Commutative Algebra, Notes 5: Topologies

Topology Induced by Filtration:

Let \mathcal{F}(G)=\{G_n\} be a descending filtration of a group G such that G_n\unlhd G for all n.  Consider the collection of all cosets \{gG_i\} of subgroups in the filtration together with the empty set; we will denote this collection by G/\mathcal{F}(G). The fact that aH\cap bK=c(H\cap K) for subgroups H,K, that G/\mathcal{F}(G) is linearly ordered (which, together with the first fact, gives aG_i\cap bG_j=cG_j (where without loss of generality G_i\geq G_j) and aG_i\cup bG_j=cG_i), and that G is maximal in the collection of cosets (which, with Zorn’s lemma, gives closure under abritrary unions) makes G/\mathcal{F}(G) a topology on G.  In particular, group multiplication can be shown to be continuous with respect to this topology–making (G,G/\mathcal{F}(G)) a topological group.  We will call this the filtrated topology (or topology induced by filtration) of G by \mathcal{F}.

Hence (linearly ordered) filtrations on a structure of “at least group-type” induce a topology.  We can also induce gradings from filtrations of groups with the assumption that elements of the filtration are normal in G.  We then define

\displaystyle G=\bigoplus_{n=0}^\infty G_n/G_{n+1}

in the same way we previously did for ideals in a ring.  As a partition, a grading can also induce a topology, where the open sets are generated by elements of the partition.  For example, in the grading of a group, let open sets be terms in the sum.  Note that unions of terms are terms, the empty set can trivially be considered a term, and finite intersections of terms are terms.

In the case of an ideal \mathfrak{a} of a ring R, the topology induced by the filtration

R=\mathfrak{a}^0\supseteq\mathfrak{a}\supseteq\cdots\supseteq\mathfrak{a}^n\supseteq\cdots

is called the Krull topology (or \mathfrak{a}adic topology) of R by \mathfrak{a}.

Spectral/Zariski Topology:

Let Spec(R) be the set of prime ideals of a unital ring R.  If \mathfrak{a} is an ideal of R, let

V(\mathfrak{a})=\{\mathfrak{p}\supseteq\mathfrak{a}:\mathfrak{p}\in Spec(R)\}.

Note that V(0)=Spec(R), V(1)=\varnothing, \cap_\alpha V(\mathfrak{a}_\alpha)=V(\sum_\alpha\mathfrak{a}_\alpha), and V(\mathfrak{a})\cup V(\mathfrak{b})=V(\mathfrak{a}\mathfrak{b}).  Hence the V(\mathfrak{a})‘s form a basis of closed sets of Spec(R).  The corresponding topology is called the spectral (or Zariski) topology of R, which we also denote by Spec(R).

[1]  Bourbaki, N.  Commutative Algebra, Ch 1-7.  Springer-Verlag.  1989.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: