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Given the Nakayama's Lemma: Let $A$ be a commutative ring with identity and $M$ be a finitely generated $A$-module. If $I\subseteq J(A)$, the Jacobson radical, such that $IM=M$, then $M=0$.

I've looked up some proofs and understood them all except this one here:

Proof: Suppose $IM=M$. Taking $\phi; M \rightarrow M$ to be the identity map . Because of the determinant trick we obtain an element $x\in A$ such that $x-1 \in I$ and $xM=0$. $I$ is cointained in the Jacobson radical. Recall that $x\in J(A)$ iff $1-xy$ is a unit for all $y\in A$. Taking $y$ to be equal $1$ we get

(*): $ 1-(1-x)$ is a unit in $A$ which implies $x$ is a unit in A and $xM=0$ which finally implies $M=0$.

I don't understand the (*) part. Why can we assume that a sum of units is a unit here? That is why can we conclude that $x$ is unit? The rest of the proof is fine with me. Thank you for your help

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  • $\begingroup$ The claim is not "sum of units is a unit." The claim is just the previous line that $1-j$ is a unit if $j\in J(A)$, and in particular $j=1-x\in I\subseteq J(A)$. $\endgroup$
    – rschwieb
    Commented Aug 28, 2019 at 14:48

1 Answer 1

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I think that the problem is just repeated use of the symbol $x$. Rewrite the proof as follows:

Suppose $IM=M$. Taking $\phi; M \rightarrow M$ to be the identity map . Because of the determinant trick we obtain an element $x\in A$ such that $x-1 \in I$ and $xM=0$. $I$ is cointained in the Jacobson radical. Recall that $z\in J(A)$ iff $1-yz$ is a unit for all $y\in A$. Taking $y$ to be equal $-1$ and $z$ to be $x-1$ we get $ 1-(-1) \cdot(x-1)$ is a unit in $A$ which implies $x$ is a unit in A and $xM=0$ which finally implies $M=0$.

Note that $x-1 \in I \subseteq J(A)$, so $1- (-1) \cdot (x-1)=x$ must be a unit.

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