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1. Spectrum of a Ring All the Rings Are Assumed to Be Commutative. Let A

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1. Spectrum of a Ring All the Rings Are Assumed to Be Commutative. Let A 1. Spectrum of a Ring 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 2 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 zero ring. Proof. If A is the zero ring, there is no proper prime ideal of A: Hence Spec A is empty. If A is a nonzero ring, A has a maximal ideal (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 2 A: (1) If T ⊂ S; then V (T ) ⊃ V (S): (2) LetphT i be the ideal generated by T: Wep 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 isp the unitp 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 fIag 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 2 V (T ); then p contains hT i1 Hence p 2 V (hT i): p p (3) The ideal I is containedp in its radical I: Hencep V (I) ⊃ V ( I)p: Suppose a prime ideal p does not lie in V ( I): Then there exists f 2 I n p: Since f 2 I; there is n > 0 so that f n 2 I: It is easy to see that f n 2 I n p because p is a prime ideal. (If not, f n 2 p implies that f 2 p:) This showsp that p does not contain I and hencep p 62 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 p p p p (5). Suppose I =p J; then V (I) = V ( I) = Vp( J) = V (J): Conversely, assume that T T V (I) = V (J): Then I = p2V (I) p = p2V (J) p = J: p p p (6). Suppose D(f) = ;: Then V (f) = Spec A: Hence (f) = pA: Notice that A = Nil(A) is the nilpotent radical of A: Since (f) ⊂ p(f) and p(f) = A; we find f 2 Nil(A); i.e. f is nilpotent. Conversely, if f is nilpotent, then f n = 0 for some n > 0: Then f n 2 p for all prime p: Hence f 2 p for all prime p: This shows that p 2 V (f) for all p 2 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 2 A so that fg = 1: Hence f is a unit. (8) Suppose p 2 V (I) [ V (J): Either p contains I or J: Suppose p contains I: Then p contains I \ J: Hence p 2 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 2 V (I) [ V (J): T S (9) p 2 V (Ia); if and only if p contains all Ia if and only if p contains Ia if and aS a only if p 2 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 n 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 Zariski topology on Spec A: The open sets D(f) are called standard open sets. Proposition 1.3. Every ring homomorphism ' : A ! A0 induces a continuous map Spec(') : Spec A0 ! Spec A sending p0 to '−1(p0): In fact, Spec(')−1D(f) = D('(f)): Proof. Suppose that T is any subset of A. Then Spec(')−1V (T ) is the set of all prime ideals p0 in A0 so that '−1(p0) 2 V (T ): Then p0 contains '(T ) for all p0 2 Spec(')−1V (T ): Hence p0 2 V ('(T )) for all p 2 Spec(')−1V (T ): In other words, Spec(')−1V (T ) ⊂ V ('(T )): Conversely, if p0 2 V ('(T )); then p0 contains '(T ) and hence '−1p0 contains T: This implies that '−1p0 2 V (T ) for any p0 2 V ('(T )): Hence V ('(T )) ⊂ Spec(')−1V (T ): We conclude that Spec(')−1V (T ) = V ('(T )): One can check that if ' : A ! A0 and '0 : A0 ! A00 are ring homomorphisms, then Spec('0 ◦ ') = Spec ' ◦ Spec '0: We conclude that: Corollary 1.2. Spec is a contravariant functor from the category of rings to the category of topological spaces..
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