# Commutative Algebra, Notes 6: Valuations

Definition 1.  Let $F$ be a field.  An absolute value on $F$ is a map $|\cdot|:F\to\mathbb{R}$ such that for all $x,y\in F$

1. $|x|\geq 0$ and $|x|=0$ iff $|x|=0.$
2. $|xy|=|x||y|.$
3. $|x+y|\leq|x|+|y|.$

We will call the pair $(F,|\cdot|)$ an absolute field.  The absolute value that sends all nonzero elements to $1$ is called the trivial absolute value.  One can also verify that the function $d(x,y)=|x-y|$ defines a metric on $F .$  Two absolute values are dependent if their induced topologies are the same, and independent otherwise.  Note that

$|1|=|1^2|=|1|^2$ and $|1|=|(-1)^2|=|-1|^2.$

Hence $|1|=1,$ and thus $|-1|=\sqrt{1}=1$ since $|x|\geq 0.$  Similarly

$|-x|=|-1||x|=|x|$

and $1=|1|=|xx^{-1}|=|x||x^{-1}|\Rightarrow |x^{-1}|=|x|^{-1}.$

Proposition 2.  Let $|\cdot|_1$ and $|\cdot|_2$ be two nontrivial absolute values on $F .$  Then $|\cdot|_1$ and $|\cdot|_2$ are dependent iff $|x|_1<1\Rightarrow |x|_2<1.$  If they are dependent, then there exist some $\lambda>0$ such that $|\cdot|_1=|\cdot|_2^\lambda.$

Theorem 3.  (Approximation Theorem) (Artin-Whaples).  Let $F$ be a field and $|\cdot|_1,...,|\cdot|_n$ be nontrivial pairwise independent absolute values on $F .$  Then there exist elements $x_1,...,x_n\in F$ such that for any $\varepsilon>0,$ there exists an $x\in F$ such that

$|x-x_i|_i<\varepsilon$

for all $1\leq i\leq n.$

An absolute field is complete if all of its Cauchy sequences converge to a point in the field.  One can show that each absolute field has a unique completion field up to isometry.

Definition 4.  An abelian group $G$ is an ordered group if it has a partial ordering such that $x\leq y\Rightarrow xg\leq yg$ for all $x,y,g\in G.$

Proposition 5.  $(G,\leq)$ is an ordered group iff $G=\{1\}\sqcup S\sqcup S^{-1}$ for a multiplicative subset $S\subset G$ where $x<1 for all $x\in S$ and $y\in S^{-1}.$

For an ordered group $G,$ we will hereafter use $G$ to denote $G[0]$ where $0g=0$ and $0 for all $g\in G.$

Definition 6.  Let $F$ be a field and $G$ be an ordered group.  A valuation on $F$ is a map $|\cdot|:F\to G$ such that

1. $|x|=0$ iff $x=0.$
2. $|xy|=|x||y|.$
3. $|x+y|\leq\max\{|x|,|y|\}.$

We will call the triple $(F,G,|\cdot|)$ a valuation field.  The image of $|\cdot|$ on $F$ is an ordered subgroup of $G.$  Two valuations are equivalent if there is an isomorphism $\lambda:F\to G$ that respects order and value.  One can also verify that $|\pm x|=|x|$ and that if $|x|<|y|,$ then $|x+y|=|y|.$

Definition 7.  A subring $R$ of a field $F$ is a valuation ring if for all $x\in F,$ $x\in R$ or $x^{-1}\in R.$

Proposition 8.  If $R\leq F$ is a valuation ring, then it induces a valuation field.

Proof.  Note that $R$ is a local ring, for, if $x,y$ are not units in $R,$ then if $x/y\in R,$  we have

$1+x/y=(x+y)/y\in R.$

So if $x+y$ is a unit in $R$, then $1/y\in R,$ which is a contradiction.  Hence $x+y$ is not a unit.  Furthermore if $x$ is not a unit, then for all $r\in R$ we have that $rx$ is not a unit.  So the nonunits form an ideal in $R,$ which is necessarily maximal (and uniquely so).  So $R$ is a local ring.  Let us denote this ideal by $\mathfrak{m}.$  Thus we can write

$F^*=\mathfrak{m}^*\sqcup U\sqcup\mathfrak{m}^{*^{-1}}$

where $U$ is the group of units of $R.$  We now give $F$ a valuation.  We will define our group $G$ as the quotient group $F^*/U.$  In turn, we define $|x|=xU$ with $|1|=U:=1.$  When then define $|0|=0,$ $|x|<1$ iff $x\in\mathfrak{m}^*,$ and $|x|>1$ iff $x\in\mathfrak{m}^{*^{-1}}.$  Observe that

$|xy|=xyU=xUyU=|x||y|$

and

$|x+y|=(x+y)U\leq\max\{xU,yU\}=\max\{|x|,|y|\}.$

Hence $(F,F^*/U,|\cdot|)$ is a valuation field.

We thus hereafter refer to a valuation ring as the restriction of the induced valuation field to the ring.

Proposition 9.  Let $F\leq K$ be fields where $F$ is a valuation field.  Then there exists an extension making $K$ a valuation field.

Proof Idea.  Let $R=\{x\in F:|x|\leq 1\}$ and $U=\{x\in F:|x|=1\}.$  We will call $R$ the valuation ring of $F .$  One first takes the morphism $\varphi:R\to R/(R-U)$ and extends it to a valuation ring $D$ in $K.$  One can construct an order preserving monomorphism

$\psi:F^*/U\to K^*/U'$

where $U'$ is the maximal ideal in $D.$  We then define $|x|_K$ as before, which is seen to agree with $|x|_F.$

Proposition 10.  Let $F\leq K$ be fields and $[K:F]=n.$  Let $F$ be a $G$-valued valuation field and $G'$ be the extension group to the induced valuation on $K.$  Then $[G':G]\leq n.$

We will call $[G':G]$ the ramification index of $F$ in $K.$

Definition 11.  A valuation is discrete if its codomain is a cyclic group.

It turns out that if $F$ is a valuation field and $\mathfrak{m}$ is the maximal ideal of its valuation ring, then there exists an element $\pi\in\mathfrak{m}$ such that $|\pi|$ is a generator of $G.$  Such an element is called a local parameter of $|\cdot|.$  One can also show that $\mathfrak{m}$ is a principal ideal generated by $\pi.$  Moreover every element $x\in F$ can be written as

$x=u\pi^r$

for some unit $u\in R$ integer $r.$  $r$ is called the order of $x$ at $|\cdot|.$  If $r>0,$ we say $x$ has a zero of order $r,$ and if $r<0,$ then we say $x$ has a pole of order $r.$

Proposition 12.  Let $F\leq K$ be fields where $K$ is a finite extension of $F .$  Further suppose that $F$ is a complete discrete valuation field (i.e. induced metric space is complete) and that $R,R'$ and $\mathfrak{m},\mathfrak{m}'$ are the corresponding valuation rings and maximal ideals after extending the valuation.  Then

$[K:F]=[G':G][R'/\mathfrak{m}',R/\mathfrak{m}].$

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

# 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.

# Commutative Algebra, Notes 3: Local Rings

Definition 1.  The Jacobson radical of a commutative ring $R$ is defined as the intersection of all maximal ideals of $R.$

Proposition 2.  Let $R$ be a ring and $J$ be the set of all nonunits of $R.$  Then the following are equivalent.

1. $J$ is an ideal.
2. $R$ has a unique maximal ideal.

Proof.  If $J$ is an ideal, then it is maximal since any bigger ideal would contain a unit, and hence equal the whole ring.  Uniqueness follows by adjoining nonunits.  Conversely, note that the product of a unit and a nonunit, as well as the product of two nonunits, must be a nonunit.  If $R$ has a unique maximal ideal, then note that if $x,y\in J$ and $x+y$ is a unit, then there is a unit $u$ such that $ux+uy=1.$  Hence the ideals generated by $x$ and $y$ are relatively prime.  Hence no maximal ideal can contain both $(x)$ and $(y),$ so uniqueness is violated.

Definition 3.  $R$ is a local ring if it satisfies the above conditions.

If $R$ is a local ring, then of course $J$ is the Jacobson radical of $R.$  Note that if $\mathfrak{p}$ is a prime ideal in $R,$ then its complement is a multiplicative subset.

Definition 4.  We define $R_\mathfrak{p}=(R-\mathfrak{p})^{-1}R.$  We may call $R_\mathfrak{p}$ the localization of $R$ at $\mathfrak{p}.$  We define the globalization of $R$ as the ring

$\displaystyle G(R)=\bigoplus_{\mathfrak{m}\subset R}R_\mathfrak{m}$

where $\mathfrak{m}$ is a maximal ideal in $R.$

So if $R$ is a local ring, then its globalization is just $R$ since the one factor $R_J$  would invert already invertible elements.

Proposition 5.  The set $\Omega(R)$ of maximal ideals of $R$ is finite iff $R/J(R)$ is a finite direct sum of fields (where $J(R)$ is the Jacobson radical of $R$).

Proof.  $R/\mathfrak{m}$ is a field for each maximal ideal $\mathfrak{m}$ in $R.$  Moreover if $|\Omega(R)|<\infty,$ then $R$ maps onto $\oplus_{i=1}^nR/\mathfrak{m}_i$ via the canonical mapping.  The kernel is clearly $J(R),$ so we have the quotient as a finite sum of fields.  Conversely, if $R/R(J)$ is a finite direct sum of fields, then it has a finite number of ideals, and hence a finite number of maximal ideals.  Since each maximal ideal in $R$ contains $R(J),$ these are just the preimages of the maximal ideals of $R/R(J)$ under the map $R\to R/R(J),$ of which there are finitely many.

Definition 6.  A  ring $R$ is semilocal if it is a finite direct sum of local rings.

Hence if $|\Omega(R)|<\infty,$ then $R/R(J)$ is semilocal, as fields are local.  Now suppose $R=\oplus_{i=1}^nM_i$ is a semilocal ring.  So each $M_i$ is local, with a maximal ideal $\mathfrak{m}_i.$  We have a canonical epimorphism

$\displaystyle f:R\to\bigoplus_{i=1}^nM_i/\mathfrak{m}_i$

whose kernel is $J(R).$  Hence if $R$ is semilocal, then $R/R(J)$ is semilocal (and in particular semisimple (as an $R$-module), which is the traditional definition of $R$ being semilocal).

Proposition 7.  Let $\Pi$ be a finite set of prime ideals of $R.$  Let us define

$\displaystyle S_\Pi=\bigcap_{\mathfrak{p}_i\in\Pi}(R-\mathfrak{p}_i).$

Then if $R_\Pi=R[S_\Pi^{-1}],$ $R_\Pi/J(R_\Pi)$ is semilocal (i.e. $R_\Pi$ is semilocal in the traditional sense), and $R_\Pi$ has maximal ideals $\mathfrak{q}_i[S_\Pi^{-1}]$ where $\mathfrak{q}_i$ are the maximal elements of $\Pi.$

Proof.  A maximal ideal in $R_\Pi$ must be contained in $\cup_{\mathfrak{p}_i\in\Pi}\mathfrak{p}_i,$ otherwise it contains a unit.  Hence it must be contained in one of the $\mathfrak{p}_i,$ and hence it must be a $\mathfrak{q_i}.$  So $\mathfrak{q}_i[S_\Pi^{-1}]$ are the maximal ideals of $R_\Pi.$  Hence by Proposition 5 $R_\Pi/J(R_\Pi)$ is semilocal.

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