ok so let us ah just recall that if a is noetherian
respectively artinian and m is finitely generated a module then m is noetherian respectively
artinian is follows directly from the ah on to homomorphism a into m a chain of sub modules
of m is a sequence m naught ah m could m naught contained in m one so here each one is a strict
inclusion of sub modules of m and a composition series
composition series of m is a chain ah such that m i mod m i plus one is simple what do you
mean by simple it has no nontrivial subgroups other than zero and itself so there are no
subgroups sub modules of this module other than zero and itself ok ah ok
so if let me maybe i’ll write this as m minus one m n minus m n equal to zero i mean i am
talking about composition series ah that terminates terminates at zero that means m one is a maximal
sub module of m naught m two is the maximal sub module of m one and so on ok and the length
of this chain is said to be n ah ah let’s look at one or two examples
so if the module itself is a simple module then your composition series now this is a
composition series right z p is a simple z module so this is a composition series
suppose i take z four let say can you give me a composition series for z four
z four z eight z two see when we talk about z two contained in
z four etcetera the ideal generated by two two z four and
z you you should not really say that z two contained in z four now that that has a lot
of scope for confusion ok ah so this is a composition series of z four ah another composition
series of let say z twelve z twelve then six z twelve three z twelve sorry ah other way
round right three z twelve six z twelve zero what is ah yeah can you give me another composition
series for z twelve two z twelve four z twelve what is eight z twelve eight z twelve is will
again be four z twelve there are no proper sub modules of [za/z] four z twelve right
four z twelve is zero four eight it does not have a proper sub module
similarly i have one more right i can say two z twelve six z twelve zero there are composition
series so the composition series of z twelve all
of them have length three right what about take the z module z itself does it have a
composition series i can start with you know if i want to construct a composition [seri/series]
i [shou/should] i should find a maximal sub module so let say two z a maximal sub module
of this four z a maximal sub module of this eight z and so on it never terminates so z
does not have a composition series if a mod ok i will come back so the first thing that
one you know tends to think about suppose suppose you take a module which has a composition
series ok and suppose it has different composition series is it true that all of them have same
length in this case eight z twelve has same so this is ah indeed true let m be a module
with a with composition series of length n
then every composition series of of m has length n moreover every composition
series ah every chain can be extended to a composition series ok so m can have ah many composition series
so let’s denote let l of m denote the minimum of lengths of composition composition series
of m i want to say that every composition series is of this length to start with i know
that you know that n is bigger than equal to m right so observe that length of m is
less than equal to n let n be a sub module of can we say that l of n is less than l of m
first of all will this have a composition series at all will n have a composition series
how can we ah construct a composition series for n yeah take a composition series for m and then
intersect with n each n that will give me a composition series for n so therefore a
n will have a composition series now what will be the length of n will it be strictly
less than m i mean will this always hold true ok let us look at this so let ah m equal to
m not contained and m one be a composition series of m now i take n i to be n intersection
m i now what can we say about see then we have then there is ah there exists a chain
n n naught is n because that is m naught is m n is a sub module therefore m intersection
n is ah n naught i mean n itself this is contained in n one n in equal to zero so here see in
this one we will have to put less than i [mean] subset or equal to because we don’t know how
there can be two of them which when intersecting with n can give rise to only one sub module
so we will just put like this now what can you say about ah n i intersects
n i mod n i plus one what is this this is m i intersection n mod m i plus one intersection
n now this this injects into m i mod m i plus one right this one if i take any x bar here
so x plus n i plus one if i map to x plus m i plus one this will be an injective a module
homomorphism so therefore this injects into m i mod m i plus one but then this is a simple
a model so therefore this is either equal or zero since m i mod m i plus one is simple
either n i n i mod n i plus one is equal to m i mod m i plus one or n i is equal to n
i plus one so if n i is equal to n i plus one i just remove it ok so by deleting the
corresponding ah i mean repeated terms what i get is a composition series by deleting
the repeating terms so let me call this chain one
we get a composition series for composition series of ah n therefore length of n is less
than or equal length of m right now suppose i take what is you know what is mean by see
our aim is to show that it is strictly less n if n is a proper sub module then this is
strictly less what is meant by this these two are equal if l of n is equal to l of m
what does that say that says that here none of them is repeating that means nowhere this
happens or in other words n one what is n ah sorry
what is n n minus one look at what is n n minus one this is n intersection m n minus
one so this is not equal to so n m ah n m minus one mod n m is equal to m n minus one
mod m n but n m is what is n m that is zero therefore n m minus one is m n minus one right
similarly n m minus two is now n m minus one is m n minus one therefore pulling back ultimately
we get and that implies n m minus two is m n minus
two so on ultimately we get n is equal to m
so therefore therefore if n is a proper sub module of m then length of n is strictly less
than length of n see this has a very interesting consequence let m equal to m naught contained
in m one be a composition series of m then length of m is bigger than or equal to length
of m one plus one right from the previous discussion this is bigger than or equal to
length of m two plus two length of m n plus n but m n is zero therefore
this is n so l of m is bigger than or equal to n if
i take any composition series l of m is bigger than equal to n but l of m by definition what
is it minimum of lengths of composition series which means if you take any composition series
that has length l of n ok ok now to prove that last part suppose i take
let m k ah sorry m naught is m ah m one be a chain of ah sub modules of m
then we know that length of m k minus one this is less than length of m so when you
say that it means m k minus one has a composition series and it has finite length you know this
is a well defined so i can find a composition series for m k minus one ok
now if m k mod m k minus one is not simple that means m k is not a maximal sub module
of m ah maximal sub module of ah m k minus one is not a maximal sub module of m k minus
two and length of m k minus two is again finite because it is [le/less] less than length of
m naught therefore it has a composition series ah take the i mean m k is not a maximal sub
module therefore you can you know extend it keep doing this and find the composition series
for i mean what we are doing is basically we are inserting modules whenever it is not
a maximal sub module ok this the fact that length of any if m has a composition series
then any proper sub module has a composition series and the length is less than the length
of the composition series of m makes us i mean enable us to complete this process ok
so less than length of m then m k minus one has a composition series if m k minus two m k minus one is not simple
then m k minus one is not a maximal sub module
of so if i take ah now the
maximal sub module of ah oops m k minus two so choose a maximal sub module of ah m k minus
two contain m k minus one and i’ve kind of apply in direction you keep doing this so
ah let m k minus two prime be a maximal sub module of
m k minus two containing m k minus one and keep doing this so therefore this length
has ah proceeding like this so what we have seen is that if a module has
a composition series then every composition series is of same length ok so therefore if
m has a composition series then there is a unique integer associated
with ah m which is the length of the composition series that is unique the length of the composition
series of m is called the length of m
ok and if m does not have a a composition series
then we say length of m to be hm ok so we have already
seen examples where it is ah finite length so what is length of z twelve three length of z p one length of z four two
suppose i take m to be take a ring a and look at the maximal ideal m power n
ok i look at m is a maximal ideal in a and look at the module m equal to a mod m power
n ok now can you give me a composition series for
this ah i should i want a sub module of m ah m yeah which is same as m right m mod m power
n contained in i mean containing m square mod
m power n what does a mod m power n modulo m mod m power
n is this a composition series see how do i say this is a composition series if this
is i call this m one m mod m one is this simple what is m mod m one m mod m one is this is isomorphic to a mod
m right that is field therefore this is simple what about m mod m square ah m mod m power
n module of m square mod m power n what is this isomorphic to m mod m square this this
need not necessarily be isomorphic to a mod m or this is m mod m square is an a mod m
module it’s an a module at the same time it’s an a mod m module but m mod m is a field so
this is a vector space over a mod m right this could be a finite dimensional vector
space this need not be a of dimension one like this
so there could be many now this need not be simple if this is an n dimensional vector
space then suppose this is generated by x one bar up to x m bar then i can within this
i can have this x one bar up to x m bar x one x two bar up to x m bar x m bar right
to zero this itself is a composition series of m mod m square right so therefore this
this is this need not necessarily be a composition series i can i can you know in insert in between
and get a composition series but then each time what we are going to do is insert finitely
many well assuming that m is finitely generated if m is finitely generated each time i will
have a i will have finite number of ah modules in between and this will be a
module with composition series with finite length ok
so so this implies that ah if m is a finitely generated a module ah finitely generated ideal
more generally if i assume a is noetherian ok then length of a mod m power n is finite
a mod m power n has a composition series that composition series need not be what i wrote
there that now a mod m ah a mod m power n containing m mod m power n containing m square
mod m power n and so on that need not be a composition series but we can insert in between
there could be many but we can insert and ultimately get a finite composition series
so this is another important ah property of composition series ah modules with composition
series m has a composition series if and only if m satisfies a c c ascending
chain condition and descending chain condition both if a composition series m has a composition
series then it should satisfy ascending chain condition and descending chain condition so
the i mean one way is immediate if m has a composition series any chain can be extended
to a composition series that’s what ah the previous result says which means any chain
has to be finite whether it is ascending or descending whatever you wherever you start
with it has to be it has to terminate right so therefore this
if m has a composition series then ah any chain of sub modules of m
has ah finite length therefore m satisfies a c c and d c c any chain will have only finitely
many elements so ascending chain condition and descending chain condition both are satisfied
now suppose i conversely let m be m satisfy ascending chain condition and descending chain
condition ok i want to say that it has a composition series so we construct a composition series
ok now let m one be a maximal sub module of m i can say this because m satisfies ascending
chain condition what is meant by ascending chain condition it is any increasing chain
terminates equivalently any collection of sub modules of m has a maximal element right
and equivalently every sub module is finitely generated these are three equivalent conditions
now i am start i know that m satisfy ah m satisfies ascending chain condition therefore
i can choose a maximal sub module of m ok now i look at so therefore i now i look at
m one i look at the collection of all sub modules of m one which are anyway sub modules
of m sub modules of m one again because of m satisfying ascending chain condition this
will have a maximal element ok m two be a ah ok so we can since m has ascending
chain condition m one such an m one exists ok then ah now sigma one be the collection
of sub modules of ah m one this is ah collection of proper sub module of m one
let m two be a in fact i should have chosen m one like this let sigma be the collection
of all proper sub modules of m let m one be a maximal element now let m two be a maximal
element in sigma one then i have so continuing like this i get a chain of sub modules we get m contained in m one contained in m
two and so on since m satisfies descending chain condition there exists k such that m
k is zero because if it is nonzero i can you know if it has a nonzero proper sub module
maximal sub module i still can go ah further down so therefore the chain terminates ah
m has a a composition series so having composition series is equivalent
to saying that it has both ascending chain condition and descending chain condition ok
so this is like you know length is pretty much similar to the you know dimension of
vector space it’s ring theoretically very close if i have a sub module you know if i
have a subspace w of v if it’s a proper subspace then dimension of v is strictly bigger than
dimension of w moreover we have we know this dimension of v is equal to this is rank nullity
theorem you can say dimension of w plus dimension of v mod w right i mean how does is come so
if i look at i have this exact sequence what this what
rank nullity theorem says is that dimension is additive on the short exact sequence dimension
of w minus dimension of v plus dimension of this is zero ok so is this true for the case
of ah modules so suppose i have a you know m prime to m to m w prime to zero modules
of all of them have modules of ah i mean all of them are modules of ah finite length
now suppose i have a say i want to say that is this is it true that length of m mod length
of m is equal length of m prime plus length of m w prime i want we have to produce one
composition series of which satisfy this property composition series of m with length equal
to this plus this can you think of one so here you can think of m prime as a sub module
of m and m double prime as a as m mod m prime right so you can think of this m prime as
a sub module of m and m double prime equal to m mod m prime m mod m prime has a composition
series so let’s look at m double prime see this will
be this has a composition series any sub module of this is of the form some you know sub module
of m mod containing m prime modulo m prime if i have a sub module k of this then k i
can think of this as some n mod m prime where n is a sub module of m containing m prime
right so i have i can write like this m prime which
is m mod m prime contained in m one mod m one mod m prime m r minus one mod m prime
and zero what is zero this is m prime mod m prime right this is a composition series
of m w prime now suppose i have a composition series of m prime m prime contained in m one
prime m r m s so this is m s minus one prime contained
in m s prime which is zero now can you tell me a composition series for
m this composition series of m double prime this is composition series of m m prime m right m m one contained in m r minus one
m r which is same as m prime contained in m one prime m s minus one prime m s prime
equal to zero now look at each each ah quotient m mod m
one m mod m one is same as this modulo this but this is a composition series therefore
this modulo this is simple this modulo this is m mod m one therefore this is simple similarly
at each at each stage the corresponding quotient is same as the corresponding quotient therefore
each time to simple therefore this is a so length is additive on short exact sequences

## Only registered users can comment.

1. Rohit Verma says:

Amazing sir.

2. kiwanoish says:

Great as always! Perhaps a bit confusing to use 'chain' here again, as this is very much associated with chains in posets.