Affine blowup algebras: normal form, nonzerodivisors, reducedness, domains

An element of the localization $R(I{)}_{at}$ is a class $z\mathrm{/}(at{)}^n$ with $z\in R(I)$ and $n\ge 0$; writing $z={\sum }_i{b}_i{t}^i$, its degree-zero component is $b_n{t}^n\mathrm{/}(at{)}^n$ with $b_n\in I^n$. Hence the degree-zero part $D=(R(I){)}_{(a)}$ consists of the classes of $b{t}^n\mathrm{/}(at{)}^n$ with $n\ge 0$ and $b\in I^n$, so that $A[I\mathrm{/}a]=D$; for $b\in I^n$, $c\in I^m$ one has $b{t}^n\mathrm{/}(at{)}^n=c{t}^m\mathrm{/}(at{)}^m$ in $D$ if and only if $a^k(a^{m}b-a^{n}c)=0$ for some $k\ge 0$, because the defining relation is $(at{)}^k((at{)}^{m}b{t}^n-(at{)}^{n}c{t}^m)=0$ and $t$ is a nonzerodivisor of $A[t]$.

The assignment $\iota \text{\hspace{0.17em}⁣}:D\to A_a$, $b{t}^n\mathrm{/}(at{)}^n\mapsto b\mathrm{/}{a}^n$, is a well-defined injective ring homomorphism: well-defined and injective because $b\mathrm{/}{a}^n=c\mathrm{/}{a}^m$ in $A_a$ holds exactly when $a^k(a^{m}b-a^{n}c)=0$ for some $k\ge 0$, i.e. exactly under the equality criterion of step 1.1; compatible with addition because $(b{t}^n)\mathrm{/}(at{)}^n+(c{t}^m)\mathrm{/}(at{)}^m=(a^{m}b+a^{n}c)t^{n+m}\mathrm{/}(at{)}^{n+m}$ and with multiplication because the product is $bc{t}^{n+m}\mathrm{/}(at{)}^{n+m}$. Its image is the $A$-subalgebra $B=A[b\mathrm{/}a:b\in I]\subseteq A_a$ generated by the fractions $b\mathrm{/}a$: every $b\mathrm{/}a$ is the image of $bt\mathrm{/}(at)$, and conversely every $b\in I^n$ is a finite sum of products of $n$ elements of $I$, so $b\mathrm{/}{a}^n$ is a sum of products of these generators (with coefficients from $A$). Hence $A[I\mathrm{/}a]\cong B$ as $A$-algebras and all computations may be performed in $B\subseteq A_a$.

The element $a$ is invertible in $A_a$, hence a nonzerodivisor on $A_a$, hence a nonzerodivisor on the subring $B$; in particular the image of $a$ in $A[I\mathrm{/}a]$ is a nonzerodivisor. Moreover $aB\subseteq IB$ because $a\in I$, and $IB\subseteq aB$ because for $b\in I$ and $c\mathrm{/}{a}^n\in B$ with $c\in I^n$ one has $b\cdot c\mathrm{/}{a}^n=a\cdot bc\mathrm{/}{a}^{n+1}$ with $bc\in I^{n+1}$; hence $IA[I\mathrm{/}a]=aA[I\mathrm{/}a]$.

The image of $A$ lies in $B$, since $c=ac\mathrm{/}a$ for $c\in A$ and $ac\in I$; thus $A_a=A[1\mathrm{/}a]\subseteq B_a$, and the inclusion $B\subseteq A_a$ induces $B_a\subseteq (A_a{)}_a=A_a$. Hence $B_a=A_a$, that is, $(A[I\mathrm{/}a]{)}_a=A_a$.

Let $\phi \text{\hspace{0.17em}⁣}:C=A[x_1,\dots ,x_r]\mathrm{/}(a{x}_i-a_i)\to B$ send $x_i$ to $a_i\mathrm{/}a$. Its image contains $A$ and every $a_i\mathrm{/}a$; since $I$ is generated by the $a_i$, every $b\in I$ is a finite sum ${\sum }_i{c}_i{a}_i$, whence $b\mathrm{/}a={\sum }_i{c}_i(a_i\mathrm{/}a)$ lies in the image; as the image is an $A$-subalgebra of $A_a$ containing all $b\mathrm{/}a$ it contains $B$. So $\phi$ is surjective.

If $A$ is reduced, then so is the localization $A_a$, and $B\subseteq A_a$ is a subring of a reduced ring, hence reduced: if $x^N=0$ in $B$ then $x=0$ already in $A_a$. If $A$ is a domain and $a\mathrm{\ne }0$, then $A_a\subseteq \mathrm{Frac}(A)$ is a nonzero subring of a field (nonzero because $a\cdot a^{-1}=1$ with $a\mathrm{\ne }0$), hence a domain, and so is its subring $B$.

Localizing $C$ at the multiplicative set generated by $a$ gives $C_a=A_a[x_1,\dots ,x_r]\mathrm{/}(a{x}_i-a_i)$ in which $a$ is a unit and hence $x_i=a_i{a}^{-1}$; the $A_a$-algebra homomorphism $C_a\to A_a$ sending $x_i\mapsto a_i\mathrm{/}a$ is therefore surjective and has zero kernel, because evaluation at $x_i=a_i\mathrm{/}a$ identifies the quotient $A_a[x_i]\mathrm{/}(x_i-a_i\mathrm{/}a)$ with $A_a$. Since the image of $C$ in $A_a$ is $B$ by step 3.3, $\ker (C\to B)=\ker (C\to C_a)$, and by the defining relation of the localization an element $f\in C$ lies in this kernel exactly when $a^{N}f=0$ for some $N\ge 0$, i.e. exactly when $f$ is $a$-power torsion.

The description $B=\{b\mathrm{/}{a}^n:n\ge 0,b\in I^n\}$ of step 2.1 uses only the pair $(I,a)$: no generating family is chosen, and a class in $D$ is compared with another by the equality criterion of step 1.1, so the use of a particular fraction representative is immaterial. A finite generating family $I=(a_0,\dots ,a_r)$ with $a=a_0$ only produces the presentation $C$ of step 3.3, whose kernel is described in step 4.1; the image $B$ and hence the algebra are the same for every such family, with the identity of $B$ as the canonical isomorphism. Finally, for $a^{\mathrm{\prime }}\in I$ with $a^{\mathrm{\prime }}\mathrm{\ne }a$ the algebras $B\subseteq A_a$ and $B^{\mathrm{\prime }}\subseteq A_{a^{\mathrm{\prime }}}$ lie in different localizations of $A$; the statement asserts no identification between them, and the transition maps between the corresponding charts are supplied separately by the standard-chart theorem.