Category Archives: IS: Commutative Algebra

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


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


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<y 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<g 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




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


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


 [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


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 Mgraded 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 Mgraded Ralgebra 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 Ralgebra if

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


\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}.


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.

Induced Graduations:

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


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:


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.

Commutative Algebra, Notes 2: Localization

We will mostly adopt the Bourbaki definitions for the preliminary material.  An ideal \mathfrak{p} in a ring R is prime if A/\mathfrak{p} is an integral domain.  Two ideals \mathfrak{a},\mathfrak{b} are relatively prime if \mathfrak{a}+\mathfrak{b}=R.  A multiplicative subset of a set X is a submagma S of X.  In a ring, we assume it is a submagma with respect to multiplication of course.  We then have the canonical extension of R by formal inverses of S written S^{-1}R:=R[S^{-1}] (although I will use the notation R[S^{-1}] to reinforce the conceptualization of a ring extension).

Now suppose M is an R-module.  A natural question to ask is “can we induce some R[S^{-1}]-module”?  The natural choice will be the R[S^{-1}]-module M\otimes_R R[S^{-1}] where we define

r(m\otimes r')=m\otimes rr'

for r,r'\in R[S^{-1}].

Proposition 1.  Let R be a ring and S be a multiplicative subset.  Then there exists a ring \hat{R} and homomorphism f:R\to\hat{R} such that

  1. Elements of f(S) are invertible in \hat{R}.
  2. For every homomorphism g:R\to R' such that elements in g(S) are invertible in R', there exists a unique homomorphism h:\hat{R}\to R' such that g=h\circ f.

It naturally turns out that \hat{R}=R[S^{-1}] and f(r)=rs/s for some s\in S.  Similarly we have:

Proposition 2.  Let R be a ring, S be a multiplicative subset, and M be an R-module.  Then there exists an R[S^{-1}]-module \hat{M}  and homomorphism f:M\to\hat{M} such that

  1. For every s\in S, the map s:\hat{M}\to\hat{M} defined by \hat{m}\mapsto s\hat{m} is bijective.
  2. For every R-module N such that the map n\mapsto sn is bijective for every s\in S and if we have a homomorphism g:M\to N, there exists a unique homomorphism h:\hat{M}\to M such that g=h\circ f.

Here we can show that \hat{M}=M\otimes_R R[S^{-1}] and f(m)=m\otimes 1.  R[S^{-1}] is called the ring of fractions of R by S, while M\otimes_R R[S^{-1}], often denoted S^{-1}M for short, is called the module of fractions of M by S.

Definition 3.  Let \mathfrak{a} be an ideal in R.  We define the radical of \mathfrak{a} in R as

\sqrt[R]{\mathfrak{a}}=\sqrt{\mathfrak{a}}=\{r\in R:r^n\in\mathfrak{a}\mbox{~for some~}n\in\mathbb{N}\}.

We call \sqrt[R]{0}=\sqrt{0} the nilradical of R.  R is a reduced ring if \sqrt{0}=0.

Let A be an R-algebra and S be a multiplicative subset of R.  Then since A is an R-module, we can make A into an R[S^{-1}]-module

S^{-1}A=A\otimes_R R[S^{-1}].

But we can also turn it into an R[S^{-1}]-algebra.  We need a homomorphism

f:\left(A\otimes_R R[S^{-1}]\right)\otimes_{R[S^{-1}]}\left(A\otimes_R R[S^{-1}]\right)\to A\otimes_R R[S^{-1}].

But the domain collapses to (A\otimes_RA)\otimes_RR[S^{-1}] and then to A\otimes_RR[S^{-1}] under the product of A.  So we have the canonical product we need.

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