Linear applications simple doubt

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Is an injective endomorfism automatically bijective?
 

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  • #4
WWGD
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Doesn't this follow from rank-nullity theorem?
 
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mathwonk
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for finitely generated modules over other rings this can fail, e.g. if Z is the integers, the injective endomorphism Z-->Z taking n to 3n is injective but not surjective. Interestingly however, a surjective endomorphism of a finitely generated module is always injective as well.
 
  • #6
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Doesn't this follow from rank-nullity theorem?
Yes. But the Fredholm alternative also hold in some infinite-dimensional cases. There, you can't use rank-nullity.
 

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