Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

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.

share|improve this answer
    
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.

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.