jategaonkar’s ring and its injective modules

Jategaonkar’s Ring and its Injective Modules

T. G. Kucerahttp://server.math.umanitoba.ca/~tkucera/

Department of Mathematics, University of ManitobaWinnipeg, Manitoba, Canada R3T 2N2

The University of Manchester, UKJanuary 25, 2018

T. Kucera (UofM) Jategaonkar’s Ring and its Injective Modules January 2018 1 / 59

Jategaonkar’s ring

Sources:A. V. Jategaonkar,

Left principal ideal domains,J. Algebra 8, 148–155 (1968).

A counter-example in ring theory and homological algebra,J. Algebra 12, 418–440 (1969),.

My stuffhttp://server.math.umanitoba.ca/~tkucera/


Outline

1 Background and motivation

2 Construction of the rings

3 A brief overview of injective modules

4 Construction of the indecomposable injective modules


Background and motivation


Counter-examples and examples


Counter-examples and examples


A classic example of a structure theorem

Source:E. Matlis,Injective Modules over Noetherian Rings Pac. J. Math 8, 511–528(1958).

Let R be a commutative Noetherian ring.

Let E an indecomposable injective R-module.


Matlis’s Theorems (i)

Theorem1 E is the injective envelope of R/P for some prime ideal P .2 E is an indecomposable injective RP-module,

and as such E = E(RP/PRP) .3 Let An = { e 2 E : Pne = 0 } . Then:

An E , An An+1, E =[

n2w

An

4 { x 2 E/An : Px = 0 } = An+1/An5 (An)n2w is the socle series of E as an RP-module.


Matlis’s Theorems (ii)

Theorem6 Let K be the quotient field of R/P . Then:

A1 ⇠= K , andAn+1/An is a finite dimensional vector space over K .

7 End(E) ⇠= RP , and E is indecomposable injective as anRP-module with the same hierarchy (An)n2w .

8 P(n) denotes the n-th symbolic prime power of P , the P-primarycomponent of Pn . Set t to be the least integer such that in anyrepresentation of P(n+1) as an irredundant intersection ofirreducible P-primary ideals, there are t components notcontaining P(n) .Then the dimension of An+1/An over K is t .


Matlis’s Theorems (addendum)

Theorem9 socn(E) = An


Indecomposable injective modulesover a non-commutative noetherian ring


Construction of the rings


Construction of the rings — Overview

Sections 2 and 3 of the paper are eleven tightly packed technicalpages to develop the theory necessary, and the “bootstrap”construction that produces the desired example.

Ore extensionsIterated Ore extensionsThe “bootstrap” construction of the ringProperties of the ring


Ore extensions

Let D be a domain (no proper zero divisors, not necessarilycommutative), x an indeterminate, and r : D ! D a monomorphism.Then there is a unique ring R = D[x , r] extending D such that:

1 R is generated by D [ { x } as a ring;2 every r 2 R has a unique expression in the form r = Âi<n dix i for

some n 2 N and elements di 2 D ;3 and for all d 2 D , xd = r(d)x .

Reference:Ore, O. Theory of non-commutative polynomials, Ann. Math. 34(1933) 480–508.


Properties of Ore extensions (1)

DefinitionAn left Ore set in D is a multiplicatively closed set M ⇢ D containing 1but not 0 , and such that for all a 2 M , 0 6= d 2 D there are a0 2 M ,d 0 2 D , such that a0d = d 0a .

DefinitionD⇤ = D \ { 0 } .A left Ore domain is a domain D in which D⇤ is a left Ore set.



TheoremIf M is a left Ore set in a domain D then there is a domain DMextending D such that every element of M is a unit of DM and everyelement of DM can be expressed [in at least one way ] as a�1d witha 2 M and d 2 D

If D is a left Ore domain, then DD⇤ is the left quotient skew field of D .



PropositionLet R = D[x , r] be a domain.Let M ⇢ D be a left Ore set in D such that r[M ] ⇢ M .Then M is a left Ore set in R .

Theorem1 If D is a left Ore domain, so is D[x , r] .2 D[x , r] is a left principal ideal domain

iffD ia left principal ideal domainandr[D⇤] is contained in the units of D .


Iterated Ore extensions

Let D be a subdomain of a domain R .

R is an iterated Ore extension of D if:1 there is an ascending chain (Rb)ba of intermediate domains,2 with R0 = D and for 0 < b a , Rb =

Sg<b Rg ,

and3 Rb = Rb[xb, rb] .4 R = Ra .

We write R = D[xb, rb : b < a] .


Iterated Ore extensions—standard form (1)

A monic standard monomial is a term q = xn1b1

· · · xnkbk

withb1 < · · · < bk < a and ni a natural number.

We allow k = 0 , in which case q = 1 , and otherwise we insist that atleast one ni > 0 .

Let Q be the set of all monic standard monomials.



There is a natural ordering of Q in order type wa as follows:

1 is the least element;otherwise set q1 � q2 (where q1 = xn1

b1· · · xnk

bkand q2 = xm1

b1· · · xmk

bk)

iff

nl = ml for t < l k , and nt < mt .

1 < x0 < · · · < x0n < · · · <

< x1 < x0x1 < · · · < x0nx1 < · · · <

< x12 < · · · < x1

n < · · · << x2 < x0x2 < · · · < x0

nx2 < · · · << x1x2 < · · · (etc.)



Every element of R can be written uniquely as finite sum

r = Âi<n

di qi

where q0 � q1 � · · · , and di 2 D .

Heart of the induction:Take xbt 2 Rb , so t 2 Rb .This is equal to r(t)xb with r(t) 2 Rg , some g < b .


Iterated Ore extensions—properties

PropositionLet R = D[xb, rb : b < a] be a domain.Let M ⇢ D be a left Ore set in D such that rb[M ] ⇢ M for all b < a .Then M is a left Ore set in R = D[xb, rb : b < a] .

Theorem1 If D is a left Ore domain, so is R = D[xb, rb : b < a] .2 R = D[xb, rb : b < a] is a left principal ideal domain

iffD is a left principal ideal domainandrb[Rb

⇤] is contained in the units of D for all b < a .


The “bootstrap” construction of the ring

Theorem[Jategaonkar (1969), Theorem 3.1]Let k be a skew field and a > 0 an ordinal.There is

a skew field K extending kand an iterated Ore extension Ra = K [xb, rb : b < a]

such that for all b < a , rb[Rb] ✓ K .

Furthermore, if |k | @0 then Ra may be chosen so that |Ra| @0 |a| .

Ra and K are constructed together by successive twistings andlocalizations along the sequence of Rb .


The localization

Let S ⇢ R consist of all those elements whose standard form has anon-zero constant term.

TheoremS is a left Ore set in Ra .Every element of a 2 Ra can be written in the form a = sq whereq 2 Q .

I write Ra for RaS .


Properties of the ring Ra

Theorem1 Ra is a principal left ideal domain.2 For each 1 6= q 2 Q , Raq is a non-zero proper two-sided ideal of

Ra .3 The elements 1 + xb , b < a , are right linearly independent in Ra .

Hence Ra does not have right Krull dimension.


Properties of the ring Ra

Theorem1 Every left ideal of Ra is a two-sided ideal and has the form Raq ,

q 2 Q . Ra is left FBN.2 The Jacobson radical of Ra is Rax0 .3 The ordering � on Q induces a reverse well ordering of the left

ideals [two-sided ideals] of order type wa + 1 .4 Jwb

= Raxb , and these are the non-zero prime ideals of Ra .5 Jwa

= 06 xb , 0 < b < a has no factorization into irreducibles.7 etc., etc.

xb =�rb(xn

g)��1 xbxn

g for all g < b and for all n > 0 .


A brief overview of injective modules


Systems of linear equations (1)

DefinitionA finite system Av + b = 0 of linear equations is (formally ) consistentiff for all compatible row-vectors r from R ,

rA = 0 implies that r · b = 0

An arbitrary system of linear equations is (formally ) consistent iff everyfinite subsystem is consistent.



DefinitionLet S be a system of linear equations over M in variables { vi : i 2 I } .

Let N � M .A solution of S in N is

– an assignment of values n : I �! N– such that N |= S[n] ,– that is,

for any equation Âi2I0 rivi + b = 0 in S , N |= Âi2I0 rini + b = 0 .

S is satisfiable if for some N � M , S has a solution in N .

S is satisfiable in M if S has a solution in M .



TheoremA system of linear equations over M is consistent

iffit is satisfiable.


Injectivity

TheoremThe following are equivalent:

1 RE is an injective module.2 Every consistent system of linear equations over RE

is satisfiable in E .3 Every consistent system of linear equations over RE

in one free variableis satisfiable in E .(Baer’s Criterion)


Injective envelopes

Every module RM has a unique-up-to-isomorphism over M

injective envelope RE = E(M) ,

characterized variously as follows:

1 A minimal injective extension of M ,2 A maximal essential extension of M ,3 An injective essential extension of M .

M ⇢ E is essential if for all M 0 ✓ E , M 0 \ M 6= 0 ;equivalently if for all 0 6= a 2 E , there is r 2 R such that 0 6= ra 2 M .


Indecomposable injective modules

“indecomposable” means “direct sum indecomposable”

A left ideal I is irreducible if I = I0 \ I1 implies I = I0 or I = I1 .

A left ideal I is critical in RM if it is maximal amongst the annihilators ofnon-zero elements of M .

TheoremRE is an indecomposable injective left R-module

iffE = ER(R/I) for some irreducible right ideal of R .

Without loss of generality, I may be assumed to be critical in E .


Injective modules over a left noetherian ring

TheoremEvery injective left module over a left noetherian ring can be writtenessentially uniquely as a direct sum of indecomposable [injective]modules.

more details. . .My notes on the systems-of-linear-equations approach to injectivity,including the abstract construction of the injective envelope, areavailable on my web-site.


The Macaulay–Northcott construction

Let K be a field, R = K [x1, . . . , xt ] the usual polynomial ring.Let P = hhx1, . . . , xt ii , a maximal prime ideal.

Then E = E(R/P) can be described as follows:

The underlying K -vector space is represented asE = K [X1

�1, . . . ,Xt�1] ,

the “inverse polynomials” in indeterminates X1�1, . . . ,Xt

�1 .

The action of R on E is determined by

(ax1m1 · · · xt

mt )�bX1

�n1 · · ·Xt�nt

�

=

⇢(ab)X1

m1�n1 · · ·Xtmt�nt if each mi � ni 0

0 otherwise


Macaulay–Northcott: sources

Macaulay, F. S.,The algebraic theory of modular systems,Cambridge Tracts 19, 1916.

Northcott, D. G.,Injective Envelopes and Inverse Polynomials,J. London Math. Soc. (2) 8, 290–296, 1974.

Kucera, T. G.,Explicit descriptions of injective envelopes: generalizations of a resultof Northcott,Communications in Algebra 17(11), 2703–2715, 1989.


Construction of the indecomposable injective modules

Kucera, T. G.,Explicit descriptions of the indecomposable injective modules overJategaonkar’s rings,Communications in Algebra, 30(12), 6023–6054, 2002.


Construction of the indecomposable injectives —Overview

More properties of the ring Ra

Inverse polynomials and E0

Eb , 0 < b a .


More properties of Ra

Ra is a principal left ideal domain, so for left modules,divisible implies injective.

Ra is left FBN, so the indecomposable injective left Ra-modulesare in one-to-one correspondence with the prime ideals.

So the indecomposable injective left Ra-modules are precisely themodules Eb = E(Ra/Raxb) , b < a , and Ea = E(R) .


degree and order

We have already noted the well-order � on Q .Thus each q has a well-defined

degreewhich is an ordinal less than wa .

So every element of Ra (and hence of Ra) has a well-defined degree,the maximum of the degrees of its terms.

For any a 2 K and q 2 Q , the order of aq is �1 if q = 1 , otherwise it isthe largest index of a variable occurring in q .

For q = xn1b1

· · · xnkbk

, define

rq = rn1b1� · · · � rnk

bk.


degree and order lemmas

1 The successor of q in the order � is x0q .

2 Let 0 6= r , s 2 Ra .Then deg(rs) = deg(r ) + deg(s) (ordinal addition!)

3 Let q0, q1 2 Q .Then ord(q0q1) = max { ord(q0), ord(q1) } .

4 Let q0, q1 2 Q .If ord(q0) < ord(q) then qq0 = rq(q0)q .


Inverse polynomials and E0

So what should E0 = E(Ra/J) look like?


What should E0 = E(Ra/J) look like?

E0 is an essential extension of Ra/J ⇠= K .E0 will be a vector space over K , so maybe we can find a basis.Since K is a division ring,for every element 0 6= e 2 E0 we can find an equation re = 1 2 K .Since Ra is a PLID, it is enough to show that E0 is divisible:every equation rv = e , 0 6= r 2 Ra , e 2 E0 , has a solution in E0 .Taking a wild leap of faith:

we introduce formal monomials q�1 , q 2 Q , to build E :

if q = xn1b1

· · · xnkbk

, then q�1 = X�nkbk

· · ·X�n1b1

,

we try taking the element q�1 , as a “canonical solution” to theequation qv = 1 .


Properties of canonical solutions (1)

What should the space of all solutions to all equations qv = a , a 2 K ,q fixed, look like?

q . q�1 = 1q . bq�1 = a ???

(qb) . q�1 = a ???rq(b)q . q�1 = a ???

rq(b) = a ???

rq is not a surjection. . .

After 20 years I no longer remember how I figured this out:


The underlying K -vector space of E0

rq is an infinite dimensional embedding of K into itself.

Let Kq denote the K -vector space with underlying set K andK -scalar operation defined by a . c = rq(a)c .

Let E0 be the left K -vector spaceL

q2Q Kqq�1 .


Properties of canonical solutions (2)

LemmaLet q, h 2 Q and a 2 Ra/J ⇠= K .Let e be any solution in E(Ra/J) to q . v = a .

Then:1 h . a = ha if deg(h) = 0 and h . a = 0 if deg(h) > 0 .2 If ord(h) < ord(q) , then h . e is a solution to q . v = rq(h)a3 If ord(h) = ord(q) , say h = hxg

m , q = qxgn , and m n , then

h . e = h . e , where e is a solution to qxgn�m . v = a .

4 If ord(h) = ord(q) as in (3) and m > n or if ord(h) > ord(q) thenh . e = 0 .

5 If ord(h) = ord(q) and m < n as in (3) then h . e is a solution toqxg

n�m . v = rq(rn�mg (h))a .


E0 as an Ra-module

Definition

h . aq�1 =8<

:

rq(h)aq�1 ord(h) = �1 or ord(h) < ord(q) ,h . aX m�n

g q�1 ord(h) = ord(q) , h = hxmg and q = qxn

g ,m n ,0 ord(h) = ord(q) ,m > n , or ord(h) > ord(q) .

Extend this to an action of Ra on E0 by distributivity.

CorollaryIn the second case of the definition, if ord(h) = ord(q) and m < n then

h . aq�1 = rq

�rn�m

g (h)�aX m�n

g q�1 .


It’s a module!

LemmaThe action of Ra on E0 is associative, and hence makes E0 into a leftRa-module.

LemmaE0 is an essential extension of Ra/J .


Is it divisible?

Yes! Seven lemmas showing how to divide in E0 .

Lemma (Lemma 4.6)Scalar multiplication cannot increase the degree. More precisely, leth, q 2 Q . Then

1 deg(h . q�1) deg(q�1) .2 If ord(h) = �1 or ord(h) < ord(q) then deg(h . q�1) = deg(q�1) .3 If ord(h) � 0 and ord(h) � ord(q) then deg(h . q�1) < deg(q�1) .4 If deg(h) > deg(q) , then h . q�1 = 0 .


Six more lemmas. . .

Lemma (Lemma 4.7)For any s 2 S the unique solution to s.v = 0 in E0 is v = 0 .

Lemma (Lemma 4.8)If 0 6= e with a0q�1

0 being the term of e of largest degree, thenann(e) = hhx0q0ii .

Lemma (Lemma 4.9)For h 2 Q , the solutions in E0 to the equation h . v = 0 are exactly theelements of Âq<h Kqq�1 , that is, the elements of lesser degree than h .


Three more lemmas. . .

Lemma (Lemma 4.10)To solve all equations of the form r . v = e ( r 2 Ra , e 2 E0 ) in E0 itsuffices to be able to solve all equations of the forms s . v = aq�1 andh . v = aq�1 , where s 2 S , q 2 Q , h 2 Q , and the situations a = 0 ,q = 1 are allowed.

Lemma (Lemma 4.11)Every equation of the form h . v = aq�1 , h 2 Q , has a solution in E0 ;in particular a canonical solution (when a 6= 0 )is found as follows:Write aq�1 = aX�n

g q�1 and split up the standard form of h ash = h0xm

g h1 (with any of h0 = 1 , m = 0 and h1 = 1 allowed). Thenv = [rq(h0)]�1ah�1

1 X�n�mg q�1 is a solution.


The last lemma, and a theorem

Lemma (Lemma 4.12)Every equation of the form s . v = aq�1 , s 2 S , has a solution in E0,and the degree of the solution is deg(q) .

Theorem (Theorem 4.15)E0 is the injective envelope of Ra/J .


Exercise

Exercise 4.16Find all solutions to the following system of linear equations overE(Ra/J) .

x0 . u + x1 . v = X�11

u + (x0 + x1) . v = 1 + X�11


All the other indecomposable injective modules

TheoremThe injective envelope Ea of Ra is the quotient skew field of Ra .


Intermediate pieces

Recall that S ⇢ Ra is the left Ore set consisting of all those elementswhose standard form has a non-zero constant term, and Ra = RS .

Let Sb = S \ Rb . This is an Ore set in Rb .Then Ra/Raxb

⇠=�Rb

�Sb

as a left Ra-module.

The scalar action of Ra on�Rb

�Sb

is just:“ordinary multiplication in Ra ,followed by setting to 0 any indeterminate xg , g � b .”

As a ring, each Rb is a left Ore domain, so we may consider�Rb

�Sb

to

be embedded in the left quotient division ring K b of Rb .


Eb , 0 < b < a (1)

DefinitionLet Qb = { q 2 Q : no xg , g < b occurs in q } (and so 1 2 Qb ).For q 2 Qb , let K b

q be the K b-vector space on K b with scalar action

t�1r . k = [rq(t)]�1rq(r )k , for 0 6= t , r 2 Rb .

Let Eb =L

q2QbK b

q q�1 .

Define an action of Ra on Eb as before:

h . aq�1 =8<

:

rq(h)aq�1 ord(h) < ord(q) ,h . aX m�n

b q�1 ord(h) = ord(q) , h = hxmg and q = qxn

g ,m n ,

0 ord(h) = ord(q) ,m > n , or ord(h) > ord(q) .


Eb , 0 < b < a (2)

TheoremEb is the injective envelope of Ra/Raxb .


Complexity of the indecomposable injectives (1)

Proposition(socle series)

1 socn(E0) = Âm<n Kxm0

X�m0 .

2 For all g � w , socg(E0) = Âm<w Kxm0

X�m0 .

Proposition(elementary socle series)socd(E0) = Âq,deg(q)<d Kqq�1 for d wa .In particular, for g < a , socwg

(E0) = Âq,q�xgKqq�1 .


Complexity of the indecomposable injectives (2)

TheoremLet 0 < b < a .

For each g , b g < a ,

E0/socwb(E0)

has an infinite direct sum of copies of Eg

as a direct summand.


jategaonkar’s ring and its injective modules

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