Integrality and integral closure commute with localisation

If $b$ satisfies $b^n+a_{n-1}{b}^{n-1}+\cdots +a_0=0$ with $a_i\in A$, then the same identity in $S^{-1}B$ reads $(b\mathrm{/}1{)}^n+(a_{n-1}\mathrm{/}1)(b\mathrm{/}1{)}^{n-1}+\cdots +a_0\mathrm{/}1=0$. By [L1], this makes $b\mathrm{/}1$ integral over $S^{-1}A$.

Conversely, assume $b\mathrm{/}1$ is integral over $S^{-1}A$. Choose a monic relation $(b\mathrm{/}1{)}^n+(a_{n-1}\mathrm{/}{s}_{n-1})(b\mathrm{/}1{)}^{n-1}+\cdots +a_0\mathrm{/}{s}_0=0$ with $a_i\in A$ and $s_i\in S$. Let $t:=s_0\cdots s_{n-1}$ and put $y:=tb$. Multiplying the relation by $t^n$ gives $(y\mathrm{/}1{)}^n+c_{n-1}(y\mathrm{/}1{)}^{n-1}+\cdots +c_0\mathrm{/}1=0$ with each $c_i\in A$. Hence [L4] gives some $u\in S$ with $u(y^n+c_{n-1}{y}^{n-1}+\cdots +c_0)=0$ in $B$. For $s:=ut$ and $z:=sb=uy$, multiplying that equation by $u^{n-1}$ yields $z^n+c_{n-1}u{z}^{n-1}+\cdots +c_0{u}^n=0$, a monic equation over $A$. So $sb$ is integral over $A$.

Now assume $A$ is a domain, $S\subseteq A\setminus \{0\}$, $K\mathrm{/}\mathrm{Frac}(A)$ is a field extension, and $\bar{A}$ is the integral closure of $A$ in $K$. If $x=a\mathrm{/}s$ with $a\in \bar{A}$ and $s\in S$, then $a$ is integral over $A$, so step 1.1 makes $x$ integral over $S^{-1}A$.

Conversely, let $x\in K$ be integral over $S^{-1}A$. Step 1.2 gives $s\in S$ with $sx$ integral over $A$, so [L2] gives $sx\in \bar{A}$. Therefore $x=(sx)\mathrm{/}s\in S^{-1}\bar{A}$. Combining this with step 2.1 proves that the integral closure of $S^{-1}A$ in $K$ is exactly $S^{-1}\bar{A}$.