Let $A$ be an algebra over a ring $R$ with $1$. Then $J(A)$, the Jacobson radical of $A$ is intersection of all the maximal ideals of $A$. The following property of $J(A)$ is well known:

If $a\in J(A)$ then $1-a$ is invertible.

I saw following proof in a book (and I partly did in some different way, and observed that proof of book seems lengthy). For fixed $a\in A$,

  1. Consider the ideal generated by $1-a$, i.e. $(1-a)A$.

  2. If it is proper, then it will be contained in a maximal ideal $M$; but then $J(A)\subseteq M$.

  3. So $1-a\in M$ and $a\in J(A)\subseteq M$ implies $1\in M$, contradiction.

  4. Thus $(1-a)A=A$.

  5. Hence there is $1-b\in A$ such that $(1-a)(1-b)=1$.

  6. Hence $b=a(b-1)\in J(A)$.

  7. By same reasoning as before, $1-b$ has a right inverse say $c$.

  8. Now $1-a=(1-a)1=(1-a)(1-b)c=1.c=c$ which has left inverse $1-b$.

In (4), though, we have $(1-a)A=A$. Can't we directly say that for $1\in A$ (RHS) there is $x\in A$ such that $(1-a)x=1$, so $1-a$ is invertible.

  • 3
    $\begingroup$ Yes, you are right. But this only shows that 1-a has a right inverse. $\endgroup$
    – Hagen Knaf
    Dec 8, 2016 at 7:02

1 Answer 1


Hagen Knaf's comment answers this question. If $R$ were a commutative ring, the proof would be done at (4). But if not we need to prove that $1-a$ both a left- and a right-inverse.


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