Fitting lemma


The Fitting lemma, named after the mathematician Hans Fitting, is a basic statement in abstract algebra. Suppose M is a module over some ring. If M is indecomposable and has finite length, then every endomorphism of M is either bijective or nilpotent.


As an immediate consequence, we see that the endomorphism ring of every finite-length indecomposable module is local.


A version of Fitting's lemma is often used in the representation theory of groups. This is in fact a special case of the version above, since every K-linear representation of a group G can be viewed as a module over the group algebra KG.



Proof


To prove Fitting's lemma, we take an endomorphism f of M and consider the following two sequences of submodules:


  • The first sequence is the descending sequence im(f), im(f 2), im(f 3),…,

  • the second sequence is the ascending sequence ker(f), ker(f 2), ker(f 3),…

Because M has finite length, the first sequence cannot be strictly decreasing forever, so there exists some n with im(f n) = im(f n+1). Likewise (as M has finite length) the second sequence cannot be strictly increasing forever, so there exists some m with ker(f m) = ker(f m+1). It is easily seen that im(f n) = im(f n+1) yields im(f n) = im(f n+1) = im(f n+2) = …, and that ker(f m) = ker(f m+1) yields ker(f m) = ker(f m+1) = ker(f m+2) = …. Putting k = max(m,n), it now follows that im(f k) = im(f 2k) and ker(f k) = ker(f 2k). Hence, ker(fk)∩im(fk)=0displaystyle mathrm ker left(f^kright)cap mathrm im left(f^kright)=0mathrm kerleft(f^kright)cap mathrm imleft(f^kright)=0 (because every x∈ker(fk)∩im(fk)displaystyle xin mathrm ker left(f^kright)cap mathrm im left(f^kright)xin mathrm kerleft(f^kright)cap mathrm imleft(f^kright) satisfies x=fk(y)displaystyle x=f^kleft(yright)x=f^kleft(yright) for some y∈Mdisplaystyle yin Myin M but also fk(x)=0displaystyle f^kleft(xright)=0f^kleft(xright)=0, so that 0=fk(x)=fk(fk(y))=f2k(y)displaystyle 0=f^kleft(xright)=f^kleft(f^kleft(yright)right)=f^2kleft(yright)0=f^kleft(xright)=f^kleft(f^kleft(yright)right)=f^2kleft(yright), therefore y∈ker(f2k)=ker(fk)displaystyle yin mathrm ker left(f^2kright)=mathrm ker left(f^kright)yin mathrm kerleft(f^2kright)=mathrm kerleft(f^kright) and thus 0=fk(y)=xdisplaystyle 0=f^kleft(yright)=x0=f^kleft(yright)=x) and ker(fk)+im(fk)=Mdisplaystyle mathrm ker left(f^kright)+mathrm im left(f^kright)=Mmathrm kerleft(f^kright)+mathrm imleft(f^kright)=M (since for every x∈Mdisplaystyle xin Mxin M, there exists some y∈Mdisplaystyle yin Myin M such that fk(x)=f2k(y)displaystyle f^kleft(xright)=f^2kleft(yright)f^kleft(xright)=f^2kleft(yright) (since fk(x)∈im(fk)=im(f2k)displaystyle f^kleft(xright)in mathrm im left(f^kright)=mathrm im left(f^2kright)f^kleft(xright)in mathrm imleft(f^kright)=mathrm imleft(f^2kright)), and thus fk(x−fk(y))=fk(x)−fk(fk(y))=fk(x)−f2k(y)=0displaystyle f^kleft(x-f^kleft(yright)right)=f^kleft(xright)-f^kleft(f^kleft(yright)right)=f^kleft(xright)-f^2kleft(yright)=0f^kleft(x-f^kleft(yright)right)=f^kleft(xright)-f^kleft(f^kleft(yright)right)=f^kleft(xright)-f^2kleft(yright)=0, so that x−fk(y)∈ker(fk)displaystyle x-f^kleft(yright)in mathrm ker left(f^kright)x-f^kleft(yright)in mathrm kerleft(f^kright) and thus x∈ker(fk)+fk(y)⊆ker(fk)+im(fk)displaystyle xin mathrm ker left(f^kright)+f^kleft(yright)subseteq mathrm ker left(f^kright)+mathrm im left(f^kright)xin mathrm kerleft(f^kright)+f^kleft(yright)subseteq mathrm kerleft(f^kright)+mathrm imleft(f^kright)). Consequently, M is the direct sum of im(f k) and ker(f k). Because M is indecomposable, one of those two summands must be equal to M, and the other must be equal to 0. Depending on which of the two summands is zero, we find that f is bijective or nilpotent.[1]



Notes



  1. ^ Jacobson (2009), p. 113–114.



References



  • Jacobson, Nathan (2009), Basic algebra, 2 (2nd ed.), Dover, ISBN 978-0-486-47187-7.mw-parser-output cite.citationfont-style:inherit.mw-parser-output qquotes:"""""""'""'".mw-parser-output code.cs1-codecolor:inherit;background:inherit;border:inherit;padding:inherit.mw-parser-output .cs1-lock-free abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/6/65/Lock-green.svg/9px-Lock-green.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-lock-limited a,.mw-parser-output .cs1-lock-registration abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/d/d6/Lock-gray-alt-2.svg/9px-Lock-gray-alt-2.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-lock-subscription abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/a/aa/Lock-red-alt-2.svg/9px-Lock-red-alt-2.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registrationcolor:#555.mw-parser-output .cs1-subscription span,.mw-parser-output .cs1-registration spanborder-bottom:1px dotted;cursor:help.mw-parser-output .cs1-hidden-errordisplay:none;font-size:100%.mw-parser-output .cs1-visible-errorfont-size:100%.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registration,.mw-parser-output .cs1-formatfont-size:95%.mw-parser-output .cs1-kern-left,.mw-parser-output .cs1-kern-wl-leftpadding-left:0.2em.mw-parser-output .cs1-kern-right,.mw-parser-output .cs1-kern-wl-rightpadding-right:0.2em

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