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For an integral normal variety that is Noetherian, removing a closed subset of codimension at least two does not create new regular functions. Affine-locally, a normal Noetherian domain is the intersection of its localizations at height-one primes inside its fraction field. The complement still contains every height-one point, so a regular function on it lies in each such localization and therefore in the original ring. On projective space this also follows directly by excluding irreducible polynomial factor of the denominators on each polynomial affine chart.

A discrete valuation ring is a principal ideal domain with exactly one nonzero maximal ideal. Every nonzero element of its fraction field is a unit times a unique integer power of a uniformizer.

For a normal Noetherian ring $A$ that is an integral domain, the integral closure in a finite separable field extension of its fraction field is finite as an $A$-module. An integral field basis by denominator clearing gives a finite lattice. Integral traces place the closure in its finite trace-dual lattice; Noetherianity makes that submodule finitely generated.

A fractional ideal of a number field is a nonzero ${\mathcal{O}}_K$-submodule $I\subset K$ for which $dI\subseteq {\mathcal{O}}_K$ for some nonzero $d\in {\mathcal{O}}_K$.

A generic point is a point whose closure is the whole space. An integral scheme has a unique generic point; on an affine scheme associated to an integral domain, it corresponds to the zero prime ideal. Its local ring is the fraction field of the domain. On every nonempty affine chart of an integral scheme these fields identify with the same function field.

For an integral domain $R$, an integer-valued polynomial is a polynomial over its fraction field that maps $R$ into itself. Its coefficients need not lie in $R$: $X(X-1)/2$ maps $\mathbb{Z}$ into $\mathbb{Z}$. These polynomials form a ring containing $R[X]$.

For a domain $R$ with fraction field $K$, its integral closure is the intersection of all valuation rings of $K$ containing $R$. Integral elements belong to every such ring because valuation rings are integrally closed. For a nonintegral $x$, the ideal $x^{-1}R[x^{-1}]$ is proper, since its containing $1$ would give a monic equation for $x$. Localize at a maximal ideal containing $x^{-1}$ and apply the valuation domination lemma. The resulting ring contains $R$ but not $x$, proving the reverse inclusion.

If $A$ is an integral domain with fraction field $K$ and $L/K$ is finite, a $K$-basis of $L$ can be rescaled to consist of integral elements over $A$. For a monic algebraic equation, choose $a\in A$ clearing the coefficient denominators; replacing $u$ by $au$ multiplies the coefficient of degree $r-j$ by $a^j$, putting it in $A$.

If a field $L$ is integral over a subdomain $A$, every nonzero $a\in A$ has an inverse in $L$ satisfying a monic equation over $A$. Multiplying that equation by $a^{r-1}$ expresses $a^{-1}$ as a polynomial in $a$ with coefficients in $A$. Hence $a^{-1}\in A$, and $A$ is itself a field. The embedding and integrality assumptions are essential; a fraction field is not generally integral over its source domain.

A fraction of invariant elements is fixed by every automorphism. If that fraction is integral over $R^G$, its monic equation also makes it integral over $R$. A normal domain therefore places it in $R$, and fixedness places it in $R^G$. This proves integral closedness of the invariant subring in its own fraction field. The proof works for any group and in every characteristic; no averaging or finiteness of the group is needed for normality.

Write $a=t_1$, $b=t_2$, $c=t_3$, and $A=\mathbb{C}[a,b,c]/(ac-b^2)$. It is the quadratic cone invariant ringwhere the involution negates both $u$ and $v$. The invariant monomials have even total degree and are generated by $u^2,uv,v^2$; reduction by $ac=b^2$ proves the asserted presentation.

To verify the normal variety property, suppose $q\in \mathrm{Frac}(A)$ is integral over $A$. The same monic equation makes $q$ integral over $\mathbb{C}[u,v]$. This polynomial ring is an integrally closed domain, so $q\in \mathbb{C}[u,v]$. Since $q$ belongs to the fraction field of $A$, it is invariant under the involution, and hence belongs to $A$. Thus $X$ is normal. The gradient of its equation is $(c,-2b,a)$, so the Jacobian criterion shows that the origin $o$ is its only singular point of an algebraic variety. In particular $X$ is not smooth.

Take the prime Weil divisor $D=V_X(a,b)$ and $x=o$. We show that $D$ is not a Cartier divisor at $o$. In the local ringits prime ideal is $I=(a,b)R$. The quotient $I/\mathfrak{m}I$ has dimension of a vector space $2$ over $\mathbb{C}$: $a,b$ generate it, and they are independent since $\mathfrak{m}I\subset {\mathfrak{m}}^2$ while the relation $ac-b^2$ has no linear term. By Nakayama lemma, $I$ cannot be generated by one element.

On a normal variety, the ideal of an effective prime Weil divisor is the divisorial ideal of functions with order at least one along that divisor and order at least zero along every other prime Weil divisor. If $D$ were a Cartier divisor at $o$, a local defining rational function $h$ would identify this ideal with $hR$, because an integrally closed domain is the intersection of its height-one localizations at a prime ideal inside its fraction field. That would make $I$ principal, contradicting the calculation. Therefore $D$ has a nonzero class in the local divisor class group.

Now let $D^{\prime }=D+\mathrm{div}(q)$ be any linearly equivalent Weil divisor. If $o$ were outside its support of a Weil divisor, there would be an open subset containing $o$ on which $D^{\prime }$ is zero. On that neighbourhood $D=-\mathrm{div}(q)$ would be a principal Weil divisor, hence a Cartier divisor, which is impossible. Therefore this point is unavoidable in every representative:The argument does not assume that $D^{\prime }$ is effective. For comparison, $\mathrm{div}(a)=2D$ because on the chart $c\ne 0$, $a=b^2/c$; the obstruction has order two, as in the Divisor class group of an A-type surface singularity.