# Let $I$ be an ideal of nilpotent elements. Show that if $a$ maps to a unit in $A/I$, then $a$ is a unit in $A$.

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.

• 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
Commented Feb 10, 2023 at 19:42
• @Mark In my head I multiplied both sides by $i^{-(n-1)}$, but now I see that doesn't work. Commented Feb 10, 2023 at 20:39

\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*}

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.

• The question is about commutative rings.
– Mark
Commented Feb 10, 2023 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.) Commented Feb 10, 2023 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
Commented Feb 10, 2023 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. Commented Feb 10, 2023 at 20:19