# 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.

# Commutative Algebra, Notes 4: Graduations and Filtrations

Definition 1.  Let $M$ be a monoid.  An $M$graded ring is a ring $R$ where

$\displaystyle R=\bigoplus_{x\in M}A_x$

and $A_xA_y\subseteq A_{xy}$ where $A_\alpha$ are abelian groups.  A graded ring will mean an $\mathbb{N}$-graded ring.  $A$ is an $M$graded $R$algebra if it is $M$-graded as a ring.  If $A$ is a graded $R$-algebra and $R$ is graded, then we say $A$ is graded with respect to $R$ if $R_iA_j\subseteq A_{i+j}$ and $A_iR_j\subseteq A_{i+j}.$

Example 2.  The tensor algebra $T(M)$ of a module is trivially graded with its concatenation product.  A group ring $R[G]$ is a $G$-graded ring, which can be seen with its decomposition $R[G]=\oplus_{g\in G}Rg.$  A $\mathbb{Z}_2$-graded ring (algebra) is called a super ring (super algebra).  Note that $\mathbb{Z}$-graded rings and graded rings induce a $\mathbb{Z}_2$ grading as follows

$\displaystyle R=R_0\oplus R_1=\bigoplus_{2n}R_n\oplus\bigoplus_{2n+1}R_n.$

Note we could slightly generalize the above definition of $A$ being graded with respect to $R$ as follows:  suppose $R$ is $M$-graded and $A$ is $N$-graded for monoids $M,N$ and that there exists a monoid homomorphism $\varphi:M\to N.$  Then we can say that $A$ is an $(M,N)$graded $R$algebra if

$\displaystyle R_mA_n\subseteq A_{\varphi(m)+n}$

and

$\displaystyle A_nR_m\subseteq A_{n+\varphi(m)}.$

A similar definition exists for an $(M,N)$-graded $R$-module.  Note an $N$-graded $R$-algebra has an $(M,N)$-grading where $M$ is the trvial monoid grading $R$ and $\varphi$ is of course the trivial morphism.  An element of an $(M,N)$-graded $R$-algebra is called homogeneous (of order $n$) if it has the form $(\cdots,0,a_n,0,\cdots)$ for some $n\in N.$  Let $A,B$ be $(M,N)$-graded $R$-algebras and $f:A\to B$ be an $R$-algebra homomorphism.  Then $f$ is called a graded homomorphism if $f(A_n)\subseteq B_n.$  In more generality, we could have $B$ be an $(M,N')$-graded $R$-algebra together with a module homomorphism $\varphi:N\to N'$ and require $f(A_n)\subseteq B_{\varphi(n)}.$

Definition 3.  A subset $S\subseteq A$ is homogeneous if for every element of $S,$ the component elements (which are homogeneous) are in $S.$  If $S$ is an ideal, then we call it a homogeneous (or graded) ideal.  Note it has a grading as it can be written as a direct sum of the ideals generated by the homogeneous elements.

Proposition 4.  Let $A$ be an $(M,N)$-graded $R$-algebra and $\mathfrak{a}$ be a homogeneous ideal in $A.$  Then $A/\mathfrak{a}$ is an $(M,N)$-graded $R$-algebra.

Proof.  Since $\mathfrak{a}$ is homogeneous, it is graded (since elements give component elements).  So we have

$\displaystyle\mathfrak{a}=\bigoplus_{n\in N}\mathfrak{a_n}=\bigoplus_{n\in N}(\mathfrak{a}\cap A_n)$

Cosets of $\mathfrak{a}$ in the quotient thus have the form

$\displaystyle A+\mathfrak{a}=\bigoplus_{n\in N}A_n+\bigoplus\mathfrak{a}_n=\bigoplus_{n\in N}\left(A_n+\mathfrak{a}_n\right),$

which gives us a decomposition.  Hence

$R_m(A/\mathfrak{a})_n=R_m(A_n/\mathfrak{a}_n)\subseteq A_{\varphi(m)+n}/\mathfrak{a}_{\varphi(m)+n}=(A/\mathfrak{a})_{\varphi(m)+n}.$

Filtrations:

Definition 5.  Let $(P,\leq)$ be a poset.  A subset $F$ is called a filter if the following hold

1. $x,y\in F\Rightarrow\exists z\in F$ with $z\leq x$ and $z\leq y.$
2. If $x\in F,$ $y\in P,$ and $x\leq y,$ then $y\in F.$

Definition 6.  Let $A$ be a structure and $F$ be a filter.  An descending (ascending) filtration on $A$ with respect to $F$ is a collection $\{A_x\}_{x\in F}$ of substructures of $A$ such that $x\leq y\Rightarrow A_x\supseteq A_y$ ($A_x\subseteq A_y$).  A filtration on $A$ will mean a filtration with respect to the filter $(\mathbb{N},\leq)$ (either ascending/descending).  We can similarly define filtrations of modules and filtrations of modules that respect the filtration of their ring.

Definition 7.  If $\mathfrak{a}$ is an ideal in $R$ and $E$ is an $R$-module with a descending filtration, then the filtration is called an $\mathfrak{a}$filtration if $\mathfrak{a}E_n\subseteq E_{n+1}$ for all $n.$ It is called $\mathfrak{a}$stable if $\mathfrak{a}E_n=E_{n+1}$ for all $n\geq m$ for some $m.$

Hence we can view multiplication by $\mathfrak{a}$ as increasing/decreasing the degree (depending upon preferred terminology) of elements in $E_n.$

Let $\mathfrak{a}$ be an ideal of $R.$  Then $R$ has an $\mathfrak{a}$-filtration:

$R=\mathfrak{a}^0R\supseteq\mathfrak{a}R\supseteq\cdots\supseteq\mathfrak{a}^nR\supseteq\cdots.$

We can define the Rees algebra (which Lang calls the “first associated graded ring”) as

$\displaystyle R[\mathfrak{a}t]=\bigoplus_{n=0}^\infty\mathfrak{a}^nt^n.$

This is clearly a graded $R$-algebra.  We could also consider

$\displaystyle gr_{\mathfrak{a}}(R)=\bigoplus_{n=0}^\infty\mathfrak{a}^n/\mathfrak{a}^{n+1}.$

This is easily verified as a graded $R$-algebra with a product defined componentwise:

$(ab)_{m+n}=a_mb_n.$

Definition 8.  Let $E$ be a graded $R$-module with grading $E=\oplus_n E_n.$  We define the Hilbert polynomial by $H_E(n)=\dim_R(E_n).$  We define the $\textbf{Poincar\'{e} series}$ of $E$ as

$\displaystyle P_E(t)=\sum_{n=0}^\infty H_E(n)t^n.$

[1] Lang, Serge.  Algebra.  Revised Third Edition.  Springer-Verlag.  2000.

[2] Dummit, David and Richard Foote.  Abstract Algebra.  Third Edition.  John Wiley and Sons, Inc.  2004.