inverses in $R/I$ where $I$ is a nilpotent ideal Given an element $x \in R$ where R is a ring $I$ is a nilpotent ideal of $R$,
i am trying to find inverses in the quotient R/I and thought about things in the general case, what would determine the inverses for elements $x \in R/I$ 
would i be right in saying that it needs to have an inverse in R as well if it is to have one in the quotient?
I'm not sure how i'd show this though, any help would be great!
thank you
 A: 
would i be right in saying that it needs to have an inverse in R as well if it is to have one in the quotient?

Interestingly, yes. 
If $x+I$ is a unit, there exists $y$ such that $xy-1\in I$ (this is just  the translation of what it means to be a unit in $R/I$.)
Then $xy-1=n$ for some nilpotent element, and rearranging we get $xy=n+1$. Since the right hand side is the sum of a nilpotent element and a unit, it is a unit as well. Then you can easily show that $xy$ being a unit in $R$ implies both $x$ and $y$ are units in $R$.

So, this even works if $I$ is merely nil and maybe not nilpotent. 
It doesn't work if $I$ is not nil, though. For example, you can look at $R=F[x,y]$ and $I=(xy-1)$. Then $x+I$ is a unit in $R/I$, but not in $R$.
And as you probably already know, if $u$ is a unit of $R$, then $u+I$ is a unit of $I$ for any proper ideal $I$.
A: Assuming $R$ is commutative with unity, suppose $x$ is a unit in $R/I$. Then $(x+I)(y+I) = 1+I$, so that $xy-1\in I$, so that $xy-1$ is nilpotent. This means that $xy-1+1 = xy$ is a unit (the sum of a unit and a nilpotent element is a unit), so that $x$ is a unit.
A: If $I$ is not nilpotent, you certainly don't have this. Consider the morphism of rings $\mathbb{Z}/6 \to \mathbb{Z}/3$. The kernel of this is the ideal $\langle 3 \rangle$ (which is not nilpotent).
Now, $2 \in \mathbb{Z}/6$ is not invertible. However, its image in the field $\mathbb{Z}/3$ is. Thus an element in the quotient with an inverse does not have a pre-image which has an inverse.
In fact, now that I think about it, a simpler example exists (again, without $I$ nilpotent). Consider the surjection $\mathbb{Z} \to \mathbb{Z}/p$ for some prime $p$. Since the latter is a field...
