1. Spectrum of a All the rings are assumed to be commutative.

Let A be a ring. The spectrum of A denoted by Spec A is the set of prime ideals of A. For any subset T of A, we denote V (T ) the set of all prime ideals containing T. For f ∈ A, we denote D(f) the complement of V (f). Proposition 1.1. The spectrum of A is empty if and only if A is the . Proof. If A is the zero ring, there is no proper prime of A. Hence Spec A is empty. If A is a nonzero ring, A has a (by Zorn’s lemma) (or every ideal is contained in a maximal ideal we can choose the zero ideal). Then Spec A is nonempty.  Hence we assume that A is a ring with identity 1 and hence A is not a zero ring.

Proposition 1.2. Let A be a ring. Suppose that T,S are subsets of A and I,J are ideals of A, f ∈ A. (1) If T ⊂ S, then V (T ) ⊃ V (S). (2) Let√hT i be the ideal generated by T. We√ have V (T ) = V (hT i). (3) If I is the radical of I, then V (I) = V ( I). (4) V (I) = ∅ if and only if I is√ the unit√ ideal. (5) V (I) = V (J) if and only if I = J. (6) D(f) = ∅ if and only if f is nilpotent. (7) If D(f) = Spec A, then f is a unit. (8) V (I) ∪ V (J) = V (I ∩ J). T S (9) If {Ia} is a family of ideals of A, then a V (Ia) = V ( a Ia). (10) D(fg) = D(f) ∩ D(g). Proof. (1) is obvious.

(2) Since T is contained in hT i, by (1) V (T ) ⊃ V (hT i). If p is an ideal containing T, i.e. p ∈ V (T ), then p contains hT i1 Hence p ∈ V (hT i). √ √ (3) The ideal I is contained√ in its radical I. Hence√ V (I) ⊃ V ( I)√. Suppose a p does not lie in V ( I). Then there exists f ∈ I \ p. Since f ∈ I, there is n > 0 so that f n ∈ I. It is easy to see that f n ∈ I \ p because p is a prime ideal. (If not, f n ∈ p implies that f ∈ p.) This shows√ that p does not contain I and hence√ p 6∈ V (I). This is equivalent to say that V (I) ⊂ V ( I). We conclude that V (I) = V ( I).

(4). If I is the unit ideal of A, V (I) = ∅ by definition2. Conversely assume that V (I) = ∅. Suppose I is a proper ideal of A, then I is contained in a maximal ideal M of A. Then V (I) contains V (M). Since M is a maximal ideal of A, M is a prime ideal. Thus V (M) contains M and hence is nonempty. This forces V (I) to be nonempty. Therefore I must be the unit ideal.

1hT i is the smallest ideal containing T. 2A prime ideal of a ring is a proper ideal. There is no proper ideal containing the whole ring. 1 2 √ √ √ √ (5). Suppose I =√ J, then V (I) = V ( I) = V√( J) = V (J). Conversely, assume that T T V (I) = V (J). Then I = p∈V (I) p = p∈V (J) p = J. p √ √ (6). Suppose D(f) = ∅. Then V (f) = Spec A. Hence (f) = √A. Notice that A = Nil(A) is the nilpotent radical of A. Since (f) ⊂ p(f) and p(f) = A, we find f ∈ Nil(A), i.e. f is nilpotent. Conversely, if f is nilpotent, then f n = 0 for some n > 0. Then f n ∈ p for all prime p. Hence f ∈ p for all prime p. This shows that p ∈ V (f) for all p ∈ Spec A. Then V (f) = Spec A and hence D(f) = ∅.

(7) Suppose D(f) = Spec A. Then V (f) = ∅. Then (f) = (1). Therefore there is g ∈ A so that fg = 1. Hence f is a unit.

(8) Suppose p ∈ V (I) ∪ V (J). Either p contains I or J. Suppose p contains I. Then p contains I ∩ J. Hence p ∈ V (I ∩ J). Conversely, suppose p contains I ∩ J. Since IJ is contained in both I and J, IJ ⊂ I ∩ J. Hence p contains IJ. Since p is a prime, either I or J is contained in p. Hence p ∈ V (I) ∪ V (J).

T S (9) p ∈ V (Ia), if and only if p contains all Ia if and only if p contains Ia if and aS a only if p ∈ V ( a Ia).

(10) Since V (fg) = V (f) ∪ V (g),D(fg) = D(f) ∩ D(g).  We say that U is a Zariski open subset of A if there exists a set T so that U = Spec A \ V (T ). Using the above proposition, we obtain that Corollary 1.1. The family of Zariski open subsets of Spec A forms a topology. The topology defined above is called the on Spec A. The open sets D(f) are called standard open sets. Proposition 1.3. Every ϕ : A → A0 induces a continuous map Spec(ϕ) : Spec A0 → Spec A sending p0 to ϕ−1(p0). In fact, Spec(ϕ)−1D(f) = D(ϕ(f)). Proof. Suppose that T is any subset of A. Then Spec(ϕ)−1V (T ) is the set of all prime ideals p0 in A0 so that ϕ−1(p0) ∈ V (T ). Then p0 contains ϕ(T ) for all p0 ∈ Spec(ϕ)−1V (T ). Hence p0 ∈ V (ϕ(T )) for all p ∈ Spec(ϕ)−1V (T ). In other words, Spec(ϕ)−1V (T ) ⊂ V (ϕ(T )). Conversely, if p0 ∈ V (ϕ(T )), then p0 contains ϕ(T ) and hence ϕ−1p0 contains T. This implies that ϕ−1p0 ∈ V (T ) for any p0 ∈ V (ϕ(T )). Hence V (ϕ(T )) ⊂ Spec(ϕ)−1V (T ). We conclude that Spec(ϕ)−1V (T ) = V (ϕ(T )).  One can check that if ϕ : A → A0 and ϕ0 : A0 → A00 are ring homomorphisms, then Spec(ϕ0 ◦ ϕ) = Spec ϕ ◦ Spec ϕ0. We conclude that: Corollary 1.2. Spec is a contravariant from the to the category of topological spaces.