$A$ is a ring and $I$ is an ideal, an element $x \in A$ maps to an invertible element of $A/I$.

Question

I am having a hard time understanding the following question

Let $A$ be a ring and $I \subset \operatorname{nilrad}A$ an ideal; if $x \in A$ maps to an invertible element of $A/I$, prove that $x$ is invertible in $A$.

What does

$x$ maps to an invertible element of $A/I$

mean? This question originally has two parts. In the first part, we prove that if $a$ is a unit and $x$ in nilpotent, then $a+x$ is again a unit. I just could not understand the second part of the question.

Answer 1

‘$x+I$ is a unit in $A/I$’ means there exists $y\in A$ such that $xy\in 1+I$, i.e. there exists $i\in I$ such that $xy=1+i$. By the first question, since $i$ is nilpotent, $1+i$ is a unit, in other words, there exists an element $u\in A$ such that $$(1+i)u=1,\quad\text{or }\;(xy)u=x(yu)=1,$$ which shows $x$ is a unit, with inverse $yu$.

Answer 2

The map in "maps" is the canonical projection $x \mapsto x+I$ from the object $A$ to its quotient $A/I$.

Since all of $I$ in $A$ maps to the additive identity in $A/I$, no element of $I$ maps to an invertible element of $A/I$. Less specifically, any non-unit in $A/I$ is not invertible so all preimages of non-units are not invertible.