A little algebraic lemma

There is a lemma in algebra, which is very interesting.

Suppose that R is a commutative ring, and A,B,C are three R-modules, and there are two morphisms, f:A\rightarrow B,g:B\rightarrow C(in other words, we have a sequence, A\xrightarrow{f}B\xrightarrow{g}C), then we have an exact sequence,

0\rightarrow ker(f)\rightarrow ker(g\circ f)\rightarrow ker(g)\rightarrow coker(f)\rightarrow coker(g\circ f)\rightarrow coker(g)

This is a ‘lovely’ lemma , since it doesn’t require any conditions  on the original sequence, no condition on the exactness of the short sequence, for example.

The proof to this lemma is rather simple, so we won’t give it here.

I believe that there are various applications of this lemma. Yet, perhaps because this lemma is too elementary, I do not get any results when I search this lemma on the Internet(perhaps I do not get its name right? I searched using the name, ‘kernel cokernel (exact) sequence’,.:) 

I came across this lemma from Milne’s book, ‘elliptic curves’ when he defines the Selmer’s group.

Selmer’s group is an important tool in proving, in a modern way, the theorem of Mordell-Weil(which says that, any elliptc curve over any field of finite extension of the rationals is an abelian group of finite rank, or more precisely, is a group of which the torsion part is finite, and the free part is of finite rank). Yet the definition for this group is not so obvious.

Suppose that E is an elliptic curve over some number field L(finite extension of the rationals), which we write as E(L). One important step in the classical proof to the theorem of Mordell-Weil is to show that E/nE is a finite group where n is a positive integer(in the proof we just need to show that this is true for the case $latexn=2$, and then we can use the height function to finish the proof).

Start from the exact sequence,

0\rightarrow E(\bar{L})[n]\rightarrow E(\bar{L})\xrightarrow{\times n} E(\bar{L})\rightarrow 0

and use the functor, A\mapsto A^G from the category of G-abelian groups to itself, where G is a group, finite or profinite, then we can get a long exact sequence(here the group is Gal(\bar{L}/L), where \bar{L} is the algebraic closure of L),

0\rightarrow E(L)[n]\rightarrow E(L)\rightarrow E(L)\rightarrow H^1(Gal(\bar{L}/L),E(\bar{L})[n])\rightarrow H^1(Gal(\bar{L}/L),E(\bar{L}))\rightarrow H^1(Gal(\bar{L}/L),E(\bar{L}))

In this way, we get that

0\rightarrow E(L)/nE(L)\rightarrow H^1(L,E[n])\rightarrow H^1(L,E)[n]\rightarrow 0

(in the aobove we write H[n] for the n-torsion part of an abelian group H, and H^1(L,E) for H^1(Gal(\bar{L}/L),E(\bar{L}))).

What is more, there are obvious maps, j_v:H^1(L,E)\rightarrow H^1(L_v,E) where L_v stands for the completion of L with respect to a place of L(finite place or infinite place).

Due to some examples where H^1(L,E) are not finite groups, we turn to consider the group,

S^n(E/L)=ker(H^1(L,E)\rightarrow \prod_vH^1(L_v,E))

Apply the above algebraic lemma to the sequence,

H^1(L,E[n])\rightarrow H^1(L,E)[n]\rightarrow \prod_vH^1(L_v,E)[n]

then we get that

0\rightarrow E(L)/nE(L)\rightarrow S^n(E/L)\rightarrow \Phi(E/L)[n]\rightarrow 0

For the present, wo do not concern with the before last term. We see that, there is an injection from E(L)/nE(L) into S^n(E/L), so if we can show that S^n(E/L) is finite, then so will be E(L)/nE(L).

An this is indeed the case, this is the modern version of the proof to the theorem of Morderll-Weil.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s