Localization Preserves Integral Closure

Theorem
Let $A \subseteq B$ be an extension of commutative rings with unity.

Let $C$ be the integral closure of $A$ in $B$.

Let $S \subseteq A$ be a multiplicatively closed subset.

Then $C_S$ is the integral closure of $A_S$ in $B_S$, where subscript $S$ indicates the localisation at $S$.

Proof
First we show that $C_S$ is an integral extension of $A_S$.

Let $x \in C_S$, and $\iota$ be the canonical inclusion from a ring to it's localisation.

There exists $s \in S$ such that $sx \in \iota(C)$, say $sx = \iota(c)$.

Since $c \in C$ is integral, there is an equation:


 * $c^n + a_{n-1}c^{n-1} + \cdots + a_1 c + a_0 = 0$

with the $a_i \in A$. Since $\iota$ is a homomorphism, we have:


 * $(sx)^n + \iota(a_{n-1})(sx)^{n-1} + \cdots + \iota(a_1)(sx) + \iota(a_0) = 0$

Multiplying by $s^{-n}$ this gives:


 * $x^n + s^{-1}\iota(a_{n-1})x^{n-1} + \cdots + x^{1-n}\iota(a_1)x + s^{-n}\iota(a_0) = 0$

This is a monic polynomial with coefficients in $A_S$, so $x$ is integral over $A_S$.

Now let $x \in B_S$ be integral over $A_S$.

So we have an equation


 * $x^n + r_{n-1}x^{n-1} + \cdots + r_1 x + r_0 = 0$

Let $t \in S$ be such that $tx \in \iota(B)$, $u_i$ such that $u_ir_i \in \iota(A)$, $i = 0,\ldots, n-1$ and $s = tu_0\cdots u_{n-1}$.

Then we have:


 * $(sx)^n + r_{n-1}s(sx)^{n-1} + \cdots + r_1 s^{n-1}(sx) + r_0s^n = 0$

Now let $sx = \iota(y)$, $y \in B$, and $r_{n-i}s^i = \iota(q_i)$, $i=1,\ldots, n$. So:


 * $\iota(y^n + q_1 y^{n-1} + \cdots + q_{n-1}y + q_n) = 0$

Now $\operatorname{ker}\left(\iota : B \to B_S\right) = \{ x \in B : sx = 0,\text{ some } s \in S \}$. So there is $v \in S$ such that:


 * $(vy)^n + q_1v (vy)^{n-1} + \cdots + q_{n-1}v^{n-1}(vy) + q_n v^n = 0$

Therefore $vy \in B$ is integral over $A$, so $vy \in C$.

Therefore $\iota(vy) \in C_S$, and $\iota(y) \in C_S$.

Thus $x = s^{-1}\iota(y) \in C_S$.