Why do we want Dedekind rings to be integral closed? As I understand, the idea of Dedekind domains is motivated by the wish to factorize ideals into prime ideals.
Dedekind rings are supposed to:


*

*be noetherian, which makes sense because that ensures that the factorization is finite;

*have every prime ideal to be a maximal ideal, which makes sense because we want to factorize into prime ideals, so they need to be "very" big.

*integral closed.
Can anybody give me an intuitive idea about why we need that property?
 A: One of the things that you want in Dedeking rings is that you have nice ideal factorisation properties. In particular, multiplication by a nonzero ideal should be injective.
Suppose $R$ is a commutative ring. Let us look at an element $x/y$ of its field of fractions who happens to be a solution of a monic polynomial equation :
Let $x,y \in R$ and $a_k \in R$ for $k=0 \ldots n-1$ and suppose the equation
$x^n = \sum a_k x^k y^{n-k}$ is true.
Now consider $I = (x,y)$, the ideal generated by $x$ and $y$. Then, 
$I^n = (x^n,x^{n-1}y, \ldots, xy^{n-1},y^n)$. But, because of that equation, the first generator is superfluous, and so $I^n = (x^{n-1}y,\ldots, xy^{n-1},y^n)$.
If you look at this ideal carefully, you get $I^n = (y)I^{n-1}$.
Now, if you want your ideals to have nice factorization properties, and if $I$ is nonzero you would want $(y) = I$. Since $I = (x,y)$, this is equivalent to $x \in (y)$, so that there exists a $z \in R$ such that $x=yz$, or also that $x/y \in R$.
Therefore if you want nice ideal factorisations, you need $R$ to be integrally closed.
(checking that a few ideals factor as they should is also what I do when I want to check if a ring is integrally closed)
Also, wikipedia lists various alternative definitions, some of which place the focus on ideal factorization :
"every proper ideal factors into primes"
"every nonzero fractional ideal is invertible"
Those may be more to your liking.
A: Integral closure is needed to get factorization of ideals into products of prime ideals.
Consider the ring $R=k[t^2,t^3]$ where $k$ is some field and $t$ is an indeterminate. The ideal $P=(t^2,t^3)$ is prime, it consists of those polynomials in $R$ that vanish at $t=0$. It is also a maximal ideal because $R/P\simeq k$ is a field. 
We see that the ideal $I=(t^3,t^4)$ is contained in $P$. Yet it is not any power of $P$ either ($P^2=(t^4,t^5)$ is already a proper subset of $I$). It follows easily that $I$ is not a product of prime ideals of $R$.
Of course, moving to the integral closure $\overline{R}=k[t]$ in the field of fractions of $R$ fixes this problem. Then the relevant prime ideal is $\mathfrak{p}=(t)$ and both $(t^2,t^3)=(t^2)$ and $(t^3,t^4)=(t^3)$ become powers of $\mathfrak{p}$.
Similar things happen with non-maximal orders of number fields.

(Extras) I used the ring $R$ as an example because it lets me point out the following connection. Namely, the ring $R$ is isomorphic to the coordinate ring $\Gamma(C)$ of the plane curve $C:y^2=x^3$,
$$\Gamma(C)=k[x,y]/(y^2-x^3),$$
the isomorphism given by $y\mapsto t^3$, $x\mapsto t^2$. Here the integral non-closure shows up geometrically in the cusp at the origin $O=(0,0)$. By calculating the partial derivatives of $y^2-x^3$ at the origin you see that the origin is not a smooth point (= a point where implicit function theorem can be applied). It is a fact that a plane curve defined by a polynomial equation has no singular points if and only if its coordinate ring is integrally closed (when it becomes a Dedekind domain).
A: I found a prove in Larry Crove's Algebra. Integral closure gives us an inverse for every prime ideal:

