# Does the Jacobson radical Nakayama lemma imply the general one?

Jacobson radical Nakayama. Let $$J$$ be the Jacobson radical. Let $$M$$ be a finitely generated $$A$$-module satisfying $$M=JM$$. Then $$M=0$$.

Proof. Induction on size of generating set. If $$M$$ is generated by zero elements then it is zero. Assume the assertion holds for modules generated by $$n-1$$ elements and let $$M$$ be generated by $$n$$ elements $$m_1,\dots ,m_n$$. The quotient $$M/m_1$$ is generated by the images of $$m_2,\dots ,m_n$$. If $$M=JM$$ then $$\tfrac{M}{m_1}=J\tfrac{M}{m_1}$$. By the induction assumption $$\tfrac{M}{m_1}=0$$ meaning $$M$$ is generated by $$m_1$$. By assumption $$M=JM$$ we have $$m_1=\epsilon m_1$$ for some $$\epsilon \in J$$. But $$1-\epsilon$$ is invertible, so $$m_1=0$$ whence $$M=0$$.

Does the above version imply the general version below?

Nakayama. Let $$I\vartriangleleft A$$ and let $$M$$ be a finitely generated $$A$$-module satisfying $$M=IM$$. There exists an element $$1+i\in 1+I$$ satisfying $$(1+i)M=0$$.

Yes, you can deduce the more general form from the Jacobson radical form. Here's a proof; let $$S=1+I=\{1+a:a\in I\}$$. $$S$$ is a multiplicatively closed set, so consider the localizations $$S^{-1}A$$ and $$S^{-1}M$$. First I claim that the ideal $$S^{-1}I$$ is contained in the Jacobson radical of $$S^{-1}A$$; to see this, fix any maximal ideal $$P$$ of $$S^{-1}A$$. If we had $$S^{-1}I\nsubseteq P$$, then by maximality of $$P$$ we would have $$P+S^{-1}I=S^{-1}A$$, and hence there would exists $$a\in I$$ and $$1+b\in S$$ such that $$\frac{1}{1}-\frac{a}{1+b}\in P$$. Multiplying through by $$\frac{1+b}{1}$$, we get $$\frac{1+(b-a)}{1}\in P$$, a contradiction, since $$1+(b-a)\in S$$ and hence becomes a unit in $$S^{-1}A$$.
So, $$S^{-1}I$$ is contained in the Jacobson radical of $$S^{-1}A$$. On the other hand, $$S^{-1}M$$ is still a finitely generated $$S^{-1}A$$-module, and since $$IM=M$$ we also have $$(S^{-1}I)S^{-1}M=S^{-1}M$$. Hence by the Jacobson radical form of Nakayama's lemma, we have $$S^{-1}M=0$$. Let $$m_1,\dots,m_n$$ be any generators for $$M$$. The condition tells us that there exist $$c_1,\dots,c_n\in I$$ such that $$(1+c_k)m_k=0$$ for each $$k\leqslant n$$. Now the element $$c:=1-\prod_{i=1}^k(1+c_i)$$ lies in $$I$$, and $$1-c$$ annihilates $$M$$, as desired.