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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?

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3  
In the commutative case this is easy, right? a can't be contained in any prime ideal (consider A-->Frac(A/P)) and hence a is nilpotent by a standard result in ring theory. In the non-commutative case though I'm not so sure: what about (0 1;0 0) in M_2(C)? M_2(C) has no non-trivial 2-sided ideals so f must be injective and although this isn't a proof, I'm surely not far off. –  Kevin Buzzard Feb 3 '10 at 19:06
    
I think "a can't be contained in any prime ideal" should be "a is contained in every prime ideal" in your answer. –  darij grinberg Feb 3 '10 at 19:11
    
Yes, thanks darij. The argument is still OK but the "typo" is rather grotesque :-) –  Kevin Buzzard Feb 3 '10 at 20:24
    
Yes and thanks to Matt too: I mean (1 0;0 0). Oh dear what a disastrous comment! I should have made it an answer, then I could have edited it! –  Kevin Buzzard Feb 3 '10 at 20:26
    
Pedantic ring theorist speaking: All elements (even nilpotents) can map to units. Just map to the zero ring. Pedantry aside: first of all we want every ideal $I$ of $R$, which contains the left and right annihilators of $a$, to NOT contain $a$ itself. This encompasses the situation of nilpotents and full idempotents. –  Pace Nielsen Feb 3 '10 at 21:24

3 Answers 3

up vote 7 down vote accepted

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.)

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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.

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(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 $N_R\subsetneq K_R$.

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Somehow I doubt your solution. Why is "image under homomorphism never unit" equivalent to "lies in every maximal ideal"? Doesn't the homomorphism $A\to A_{\left\{1,a,a^2,...\right\}}$ (where $A_{\left\{1,a,a^2,...\right\}}$ means localization at the multiplicative subset \left\{1,a,a^2,...\right\}$) contradict this? –  darij grinberg Feb 3 '10 at 19:14
    
"Since ring homomorphisms preserve the Jacobson radical". Wikipedia says that if $f:R\to S$ is a surjective ring homomorphism, then $f(J(R))\subseteq J(S)$, which is somewhat weaker. Are you sure that your answer still works then? –  darij grinberg Feb 3 '10 at 19:25
    
And I still believe that the original question is correct for commutative rings. –  darij grinberg Feb 3 '10 at 19:26
    
Let $R$ be the subring of $S=\mathbb{Q}$ with odd denominators. Then $R$ embeds in $S$, but the Jacobson radical maps to units in $S$. –  Pace Nielsen Feb 3 '10 at 20:30

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