5

Something similar is asked here but it is not exactly the same thing.

An element $a$ of a ring $R$ is called nilpotent if $a^n=0$ for an $n \in \mathbb{N}$. Show that if $u$ is a unit and $a$ is nilpotent in $R$, then $u+a$ is a unit.

My proof:

Let $a^n=0$. The equality $$(u+a)^na^{n-1}=u^na^{n-1} \tag{1}$$ results from expanding $(u+a)^n$ together with the property that $a$ is nilpotent.

Let $u_1$ be the left inverse of $u$. Then by multiplying both sides of (1) on the left by $u_1^n$, the resulting equality is $u_1^n(u+a)^na^{n-1}=a^{n-1}$. This is where I am nervous. I thought that you could then write $u_1^n(u+a)^n=1$ by the property of the multiplicative identity in an associative ring with unity and setting the coefficents equal to each other, but I feel like this looks like applying a cancellation to the $a^{n-1}$ on both sides, which I do not mean to be doing.

The last part is to rewrite the resulting equality as $(u_1^n(u+a)^{n-1})(u+a)=1$ to show the existence of a left inverse of $u+a$ and then do a similar thing starting from (1) to show the existence of a right inverse.

Is there an error with this?

Community
  • 1
Joseph Mennuti
  • 51

2 Answers2

7

This fails, for instance, $$\begin{pmatrix}0&-1\\1&0\end{pmatrix}+\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\1&0\end{pmatrix}$$ is singular.

Olivier Bégassat
  • 21,101
-2

In a ring $R$ the nilpotent elements constitutes an ideal which is called nilradical, $N(R)$ .

$N(R)$ is the intersection af all prime ideals; so if $u$ is a unit and $a$ is nilpotent $a+u$ must be a unit.

Otherwise there would be a maximal ideal $M$ with $a+u \in M$. But a maximal ideal is prime, and so $$a \in N(R) \subseteq M$$ and so $$u = (u+a) - a \in M$$ which is impossible because $u$ is a unit.

WLOG
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