This is again a question asked to me by [this user](https://mathoverflow.net/users/3582/). He apparently quit using MO due to a busy time in personal and professional life and resulting difficulties in spending time here with patience. I am taking the liberty to ask it myself(with permission) as I consider him and his questions to be of value. 


Let $A$ be unitary ring(ie ring with identity), $a \in A$ be such that for all ring homomorphisms $f : A \rightarrow B$, $B$ a unitary non-zero ring, $f (a)$ is not a unit in $B$.


[a unit in a unitary ring is an element both right and left invertible].


Does it follow that $a$ is nilpotent? 


[in particular, $f(a)$ is neither left, nor right invertible for all $f : A \rightarrow B \neq 0$.]


A weaker version may be, if it $a \in A$ is such that $f (a)$ does neither left nor right invertible for all $f : A \rightarrow B \neq 0$ imply that $a$ is a nilpotent element?



Characterizing nilpotents in a ring by a universal property
Let $e \in A$ be a non-zero idempotent (and hence not nilpotent).  

Then if $f(e)$ is a unit, we find that $f(e) = 1,$
and so $f(e - 1) = 0.$ Thus if $e - 1$ generates (as a two-sided ideal) the entire
ring, we find that $f$ is identically zero, and hence that $B = 0$.


Thus, if we can find a non-zero idempotent $e \in A$ such that $A(1-e)A = A$,
we have a counterexample.


Note by the way that $f: = 1 - e$ is again idempotent, and so it suffices instead
to find a non-unital idempotent $f$ such that $A f A = A$.


E.g. If $A$ is simple (so that any non-zero two-sided ideal equals $A$), any non-unital and
non-zero idempotent gives a counterexample.


E.g. if $A = M\_2(k)$ for some field $k$, and $f = (1 0 , 0 0)$, we are done. (I think
this is what Kevin intended to write down in his comment.)


