Intuition behind the definition of algebraic integers I've just started studying algebraic number theory, and found myself reluctantly admitting the definition of "integrality". As the definition says, an element is called "integral" if it is a root of a monic polynomial. Thus, the integral closure of Z in $Q(\sqrt5)$ consists of elements of form $(a+b\sqrt5)/2$. But I find numbers of the $a+b\sqrt5$ more natural than $(a+b\sqrt5)/2$. So why would mathematicians use $Z[(1+\sqrt5)/2]$ instead of $Z[\sqrt5]$? My guess is that, fields like $Z[\sqrt5]$ are hard to examine whether they are ufd, but on the other hand, fields like $Z[(1+\sqrt5)/2]$ are dedekind domains, which are easier to control(since on a dedekind domain PID and UFD are the same thing.) Can anyone bring me the correct yet easy intuition to integrality?
 A: Here is how Swinnerton-Dyer  motivated the modern definition of integer:
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In order to repair the failure of unique factorization in domains above $\Bbb Z$, which was historically a major motivating factor for the development of classical algebraic number theory, it was seen that integrality was an important feature to have in the transition to ideals. 
In particular for squarefree $D\equiv1$ mod $4$ the number ring $\Bbb Z[\sqrt{D}]$ is "missing" $\frac{1+\sqrt{D}}{2}\in{\cal O}_{\Bbb Q(\sqrt{D})}$, which leads to an ideal that fails to admit a unique factorization into ideals, via $(2,1+\sqrt{D})$.
I provide more detail here.
A: For any hope of unique-factorization we must extend to the integral closure, since non-integrally closed (sub)rings are not UFDs (UFDs are integrally closed via monic case of Rational Root Test).
Here is one way to motivate the notion of an algebraic integer. Suppose that we desire to consider as "integers" some subring $\,\mathbb I\,$ of the field of all algebraic numbers. To be a purely algebraic notion, it cannot distinguish between conjugate roots, so if $\rm\,\alpha,\alpha'\,$ are roots of the same polynomial irreducible over $\rm\,\mathbb Q,\,$ then $\rm\,\alpha\in\mathbb I\iff \alpha'\in\mathbb I\,.\,$ Also we desire that $\rm\,\mathbb I\cap \mathbb Q = \mathbb Z,\,$ equivalently $\,m\mid n\,\ {\rm in} \,\ \Bbb I\iff m\mid n\,\ {\rm in}\ \,\Bbb Z,\,$ so that our notion of algebraic integer is a faithful extension of the notion of a rational integer. Now suppose that $\rm\,f(x)\,$ is the monic minimal polynomial over $\rm\,\mathbb Q\,$ of an algebraic "integer" $\rm\,\alpha\in \mathbb I.\,$ Then $\rm\,f(x) = (x-\alpha)\,(x-\alpha')\,(x-\alpha'')\,\cdots\,$ has coef's in $\rm\,\mathbb I\cap \mathbb Q = \mathbb Z.\,$ Thus the monic minimal polynomial of elements $\in\mathbb I\,$ must have coef's $\in\mathbb Z\,.\,$ Conversely, one easily shows that the set of all such algebraic numbers contains $1$ and is closed under both difference and multiplication, so it forms a ring. Further, the quotient field of $\rm\,\mathbb I\,$ is the field of all algebraic numbers. Hence a few natural hypotheses on the notion of an algebraic integer imply the standard criterion in terms of minimal polynomials.
Since this notion of integer faithfully extends the notion of rational integers, we can employ algebraic integers to deduce results about rational integers. This often results in great simplifications because many diophantine equations become simpler - being "linearized" - when we factor them in algebraic extension fields. For example, see proofs about Pythagorean triples using Gaussian integers, or classical proofs of FLT for small exponents employing algebraic integers.
