mathop{\mathrm{Spec}}(k[x]); if \varphi ({\mathfrak p}) = (0) then localize with respect to S = \{ f\in k[x] \mid f \not= 0\} and use result of lecture on localization and \mathop{\mathrm{Spec}}.] (Why do you think algebraic geometers call this affine 2-space?)
Exercise 111.6.31. Compute \mathop{\mathrm{Spec}}(\mathbf{Z}[y]). [Hint: as above.] (Affine 1-space over \mathbf{Z}.)
Comments (0)