Last update: 3 June 2013
Let be a commutative ring with 1. Let denote the set of all prime ideals of is a prime ideal is an integral domain; thus itself is not a prime ideal.) If it is sometimes convenient to write for the ideal For each subset of let If consists of a single element we write in place of
Only (iv) is not entirely trivial. Clearly Conversely, if then there exist and such that and since is prime, we have hence
It follows from (3. 1) that the sets satisfy the axioms for closed setsJn a topology on This topology is called the Zariski topology or spectral topology on and it is the only one we shall use.
If is an ideal in the radical of is the set of all such that some power of lies in it is also the intersection of all the prime ideals of which contain In particular, the radical of the zero ideal is the set of all nilpotent elements of this ideal is called the nilradical of
If is a subset of and if is the ideal generated by then
We need some more notation:
local ring of with respect to the prime ideal
maximal ideal of
residue field of field of fractions of
If denotes the class of mod. in Thus if and only if
'support' of it is an open set.
Finally, if denotes Thus Then we have the following formulas:
|(i)||(the nilradical of|
|(iv)||radical of the ideal generated by|
It follows from (iv) and (v) that gives an order-reversing one-one correspondence between closed subsets of and ideals in such that Hence, if the ring is Noetherian, is a Noetherian space. (The converse of this is false: can be Noetherian and not Noetherian. For example, let be a polynomial ring over a field in a countable infinity of indeterminuates, let be the ideal generated by and let Then is not Noetherian but has exactly one prime ideal.)
If then (i.e., is a specialization of if and only if Hence is a closed set (by abuse of language, is a closed point of if and only if is a maximal ideal of Thus is a space (every point is closed) if and only if every ideal of is maximal, i.e., However, is always a (this means that, given any two distinct points in then either there is a neighbourhood of which does not contain or else a neighbourhood of which does not contain
Next, let us look at the open sets First, from (3.1) (iv) we have
|(i)||The open sets form a base of open sets for the topology of|
|(ii)||Each is quasi-compact. In particular is quasi-compact.|
(i) If is an open set in then for some we have (3.1) (iii), hence
(ii) By virtue of (i) it is enough to show that every covering of a set by open sets has a finite subcovering. Suppose then that let be the ideal of generated by the then hence and therefore so that and therefore for some Say where is some finite subset of Then where is the ideal generated by the hence Taking complements, we have as required.
The open sets will be called basic open sets.
Let be an ideal of Then the ideals of correspond one-to-one to the ideals of which contain and therefore is canonically homeomorphic to the closed subspace of In particular, and are canonically homeomorphic nilradical of
Proposition (3.4). is irreducible is an integral domain.
From what has just been said, we may as well take Suppose is reducible; then there exist proper closed subsets in such that and therefore (by (3.2)). But and are hence there exist such that and Hence is not an integral domain.
Conversely, if is not an integral domain we have in such that and Hence (since but Consequently is reducible.
In the correspondence between closed subsets of and ideals of which are equal to their radicals, the irreducible closed subsets correspond to the prime ideals. In particular the irreducible components of correspond to the minimal prime ideals of Furthermore, gives a one-to-one correspondence between the points of and the irreducible closed subsets of i.e., every irreducible closed subset of has exactly one generic point. For if then is irreducible by (2.1) (iv). If then each of and is a specialization of the other, so that i.e. Conversely, if is an irreducible subset of corresponds to a prime ideal of i.e.
Let be a field, an algebraically closed extension of and let be a variety as in Chapter 1; let be the coordinate ring of (a finitely generated with no nilpotent elements), and let What is the relationship between and Let us assume that is a universal domain in the sense of Weil, i.e. that has infinite transcendence degree over this is just to give us plenty of elbow room. Let then determines a homomorphism whose kernel is a prime ideal of i.e. an element of Conversely, if is any prime ideal of we can embed in (for the field of fractions of is a finitely generated field extension of hence is an algebraic extension of a pure transcendental extension of and thus we have a homomorphism with kernel Hence is a map of onto and is obtained from by identifying 'equivalent' points in i.e. points which are generic specializations of each other.
At the other extreme, if then may be identified with the set of maximal ideals of i.e. with the set of closed points of so in this case the map described above is injective (and not in general surjective).
Let be two rings and let be a ring homomorphism (which is always assumed to map identity element to identity element). If then is a prime ideal in hence a point of Thus we have a mapping
said to be associated with Let denote the embedding of in induced by then extends to a field monomorphism
|(i)||for any subset of In particular:|
(i) is straightforward and (ii) follows from (i). To prove (iii) we may assume that since and Put and let then by (3.2) (v). Also:
From (i) or (ii) above it follows that is continuous. Clearly, if is another ring, another ring homomorphism, then so that Spec is a contravariant functor from the, category of rings and ring homomorphisms to the category of topological spaces and continuous maps.
(1) If is an ideal in and the projection, then is a homeomorphism of onto
(2) Let be a multiplicatively closed subset of (i.e. is closed under finite products, so that in particular (take the empty product!)). Then we can form the ring of fractions and we have a canonical mapping hence It is a well-known and not difficult fact of commutative algebra that the prime ideals of are in one-one correspondence (under with the prime ideals of which don't meet and consequently is a homeomorphism of onto the set of all such that (In general this subset of is neither open nor closed, nor even locally closed.)
(3) In particular, may be canonically identified with the subspace of consisting of all generizations of i.e. all such that
(4) As another example, let and let be the set of all In this case is usually denoted by Then is identified with the set of all such that contains no power of i.e. such that Hence
Proposition (3.6). If is the canonical homomorphism then is a homeomorphism of onto the open set
(5) The 'characteristic morphism'. Since has an identity element, there is a canonical mapping where is the ring of integers; hence Now the points of are and the prime ideals a positive prime number), and is just the ideal generated by the characteristic of the residue field of
Proposition (3.7). Let be a ring homomorphism, the associated map.
|(i)||If is surjective, is a closed embedding (i.e. a homeomorphism of onto a closed subset of|
|(ii)||If is injective, is dominant (i.e. is dense in|
(i) is just Example 1 above.
(ii) follows from (3.5) (iii): (since is injective)
This is a typed excerpt of the book "Algebraic Geometry: Introduction to Schemes - I.G. Macdonald".