If the intersection of all non-zero ideals of an integral domain R is non-zero, show that R is a field. Let $R$ be a commutative ring with unity.  Call $R$ a *-ring if the intersection of all non-zero ideals of $R$ is non-zero.  Note that we consider $R$ itself an ideal.  If $R$ is an integral domain that is a *-ring, show that $R$ is a field.
My idea was to show that a maximal ideal $I$ is the zero ideal, but I'm not really getting anywhere.  Doesn't seem like too hard of a problem, so any hints would be amazing.
 A: Hint: Suppose that the intersection of the principal non-zero ideals is non-zero, i.e. $$0\not= a \in \bigcap_{0 \not=b \in R} bR$$  Note that $a$ is also in the principal ideal $a^2R$.  Use the fact that $R$ is a domain to conclude that $a$ is a unit.  
A: You have a minimal non-zero ideal $I$. By minimality $I=aR$ for any nonzero
$a\in I$. Fix such an $a$. Let $b$ be nonzero, then $ab\ne0$ and $I=abR=aR$.
so $a=abc$ for some $c\in R$ etc.
A: May I present a few clarifications on what I assume is the direction of reasoning that Lord Shark the Unknown intended: as introduced in the above answer $I$ is indeed the minimum in the ordered set of non-zero ideals; let us fix $a \in I\setminus \{0_A\}$
(for linguistic reasons, I prefer to denote my rings with $A, B, C...$).
Remark first of all that by minimality $I$ will be generated by any of its nonzero elements. Consider now an arbitrary $x \in A^{\times}=A \setminus \{0_A\}$; since $A$ is a domain $ax \neq 0_A$ and equally $ax \in Aa=I$; thus it follows that $I=A(ax)=Aa$ and from this, since $a \in Aax$, we infer the existence of $y \in A$ such that $a=axy$; since non-zero elements are cancellable in a domain, this last equation leads to $xy=1_A$, by which we can conclude that every non-zero element is invertible and hence $A$ is actually a field.

P.S. Actually an even stronger claim can be shown to hold true: if $A$ is a nonzero (commutative) integral domain such that the intersection $\displaystyle\bigcap_{x \in A^{\times}}(x) \neq \left(0_A\right)$ of all nonzero principal ideals be nonzero, then $A$ is a field. Indeed, let us denote the intersection in question by $I$ and consider a fixed $a \in I^{\times}\colon=I \setminus \{0_A\}$. Since in particular $I \subseteq \left(a^2\right)$, it follows that $a \in Aa^2$ and thus there exists $b \in A$ such that $a=ba^2$. Simplification by the nonzero element $a$ entails $ab=1_A$, signifying that $a \in \mathrm{U}(A)$ is a unit and subsequently $I=A$. This has the consequence that for any $x \in A^{\times}$ we have $A=(x)$, in other words $A^{\times} \subseteq \mathrm{U}(A)$ -- any nonzero element is invertible -- and hence $A$ is indeed a field.
