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Let $R$ be a commutative ring with unity, and let $a$ be a fixed element of $ R $. Suppose that for every $ r \in R $, $ ar + 1 $ is invertible in $ R $. Show that $ a $ belongs to the Jacobson radical of $ R $.

rschwieb
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rogerstack
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  • Can you share your thoughts on the problem, and explain what you've tried and what you're having trouble with? –  Sep 24 '13 at 06:27
  • Well, I tried to expand the series but the problem is might not stop anywhere. I mean, the converse is very easy to show. If $ n $ belongs to the nilradical, then, we can use the expansion of $ (1+nr)^{-1 } $ to get the inverse. I tried to use the idea of the ideal generated by $ n $, and the fact that $ 1 + (n) $ has all its elements invertible. I tried the idea that the set $ \left {1+nr | r \in R \right } $ is multiplicative, and tried something around prime ideals, but again, of no use. :/ – rogerstack Sep 24 '13 at 06:31
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    As stated it is not true. For $R$ take $K[x]$ localized at $(x)$ and set $n=x$; another (almost the same) example is $R=K[[x]]$ ($K$ a field). Something is probably missing in the statement of the problem. – user8268 Sep 24 '13 at 07:00
  • Can you explain the counterexample in more layman terms? What do you mean by $K[x]$ localized at $ (x) $. I know that $ K[x] $ is the ring of polynomials over field $ K $, but what does it mean to say that it is localized over $ (x) $? – rogerstack Sep 24 '13 at 07:09
  • A google search with a slight change of variables gave me this: http://www.math.purdue.edu/~sbasu/teaching/fall08/557/hw1.pdf

    So, the question is probably right.

     – rogerstack Sep 24 '13 at 07:19
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    I just realized that the question in the link says $ rad(A)$, so refers to the Jacobson radical of $ A $. Now, if the ring $ R $ is Artinian, then we can apply a nice argument to show our property. – rogerstack Sep 24 '13 at 07:28
  • @rogerstack Sure, in the case of a commutative Artinian ring, the nilradical and Jacobson radical coincide. But Fredrik's answer is the way to go if you don't have the Artinian hypothesis. – rschwieb Sep 24 '13 at 12:47

1 Answers1

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As you discovered in your comments, the question probably refers to the Jacobson ideal of $A$, that is, the intersection of all the maximal ideals in $A$.

Now, assume the opposite, namely that there is some maximal ideal $\mathfrak m$ such that $a \not \in \mathfrak m$. Then $\bar a$ is an invertible element in $A/\mathfrak m$. This means that there exists an $r \in A$ such that $ar-1\in \mathfrak m$. Then also $-ar+1=a(-r)+1 \in \mathfrak m$. But by assumption every such element was invertible, and so $\mathfrak m = (1)=A$. Contradiction.

Fredrik Meyer
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