§ Nakayama's lemma

I read the statement as IM=M    M=0IM = M \implies M = 0, when II is in the jacobson radical.
  1. Essentially, it tells us that if a module MM "lives by the II", then it also "dies by the II".
  2. Alternatively, we factor the equation as M(I1)=0M(I - 1) = 0. Since our ideal IIis a member of the jacobson radical, (1I)(1 - I) is "morally" a unit and thus M=0M = 0.This is of course completely bogus, but cute nontheless.
  3. We can think of a graded ring, say R[x]R[x] acting on some graded module MM (say, a subideal, M=(x2)M = (x^2)). When we compute IMIM,this will bump up the grading of MM. If IM=MIM = M, then MM could not have had non-trivial elements in the first place, since thevector of, say, "non-zero elements in each grade" which used to look like (v0,v1,v2,)(v_0, v_1, v_2, \dots) will now look like (0,v0,v1,)(0, v_0, v_1, \dots).Equating the two, we get v0=0,v1=v0=0,v2=v1=0v_0 = 0, v_1 = v_0 = 0, v_2 = v_1 = 0 and so on, collapsing the entire ring.