Integral Domains and Maximal Ideals I need to show that if $R$ is an integral domain (containing three ideals: $\{0\}$, $I$ and $R$), then $a,b\in I\Rightarrow ab=0$.
I know that since $R$ is an integral domain, $ab=0\Leftrightarrow a=0$ or $b=0$, but does this mean that $a,b\in I\Rightarrow a=0$ or $b=0$, and if so, how can I show this?
 A: You want to show that there is no integral domain $R$ that has exactly three ideals.
Suppose the ideals are $(0),I,R$. Take $a\ne0$, $a\in I$. Then $(a)=I$. If $x\in R$, $x\ne0$, then $ax\in I$ and $ax\ne0$, so $(ax)=I=(a)$.Therefore $a=axy$, so $xy=1$, implying $x$ is invertible. Hence $R$ is a field and has only two ideals.
A: You have already received good elementary answers for the "3 ideals" version of this question.  
But I'd like to add that it has several generalizations, and that this version is quite a "toy" in comparison.What follows is certainly overkill, but I think it's also a good lesson in how far things can be stretched if one is paying attention to how the arguments worked.

The arguments mentioned have an idea like this: "$x^2R$ must be strictly contained in $xR$."  Let's just take that idea and run with it in an integral domain.
Given any ring, you always have that $xR\supseteq x^2R\supseteq x^3R\supseteq\dots$ and if you are in a domain, this has to be either a) equal all throughout or b) strictly decreasing.
For if $x$ is a unit, everything is obviously equal. If $x$ is $0$, everything is obviously equal. If $x$ is not a unit and not zero, and somewhere this chain is not strictly decreasing, we'd have $x^{n+1}R=x^nR$ for some $n$.   From $x^n=x^{n+1}r$ we can cancel (since we're in a domain) $x$'s from both sides, winding up with $1=xr$, and $x$ is a unit (a contradiction). So the chain had to be strictly decreasing, if $x$ is to be a nonunit, nonzero element.
For an integral domain, this means either


*

*All elements are zero or units, and you are looking at a field with exactly 2 ideals; or

*There is a nonzero nonunit element, and you have to have infinitely many ideals. In particular, there's an infinite chain like $xR\supseteq x^2R\supseteq\ldots$.


So not only do you have that 

An integral domain with at most $3$ ideals has exactly two ideals.

you also have 

An integral domain with finitely many ideals has exactly two ideals.

But moreover, even if you suppose there might possibly be infinitely many ideals, 

An Artinian integral domain has exactly two ideals.

Or more generally 

An integral domain which satisfies the descending chain condition (DCC) on principal ideals has exactly two ideals.

But the most general of all is pretty much exactly what I started out with

An integral domain which has the DCC on chains of the form $xR\supseteq x^2R\supseteq x^3R\supseteq\ldots$ has exactly two ideals.

A ring with this condition on chains is called a strongly $\pi$-regular ring.
A: We need to show that no integral domain has exactly three ideals. Indeed, this follows from the following observation. Let $a \in R$ be non-zero and not a unit. Then the ideal generated by $a$ is strictly contained in the ideal generated by $a^2$ (since if $a^2 = ba$ for a unit $b$ we have that $a = b$ is a unit, contradiction).
A: Suppose there were
$a, b \in I \tag 1$
such that 
$ab \ne 0; \tag 2$
consider the set
$aI \subset R; \tag 3$
$aI$ is an ideal since for any two elements
$aj, ak \in aI, \; j, k \in I, \tag 4$
we have
$aj - ak = a(j - k) \in I \tag 5$
by virtue of the fact that
$j - k \in I, \tag 6$
and $I$ is an ideal in $R$; also, for $c \in R$ we have
$c(aI) = (ca)I = (ac)I = a(cI) \subset aI, \tag 7$
again since $I$ is an ideal; the arguments ca. (4)-(7) establish the assertion that $aI$ is an ideal in $R$; it is furthermore clear that
$aI \subset I; \tag 8$
now by virtue of (2) we may write
$0 \ne ab \in I \Longrightarrow aI \ne \{0\}, \tag 9$
and since $R$ has precisely the three ideals $\{0\}$, $I$, and $R$ we find in light of (8)-(9) that
$aI = I; \tag{10}$
thus
$\exists c \in I, \; ac = a \Longrightarrow a(c - 1_R) = ac - a = 0; \tag{11}$
with $R$ an integral domain and $a \ne 0$ (true by virtue of (2)) we thus find
$c - 1_R = 0 \Longrightarrow c = 1_R \Longrightarrow 1_R \in I, \tag{12}$
but now for any $r \in R$,
$r = r1_R \in I \Longrightarrow I = R, \tag{13}$
in contradiction to the assertion that $R$ has precisely the three ideals $\{0\}$, $I$, and $R$; we thus conclude that
$a, b \in I \Longrightarrow ab = 0. \tag{14}$
