# Use of units in Nakayama's Lemma's proof

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

• 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 $j=1-x\in I\subseteq J(A)$. Aug 28, 2019 at 14:48

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.