This is a well known lemma that consistently appears in textbooks, either as a statement without proof, or as an exercise. If 0 id→ A f→ B g→ C h→ 0 is a short exact sequence of finitely generated abelian groups, then rank B=rank A+rank C I've been trying to prove this unsuccessfully. What do we know? f is injective, g is surjective, Imf=kerg, Img=kerh, C=∼BA So I start with a maximally linearly independent subset {aα} of A such that the sum (with only finite non-zero entries) ∑nαaα=0 for nα∈Z, implies that nα=0 Where to go from here is a puzzle?

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This is a well known lemma that consistently appears in textbooks, either as a statement without proof, or as an exercise. If 0 id→ A f→ B g→ C h→ 0 is a short exact sequence of finitely generated abelian groups, then rank B=rank A+rank C I&#039;ve been trying to prove this unsuccessfully. What do we know? f is injective, g is surjective, Imf=kerg, Img=kerh, C=∼BA So I start with a maximally linearly independent subset {aα} of A such that the sum (with only finite non-zero entries) ∑nαaα=0 for nα∈Z, implies that nα=0 Where to go from here is a puzzle?

porschomcl · Accepted Answer

Step 1  If youre

Annie Levasseur · Accepted Answer

Step 1  This holds for integral domains R and finitely generated R-modules. If we put A=M1, B=M2, and C=M3, then the given short exact sequence is isomorphic to the following short exact sequence:  0→M1→M2→M2M1→0  If r1=rank M1, r2=rank M2, and r3=rank M3=rankM2M1, then let u1,⋯,ur1 be linearly independent in M1 and v―1,⋯,v―r3 be linearly independent in M2M1 Then, if  a1u1+⋯+ar1ur1+b1v1+⋯+br3vr3=0,  in M, then reducing this equation modulo M1 gives  b1v―1+⋯+br3v―r3=0― in M2M1  So b1=⋯=br3=0SKbyl∈ear∈dependenceofthesee≤ments,andthusPSKa1=⋯=ar1=0 by linear independence of the ui. Hence rank M2≥r1+r3  Suppose that w1,⋯,ws are linearly independent in M2 with s&amp;gt;r1+r3. Now (exercise) there is a free submodule NM1⊂M2M1 with Rr3=∼NM1

This is a well known lemma that consistently appears in textbooks, either as a statement without pro

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