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Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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UNIT-III<br />

65<br />

Theorem 1<br />

Let M, N be R-modules and let f : M → N be a non-zero R-morphism. Then<br />

1. If M is simple, f is a monomorphism.<br />

2. If N is simple, f is an epimorphism.<br />

Proof<br />

1. Ker f is a submodule of M, since f is not the zero morphism, we must have ker f = (0), because M is<br />

simple, so only submodules of M are (0) and M itself (if Ker f = M, then f(M) = (0) ⇒ f = 0, but<br />

).<br />

Hence Ker<br />

is a monomorphism.<br />

2. Imf is a submodule of N, But N is simple, so Imf<br />

but<br />

. Therefore,<br />

Corollary : (Shur's Lemma) If M is a simple R-module, then the ring End of R-morphisms.<br />

f : M –––→ M is a division ring.<br />

Proof<br />

From (1) and (2) above, every non-zero f ∈ End is an isomorphism and so is an invertible element in<br />

the ring. Hence End R ( M ) is a division ring.<br />

Fundamental structure theorem for finitely generated modules over a principal ideal domain :<br />

Before proving this we have to build some tools needed to prove above theorem :<br />

= f = ≠f<br />

( 0<br />

M<br />

) =<br />

) <br />

N( = 0.<br />

(),<br />

f0<br />

)<br />

R( n M<br />

)<br />

Suppose we have a sequence of modules with a homomorphism from each module to the next :<br />

f o f1<br />

f 2 f 3<br />

......... – → M – → M – → M – → ...........<br />

This sequence is said to exact at M 1<br />

if<br />

Im f 1<br />

= Ker f 2<br />

o<br />

The sequence is exact if it is exact at every module<br />

An exact sequence of the form<br />

α α β<br />

( 0) – → M – → M – → M – → ( 0)<br />

1 2<br />

is called a short exact sequence.<br />

1<br />

2<br />

or<br />

Here f is an epimorphism.<br />

Recall that every module over a general ring R is a homonorphic image of a free module. Every R-module M<br />

forms part of a short exact sequence.<br />

( 0) – → G – → F – → M – → ( 0)<br />

where F is free, this is called a presentation of M; If M is finitely generated, F can be taken to be of finite<br />

rank.<br />

We shall use the result (without proving it)<br />

"If R is a principal ideal domain, then for any integer n, any submodule of<br />

. If<br />

is free of rank at most n."<br />

Using this, we assume that above G is free, at least when M is finitely generated. More precisely, when M is<br />

generated by n elements, then it has a presentation<br />

then

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