# 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.