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 $(1+i)M=0$.


1 Answer 1


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.


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