Why are integral elements defined in terms of monic polynomials? Why do we wish to split the cases between non monic polynomials and monic polynomials? I.e, algebraic elements and integral elements. Motivation?
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2Why bother with the concept of algebraic integers in the first place... – Angina Seng Oct 24 '17 at 05:17
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I second @LordSharktheUnknown's comment. What do you want from the concept of integrality? – Youngsu Oct 24 '17 at 18:35
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Try with $\sqrt{2}$ then the ring $\mathbb{Z}[\sqrt{2}]$ is just $\{ a+b\sqrt{2}, (a,b) \in \mathbb{Z}\}$. With $\frac{\sqrt{2}}{2}$ then the ring $\mathbb{Z}[\frac{\sqrt{2}}{2}]$ is $\{\frac{a+b\sqrt{2}}{2^k}, (a,b,k) \in \mathbb{Z}\}$. The difference is that $\frac{\sqrt{2}}{2}$ is not the root of any monic polynomial in $\mathbb{Z}[x]$.
In other words $\mathbb{Z}[\alpha]$ is a finitely generated $\mathbb{Z}$-module iff $\alpha$ is the root of a monic polynomial.
reuns
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Any answer might be a bit arguing from hindsight. One could say that the reason algebraic integers are important is because a lot of theory has been successfully developed using algebraic integers as a concept.
But one way to look at it is that the ring of integers of a number field is a ring which is a finitely generated module over $\mathbb{Z}$, which would not be the case if you included the roots of non-monic polynomials. (@reuns comment above gives an example of this.) Finite-generation is kind of like being discrete, which is an important difference between $\mathbb{Z}$ and $\mathbb{Q}$.
Ted
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$\mathbb{Z}[x]$ and $\mathbb{Z}[\sqrt{2}/2]$ are finitely generated as rings (by $1$ and $x$) but not as $\mathbb{Z}$-module (ie. as abelian groups) – reuns Oct 24 '17 at 21:36
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