Normalization Finiteness for Affine Domains: Examples

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 $A+At$.

The substitution $\sigma$ kills $P$: $\sigma (y{)}^2-\sigma (x{)}^2(\sigma (x)+1)=t^2(t^2-1{)}^2-(t^2-1{)}^2{t}^2=0$, because $\sigma (x)+1=t^2$ and $\sigma (y)=t(t^2-1)$. By [L1] it therefore induces a unique $k$-algebra homomorphism $\phi :A\to k[t]$ with $\phi (\bar{x})=t^2-1$ and $\phi (\bar{y})=t(t^2-1)$, where $\bar{x},\bar{y}$ are the classes of $x,y$.

$\phi$ is injective. By [L2], applied in the polynomial ring $(k[x])[y]$ to the monic polynomial $y^2-x^2(x+1)$ of degree $2$, every element of $A$ has a unique representative $a(x)+b(x)y$ with $a,b\in k[x]$; hence it suffices to show that $\phi (a+by)=a(t^2-1)+t(t^2-1)b(t^2-1)=0$ forces $a=b=0$. The first summand is a polynomial in $t^2$ and the second is $t$ times a polynomial in $t^2$, so comparing the coefficients of $t^{2m}$ and of $t^{2m+1}$ in the sum gives $a(t^2-1)=0$ and $(t^2-1)b(t^2-1)=0$ separately. For a polynomial $c={\sum }_{m\le M}{c}_m{s}^m$ with $c_M\mathrm{\ne }0$, the value $c(t^2-1)={\sum }_m{c}_m(t^2-1{)}^m$ has coefficient $c_M$ at $t^{2M}$ by [L5], since $(t^2-1{)}^m$ is monic of degree $2m$ and all lower terms have degrees $<2M$; hence $c(t^2-1)\mathrm{\ne }0$. Applying this to $c=a$ gives $a=0$. Since $k[t]$ is a domain by [L3] and $t^2-1\mathrm{\ne }0$ (it is monic of degree $2$), $(t^2-1)b(t^2-1)=0$ forces $b(t^2-1)=0$, and the same argument gives $b=0$. So $\phi$ is injective, and $A$ is a domain, a $k$-subalgebra of $k[t]$.

In $\mathrm{Frac}(A)$ one has $\phi (\bar{y})=t\phi (\bar{x})$ with $\phi (\bar{x})=t^2-1\mathrm{\ne }0$, so $t=\bar{y}\mathrm{/}\bar{x}\in \mathrm{Frac}(A)$; hence $k[t]\subseteq \mathrm{Frac}(A)$ and $k(t)=\mathrm{Frac}(k[t])\subseteq \mathrm{Frac}(A)$ by [L3], while $A\subseteq k[t]$ gives the reverse inclusion, so $\mathrm{Frac}(A)=k(t)$. Moreover $\phi (\bar{x})+1=t^2$, so $t$ is a root of the monic polynomial $T^2-(\bar{x}+1)\in A[T]$ and is integral over $A$ by [L6]; and $k[t]=A+At$, because $t^2=\bar{x}+1\in A$ reduces all exponents modulo $2$, so $k[t]$ is a finite $A$-module and every element of $k[t]$ is integral over $A$.

If $z\in k(t)=\mathrm{Frac}(A)$ is integral over $A$, then 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 $k(t)$ by [L4] and [L3], $z\in k[t]$. With step 2.1 this identifies the integral closure of $A$ in $\mathrm{Frac}(A)$ with $k[t]=A+At$, a finite $A$-module.

The parametrisation: a point of the curve with $\bar{x}=0$ has ${\bar{y}}^2=0$, so $\bar{y}=0$ in the field $k$; thus the origin is the unique point with both coordinates zero. Under the parametrisation $x=t^2-1$, $y=t(t^2-1)$ the condition $x=0$ is $t^2=1$, that is $t^2-1=(t-1)(t+1)=0$, so $t=1$ or $t=-1$ by [L3] (a product of two elements of the field $k$ vanishes only if one factor does); these two values are distinct because $\mathrm{char}k\mathrm{\ne }2$ gives $1\mathrm{\ne }-1$, and both give $y=t(t^2-1)=0$. Hence the origin has exactly the two preimages $t=1$ and $t=-1$; the hypothesis $\mathrm{char}k\mathrm{\ne }2$ is used here, since in characteristic $2$ the two values coincide.

$t=t^4\mathrm{/}{t}^3$ is a quotient of two elements of $A$ with $t^3\mathrm{\ne }0$, so $t\in \mathrm{Frac}(A)$ by [L1]; hence $k[t]\subseteq \mathrm{Frac}(A)$, and $k(t)=\mathrm{Frac}(k[t])\subseteq \mathrm{Frac}(A)\subseteq k(t)$ by [L2] because $A\subseteq k[t]$. Thus $\mathrm{Frac}(A)=k(t)$.

The element $t$ is integral over $A$, being a root of the monic polynomial $T^3-t^3\in A[T]$; and $k[t]=A+At+A{t}^2$ is a finite $A$-module, because every polynomial in $t$ is a finite $k$-linear combination of powers of $t$ and the relation $t^3=t^3\in A$ reduces the exponents modulo $3$: $t^{3m+r}=t^{3m}{t}^r=(t^3{)}^m{t}^r$ with $t^3\in A$ and $r\in \{0,1,2\}$. Hence every element of $k[t]$ is integral over $A$.

As in the cusp computation: if $z\in k(t)=\mathrm{Frac}(A)$ is integral over $A$, every monic equation for $z$ over $A$ also has coefficients in $k[t]$, so $z$ is integral over $k[t]$, and since $k[t]$ is integrally closed in its fraction field $k(t)$ by [L3] and [L2], $z\in k[t]$. With step 1.2 this shows that the integral closure of $A=k[t^3,t^4,t^5]$ in $k(t)$ is exactly $k[t]=A+At+A{t}^2$, a finite $A$-module. The conductor ideal $(t^3,t^4,t^5)\subseteq A$ is a proper ideal of $A$ contained in $A$, not in the closure beyond $A$, and plays no role in the identification of the closure.