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I'm trying to straighten out the definition of an integral element... An integral element is not necessarily an integer itself, but is the root of a monic polynomial with integer coefficients? Does that sound right or am I off base?

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In this answer you can find some motivation for the definition of an algebraic integer, including a derivation of the definition from a few simple natural requirements. Unfortunately such motivation is often lacking in textbooks. The same motivation works generally for deriving the definition of an integral algebraic element over any domain (or ring), vs. the classical ground ring $\Bbb Q$. –  Math Gems Jan 20 '13 at 23:49
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Yes, you're correct.

The set $\mathbb{Z}$ of integers forms a ring, i.e. an algebraic structure which allows you to make sense of addition and multiplication; then a real (or complex) number $r$ is integral if it satisfies a polynomial with integer coefficients and with leading coefficient $1$.

Generally, if $R$ and $S$ are commutative rings and $S \subseteq R$, then $r \in R$ is integral over $S$ if it is the root of a polynomial with coefficients in $S$ and with leading coefficient $1$. This is just the special case where $R = \mathbb{R}$ or $\mathbb{C}$ or whatever, and $S=\mathbb{Z}$.

For example, $\frac{1}{2}$ is not integral (over $\mathbb{Z}$) since by the rational root theorem if it is the root of a polynomial with integer coefficients then the leading coefficient cannot be $1$. However it is integral over $\mathbb{Q}$ since it is a root of the polynomial $x-\frac{1}{2}$.

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I understand most of that, but why wouldn't 1/2 be integral over Z? Couldn't you write it as the root of 2x-1 = 0? –  user58437 Jan 20 '13 at 22:29
    
@user58437: $2x-1$ isn't monic. –  Hurkyl Jan 20 '13 at 22:30
    
oh OK got it, thanks –  user58437 Jan 20 '13 at 22:30
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Let $A$ be a commutative ring and $B$ a subring of $A$. Then $a \in A$ is integral over $B$ if there is monic polynomial $f(x) \in B[x]$ such that $f(a)=0$. As I understand it, it is the same idea as an algebraic number $\alpha$ over a field $F$ except we are dealing with rings.

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