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Let $D$ be a domain which is not a field. If there exists $b \in D - \{0\}$ such that for all $a \in D - \{0\}$, $a|b$, then is it true that $b = 0$?

This is certainly true if one has a UFD with infinitely many irreducibles (that are not associates) or a Jacobson domain which is not a field (or more generally a domain where the intersection of all non-zero prime ideals is the zero ideal), but I can't seem to be able to prove this for a domain in general or find a counterexample.

If this statement is false for an arbitrary domain, then is there some additional weak hypothesis on the domain which makes the above statement true?

I was actually trying to answer a question asked yesterday on MSE, which was the following:

If $D$ is a domain which is not a field and $Q = Frac(D)$, then $Hom_D(Q,D) = \{0\}$. Note that if $\varphi$ is any such $D$-linear map, then for all $a \in D- \{0\}$, $a\varphi(1/a) = \varphi(a/a) = \varphi(1)$. Thus, I get for all $a \in D - \{0\}$, $a|\varphi(1)$, and I want to conclude that $\varphi(1) = 0$, but cannot.

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up vote 5 down vote accepted

Suppose $b \ne 0$.

Then it is a unit since $b^2k = b$ for some $k$ so $bk = 1$ by cancellation.

Now let $a \in D$, nonzero. Then $ak = b$ so, multiplying by the inverse of $b$ we get that $a$ is a unit.

Thus $D$ is a field.

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Of course, thanks. Not my brightest moment. – Rankeya Nov 29 '12 at 3:11

Hint $\rm\,\ b^2\mid b\:\Rightarrow\: b\mid 1\:$ so $\rm\:a\mid b\mid 1\Rightarrow\:a\:$ unit, so $\rm\:D\:$ is a field.

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Note: if $R$ is a domain and not a field and $Q$ is a divisible $R$ module, then $Hom_{R}(Q, R)=0$. A similar argument as above works.

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Thanks, now that I know that b is a unit, I know how to answer the question I was trying to solve yesterday. – Rankeya Nov 29 '12 at 4:28

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