Characterizing nilpotents in a ring by a universal property This is again a question asked to me by this user. 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?
 A: 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.)
A: EDIT: Now it has a chance of making sense.
I think the equivalence
"every unital ring homomorphism $A\to B$ with $B\neq 0$ maps $a$ to a non-unit $\Longleftrightarrow$ $a$ is nilpotent"
cannot be true (though $\Longleftarrow$ holds, of course). Otherwise, it would yield that whenever $a$ is nilpotent, then so is $ua$ for any unit $u$ of $A$, but this is not satisfied in the ring $\mathbb{Z}\left\langle X,Y,Z\right\rangle / \left(X^2,YZ-1,ZY-1\right)$ (the ideal is two-sided). The counterexample is $u=Y$, $a=X$, $v=1$ (while $a=X$ is nilpotent, $ua=YX$ isn't).
The same counterexample proves that
"every unital ring homomorphism $A\to B$ with $B\neq 0$ maps $a$ to an element neither left-invertible nor right-invertible $\Longleftrightarrow$ $a$ is nilpotent"
must be wrong.
A: (Rephrased) If $R$ is a ring, let $J_R$ be its Jacobson radical, let $N_R$ be the set of nilpotent elements of $R$ (which is an ideal when $R$ is commutative) and let $K_R$ the set of elements $r$ in a the ring $R$ such that $f(r)$ is not a unit for all morphisms $f:R\to S$. Since ring homomorphisms preserve the Jacobson radical, $J_R\subseteq K_R$, and $N_R\subseteq J_R$. Since in general $N_R\subsetneq J_R$, we conclude that in general $N_R\subsetneq K_R$.
