Normalization of the cusp semigroup ring

The element $t$ is nonzero in the domain $k[t]$, and then $t^2\mathrm{\ne }0$ because a nonzero element of an integral domain is not a zero divisor by [L5]; so $t=t^3\mathrm{/}{t}^2$ is a quotient of two elements of $A$ with nonzero denominator, and $t\in \mathrm{Frac}(A)$ by [L1]. Consequently $k[t]=A[t]\subseteq \mathrm{Frac}(A)$, and since $A\subseteq k[t]\subseteq \mathrm{Frac}(A)$ with $\mathrm{Frac}(A)$ a field, $k(t)=\mathrm{Frac}(k[t])\subseteq \mathrm{Frac}(A)$ by [L2]; the reverse inclusion is $A\subseteq k[t]$. Hence $\mathrm{Frac}(A)=k(t)$.

The element $t$ is a root of the monic polynomial $T^2-t^2\in A[T]$, so $t$ is integral over $A$ by [L4]. Every element of $k[t]$ is a finite sum ${\sum }_n{a}_n{t}^n$ with $a_n\in k$, and reducing exponents modulo $2$ via $t^2=t^2\in A$ expresses it as an $A$-linear combination of $1$ and $t$: that is, $k[t]=A+At$ is generated as an $A$-module by the two elements $1,t$. Being a finite $A$-module generated by integral elements it is integral over $A$, so every element of $k[t]$ is integral over $A$.

Suppose $z\in k(t)=\mathrm{Frac}(A)$ is integral over $A$. A monic equation for $z$ over $A$ has coefficients in $A\subseteq k[t]$, so $z$ is integral over $k[t]$; since $k[t]$ is integrally closed in its fraction field $k(t)$ by [L3] and [L2], such a $z$ lies in $k[t]$. Conversely every element of $k[t]$ is integral over $A$ by step 1.2. Therefore the integral closure of $A=k[t^2,t^3]$ in $k(t)$ is exactly $k[t]$, which is the finite $A$-module $A+At$.