Zariski Topology

Let R be a ring (commutative, with 1). An ideal p ⊂ R is called prime if p 6 = R and for all xy ∈ p, either x ∈ p or y ∈ p. Let spec(R) denote the set of prime ideals of R. The Zariski topology on spec(R) is defining the sets V (E) = {p ∈ spec(R) : E ⊂ p} for any E ⊂ R to be closed. 1.1 A proof that the collection of V (E) defines a topology Let E ⊂ R and let I be the ideal generated by E. Every ideal containing E contains I, therefore every prime ideal containing E contains I. So V (E) = V (I). Therefore, {V (E) : E ⊂ p} = {V (I) : I is an ideal in p} So, we need only look at sets V (I) where I is an ideal inR. It’s also helpful to observe that if I ⊂ J, then V (J) ⊂ V (I). The empty set and the entire space are closed. By definition, no prime ideal con-taines 1, so V (1) = ∅. Also, since 0 is in every ideal, V (0) = spec(R). The union of two closed sets is closed. For any two ideals I and J, the product IJ is the ideal generated by products xy where x ∈ I and y ∈ J. Note that IJ ⊂ I and IJ ⊂ J, therefore V (I) ∪ V (J) ⊂ V (IJ). Note also that V (IJ) ⊂ V (I) ∪ V (J), since if p is a prime ideal containing IJ, p contains either I or J (otherwise, you’d have an element xy ∈ IJ ⊂ p with x / ∈ p and y / ∈ p). Thus V (I) ∪ V (J) = V (IJ), a closed set. 1 The intersection of closed sets is closed. For any collection of ideals Iα, there is a smallest ideal, denoted by Iα containing all the Iα. (There’s a construction o