If A is the zero ring, there is no proper prime ideal of 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 ∈ 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 ∈ 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 ).

(9) If {Ia} is a family of ideals of A, thenT

aV (Ia) = V (S

aIa).

(10) D(f g) = 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 i¹Hence p ∈ V (hT i).

(3) The ideal I is contained in its radical √

I. Hence V (I) ⊃ V (√

I). Suppose a prime ideal p does not lie in V (√

I). Then there exists f ∈√

I \ p. Since f ∈ √

I, there is n > 0 so that fⁿ∈ I. It is easy to see that fⁿ∈ I \ p because p is a prime ideal. (If not, fⁿ ∈ 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 V (I) = V (J ). Then√

I =T

p∈V (I)p=T

p∈V (J)p=√ J .

(6). Suppose D(f ) = ∅. Then V (f ) = Spec A. Hence p(f) = √

A. Notice that √ A = Nil(A) is the nilpotent radical of A. Since (f ) ⊂p(f) andp(f) =√

A, we find f ∈ Nil(A), i.e. f is nilpotent. Conversely, if f is nilpotent, then fⁿ= 0 for some n > 0. Then fⁿ ∈ 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 f g = 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 ).

(9) p ∈ T

aV (I_a), if and only if p contains all I_a if and only if p contains S

aI_a if and only if p ∈ V (S

aI_a).

(10) Since V (f g) = V (f ) ∪ V (g), D(f g) = 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 Zariski topology on Spec A. The open sets D(f ) are called standard open sets.

Proposition 1.3. Every ring homomorphism 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 functor from the category of rings to the category of topological spaces.