1

This is the same as this question, though a solution was never given, and my approach is different than the hint.

It's also the first exercise in Commutative Ring Theory by Matsumura.

Let $A$ be a (commutative) ring and $I \subset \operatorname{nil}(A)$ an ideal made up of nilpotent elements; if $a\in A$ maps to a unit of $A/I$ then $a$ is a unit of $A$.

Attempt:

Since $a$ maps to a unit in $A/I$, we have that for $a+I$ there exists some coset $a'+I$ such that $(a+I)(a'+I)=1+I$. By coset multiplication, we have $(aa')+I=1+I$, which implies that $aa'= 1+i$ for some $i\in I$. Since $i$ is nilpotent, there exists $n>0$ such that $i^n=0$. If $i=0$, we are done. Otherwise, we have that $n \geq 2$ so we can do the following:

\begin{align*} aa' &= 1+ i\\ aa'i^{n-1}&=(1+i)i^{n-1}\\ aa'i^{n-1} &= i^{n-1}+0\\ aa'i^{n-1} &= i^{n-1}\\ aa' &= 1, \end{align*}

showing that $a$ is indeed a unit in $A$.

Is this correct?

P.S. I would guess that there is an easier method, but I wanted to try to proceed directly from the definition and work with cosets in the quotient ring.

pyridoxal_trigeminus
  • 2,397
  • 5
  • 18
  • 3
    How did you deduce $aa'=1$ in the last step? Remember, in a ring you can't in general divide by elements. Also, note that your solution can't be correct, because it would actually tell us that any representative of the coset $(a+I)^{-1}$ is an inverse of $a$, which can't be true, because an inverse is unique. – Mark Feb 10 '23 at 19:42
  • @Mark In my head I multiplied both sides by $i^{-(n-1)}$, but now I see that doesn't work. – pyridoxal_trigeminus Feb 10 '23 at 20:39

2 Answers2

1

Since the Jacobson radical contains every nil ideal, every element of $I$ is quasiregular.

So, if $ab\equiv 1\mod I$, it amounts to $ab-1=i$ for some $i\in I$. By quasiregularity $i+1=ab$ is invertible in $R$. The same can be said for $ba$ and then you can conclude that $a$ is invertible in $R$ too.

If you want something a little concrete with how nilpotence is connected, the question about why $1-x$ is a unit if $x$ is nilpotent is one of the most-asked questions in the ring-theory tag! You can use this instead of quasiregularity to see why $ab$ is a unit.

rschwieb
  • 153,510
  • The question is about commutative rings. – Mark Feb 10 '23 at 19:54
  • @Mark This proof includes commutative rings so... ? (Maybe I should mention that I like giving proofs that apply even if the ring isn't commutative, if they are available.) – rschwieb Feb 10 '23 at 20:01
  • Anyway, this answer is correct, though maybe goes a bit too far. In the last sentence OP asked for a proof using just the definitions. (otherwise, in the commutative case you can also easily prove it using prime ideals) – Mark Feb 10 '23 at 20:05
  • @Mark I'm willing to include a link to the reason why 1+nilpotent is a unit because it's been beaten to death on math.se over the years. It seems like the simplest analogue. – rschwieb Feb 10 '23 at 20:19
1

\begin{align*} aa' &=1+i\\ aa'-1&=i\\ (aa'-1)^n&=i^n\\ (aa'-1)^n&=0\\ \sum_{k=0}^n {n\choose k}(aa')^k(-1)^{n-k}&=0\\ a\left(\sum_{k=1}^n {n\choose k}a^{k-1}a'^k(-1)^{n-k}\right)&=(-1)^{n+1}\\ a\left(\sum_{k=1}^n {n\choose k}a^{k-1}a'^k(-1)^{1-k}\right)&=1 \end{align*}

Joshua Tilley
  • 7,070