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I'm seeing what appear to be differing definitions of the integral closure of an ideal, and I want to know if they are actually different and what the accepted definition is.

We have $A \subset B$ commutative rings and I an ideal of A.

The first definition I saw is the one from Atiyah MacDonald: x is integral over $I$ if there is a monic polynomial with the rest of the coefficients in I such that x is a root.

The second definition I saw is the one from wikipedia: x is integral over $I$ if there is a polynomial with i'th coefficient in $I^i$ such that x is a root.

Obviously definition 2 implies definition 1. Also, using the standard "determinant trick"/Cayley-Hamilton, it appears we almost get 2 from 1. The difference being that we get definition 2 but for a power of x.

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1 Answers1

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Consider the following example:

$A = \Bbb Z$, $I = 2\Bbb Z$, $B = \Bbb Z[\sqrt[3]2]$, and $x$ is the element $\sqrt[3]2$ in $B$.

  • Is $x$ integral over $I$, according to your first definition?

    Yes, because we have the polynomial $f(t) = t^3 - 2$ which has $t = x$ as root.

  • Is $x$ integral over $I$, according to your second definition?

    No. Proof: assume that there is a polynomial $g(t) = t^d + a_1 t^{d - 1} + \dotsc + a_d$, with $a_i \in 2^i \Bbb Z$, such that $g(x) = 0$. We look at the $2$-adic valuation, $v_2$, of each term $a_i x^{d - i}$.

    Since $v_2(x) = \frac13$, we havee $v_2(a_ix^{d - i}) = \frac {d - i} 3 + v_2(a_{d - i}) \geq \frac {d - i} 3 + i = \frac d 3 + \frac 2 3 i$.

    This implies that for each $i > 0$, the valuation $v_2(a_ix^{d - i})$ is strictly larger than $\frac d 3$.

    However, by assumption we have $g(x) = 0$, which means $x^d = -\sum\limits_{i = 1}^d a_i x^{d - i}$. Now the right hand side has $2$-adic valuation strictly larger than $\frac d 3$, while the left hand side has $2$-adic valuation equal to $\frac d 3$, a contradiction.


Thus we conclude that the two definitions are not equivalent.

Anyway, we don't see either one of them very often in "standard" mathematical texts. What usually appears is just the case where $I = A$, i.e. the integrality over the ring itself. In that case, of course, the two definitions do agree.

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