Example of a commutative ring with identity with two ideals whose product is not equal to their intersection I need a specific example of a commutative ring with identity, and two ideals in the ring whose product is not equal to their intersection.
I know that for two such ideals I and J, IJ = I ∩ J if I + J = R. I have seen some results about what conditions guarantee a ring will have ideals whose intersections are not equal to their products, but I didn't really understand the terminology of those results well enough to come up with a specific example.
 A: There is a very logical reason why one should require $I+J=R$ to say that $IJ=I\cap J$. 
A Dedekind domain $R$ is a ring such that every non-zero proper (all the ideals in the following discussion are assumed non-zero and proper unless stated otherwise) ideal $I$ of $R$ has a unique factorization into prime ideals $I=\mathfrak{p}_1^{e_1}\cdots\mathfrak{p}_m^{e_m}$. Any PID is a Dedekind domain, as well as any number ring (the integral closure of $\mathbb{Z}$ in a finite extension of $\mathbb{Q}$) such as $\displaystyle \mathbb{Z}\left[\frac{1+\sqrt{-15}}{2}\right]$.
In these rings, the usual ideal-theoretic operations can be almost completely analogized to element-theoretic operations in a UFD. This is largely because of the fact that in a Dedekind domain $R$, one has that $I\subseteq J$ if and only if $J\mid I$ (i.e. there exists some other ideal $J'$ such that $I=JJ'$). From this one can show that the intersection and sum of ideals has a very familiar form. In particular, the intersection of ideals $I\cap J$ plays the role of least common multiple $\text{lcm}(I,J)$. Namely, if 
$$I=\mathfrak{p}_1^{e_1}\cdots\mathfrak{p}_m^{e_m}\qquad J=\mathfrak{p}_1^{f_1}\cdots\mathfrak{p}_m^{f_m}$$
(where $e_m$ and $f_m$ are integers, possibly zero) then 
$$I\cap J=\mathfrak{p}_1^{\max\{e_1,f_1\}}\cdots\mathfrak{p}_m^{\max\{e_m,f_m\}}=\text{lcm}(I,J)$$
Similarly, the sum $I+J$ plays the role of the greatest common divisor $\text{gcd}(I,J)$ so that, in particular if $I$ and $J$ are as above, then 
$$I+J=\mathfrak{p}_1^{\min\{e_1,f_1\}}\cdots\mathfrak{p}_m^{\min\{e_m,f_m\}}=\text{gcd}(I,J)$$
So, now we can rephrase your question as 

When does $\text{lcm}(I,J)$ equal $IJ$?

Well, analogizing the case of UFDs, we know what the answer should be: precisely when $\text{gcd}(I,J)=(1)$ (note that the ideal $R=(1)$ plays the role of identity element with respect to multiplication).
But, the above phrasing tells us that $\text{gcd}(I,J)=I+J$. Thus, $I\cap J=IJ$ should occur precisely when $I+J=R$. 
Not only does this give us intuition about why we should need $I+J=R$, but in fact allows us to produce many counterexamples. In particular, the above actually shows that $I\cap J=IJ$ if and only if $I+J=R$. So, taking any two non-coprime ideals $I$ and $J$ of a Dedekind domain will produce an example of ideals such that $IJ\subsetneq I\cap J$. 
In particular, if we take the Dedekind domain $\mathbb{Z}$, the way we factor ideals into prime ideals is simple. All the non-zero ideals of $\mathbb{Z}$ are of the form $(a)$ where $a\in\mathbb{N}$ and $a\geqslant 2$. But, we can factor $a$ into the product of primes $a=p_1^{e_1}\cdots p_m^{e_m}$ and then 
$$(a)=(p_1)^{e_1}\cdots (p_m)^{e_m}$$
is the factorization of $(a)$ into the product of prime ideals. So, we must produce two non-coprime ideals which, by what we just said, amounts to producing two non-coprime elements of $\mathbb{Z}$. Let's take $a=6$ and $b=15$. Then,
$$(a)\cap (b)=\text{gcd}((2)(3),(3)(5))=(2)(3)(5)=(30)$$
but
$$(a)(b)=((2)(3))((3)(5))=(2)(3)^2(5)=(90)$$
So, there's one example. 
Let's produce a slightly more sophisticated example. The ring $\displaystyle R=\mathbb{Z}\left[\frac{1+\sqrt{-15}}{2}\right]$ is a Dedekind domain, but not a PID (it has class number $2$ if that means anything to you). But, just like the above all we have to do is produce three prime ideals $\mathfrak{p}_1$, $\mathfrak{p}_2$, and $\mathfrak{p}_3$ and consider the ideals $I=\mathfrak{p}_1\mathfrak{p}_2$ and $J=\mathfrak{p}_2\mathfrak{p}_3$. So, here are three prime ideals of $R$ (I leave it to you to verify their primality):
$$\mathfrak{p}_1=\left(2,\frac{1+\sqrt{-15}}{2}+1\right),\quad \mathfrak{p}_2=\left(3,\frac{1+\sqrt{-15}}{2}\right),\quad \mathfrak{p}_3=\left(5,\frac{1+\sqrt{-15}}{2}\right)$$
So, then the ideals 
$$I=\mathfrak{p}_1\mathfrak{p}_2=\left(6,3+\frac{1+\sqrt{-15}}{2}\right),\quad J=\mathfrak{p}_2\mathfrak{p}_3=\left(\frac{1+\sqrt{-15}}{2}\right)$$
Should give us counterexamples. Indeed
$$IJ=\left(30,3\frac{1+\sqrt{-15}}{2}+15\right)$$
and
$$I\cap J=\left(1+\sqrt{-15}\right)$$
and you can check directly that $IJ\subsetneq IJ$, which is what we knew had to happen.
But, as I said, the fact that $IJ=I\cap J$ if and only if $I+J=R$ for a Dedekind domain $R$ allows one to cook up infinitely many examples.
A: Hint: Consider the ring $R=\mathbb{Z}$ and the ideal $I=2\mathbb{Z}$.
A: Hint: Come up with an example $I \subset \mathbb{Z}$ when $I^2 \neq I$. Any ideal $I$ should do, except the zero ideal and $\mathbb{Z}$.
A: Let $p$ be a prime number.
Consider the ideals $p\mathbb{Z}$ and $p^2\mathbb{Z}$ in the ring $\mathbb{Z}$.
