Let $R$ be a DVR and $S$ a finitely generated flat $R$-algebra. How can I prove that there is a subalgebra $C$ of $S$ such that there is a finite and injective morphism $R[t_1,\dots, t_d] \rightarrow C$ and such that $ \operatorname{Spec} S \rightarrow \operatorname{Spec}C$ is an open immersion?

  • $\begingroup$ YACP: Thank you for reminding me! $\endgroup$ – Heidar Svan Apr 1 '13 at 8:46

Denote by $X=\mathrm{Spec}(S)$. It is enough to show

There exists a quasi-finite and dominant morphism $\pi : X\to \mathbb A^d_R$.

Indeed, Zariski's main theorem then implies that $\pi$ factors into an open immersion $X\to X'$ follow by a finite and surjective morphism $X'\to \mathbb A^d_R$. This implies that $X'$ is affine and $O_{X'}(X')$ is the $C$ you are after.

To construct such a morphism $\pi$, I need to work with projective schemes. By the usual process of projectivization, we can write $X=D_+(f_0)$ in a projective scheme $\overline{X}=\mathrm{Proj}(B)$ with $f_0$ homogeneous and $X$ dense in $\overline{X}$. Denote by $s$ the closed point of $\mathrm{Spec}(R)$ and by $X(s)$ the Zariski closure of $X_s$ in $(\overline{X})_s$ (in general, $X_s$ may not be dense in $(\overline{X})_s$).

Let $d=\dim X_s=\dim X_\eta$ (generic fiber). As $X_s\cap V_+(f_0)=\emptyset$, we have $$\dim (X(s)\cap V_+(f_0))\le d-1, \quad \dim (\overline{X}_\eta\cap V_+(f_0))\le d-1.$$ Using graded avoidance lemma, we can find a homogeneous $f_1\in B$ such that $V_+(f_1)$ doesn't contain any generic point of $X(s)\cap V_+(f_0)$ and of $\overline{X}_\eta\cap V_+(f_0)$. Therefore $$\dim (X(s)\cap V_+(f_0)\cap V_+(f_1))\le d-2, \quad \dim (\overline{X}_\eta\cap V_+(f_0)\cap V_+(f_1))\le d-2.$$ Repeating the same argument, we find $f_1, \dots, f_d$ such that $$X(s)\cap (\cap_{0\le i\le d} V_+(f_i))=\emptyset, \quad \overline{X}_\eta\cap (\cap_{0\le i\le d}V_+(f_i))=\emptyset.$$ Replacing each $f_i$ be a suitable positive power, we can suppose the $f_i$ all have the same degree. Then define a rational map $$\pi: \overline{X} --\to \mathbb P^d_R, \quad x\mapsto [f_0(x),\dots, f_d(x)].$$ It is regular on $\overline{X}\setminus (\cap_{0\le i\le d} V_+(f))\supseteq \overline{X}_\eta\cup X(s)$. On $X(s)$, $\pi: X(s)\to \mathbb P^d_S$ is projective because $X(s)$ is projective, and $\pi^{-1}(D_+(T_i))=X(s)\cap D_+(f_i)\to D_+(f_i)$ is at the same time an affine morphism and a projective morphism. It is then finite. Similarly, $\pi$ is finite on $\overline{X}_\eta$. As $X\subset \overline{X}_\eta\cup X(s)$, we get a quasi-finite morphism $X\to \mathbb P^d_R$. By construction, the image of $X$ is in $D_+(T_0)=\mathbb A^d_R$ because $X=D_+(f_0)=\pi^{-1}(D_+(T_0)$. Finally, the morphism is dominant by comparing the dimensions.

  • $\begingroup$ Dear @QiL'8, I don't know exactly what means an open immersion, but in the classical case of normalization the extension is finite. Are these two concepts equivalent? If not, how can one translate "open immersion" in terms of commutative algebra? $\endgroup$ – user26857 Mar 23 '13 at 21:38
  • $\begingroup$ @YACP: an open immersion can be seen as a special type of localization (e.g., $S=C_f$ with $f\in C$). Over a ring, it is impossible to have finite instead of quasi-finite (more or less: quasi-finite = open in something finite), e.g. $\mathbb Z[1/2]$ will never be finite over a subring of finite type over $\mathbb Z$. $\endgroup$ – user18119 Mar 23 '13 at 23:47
  • $\begingroup$ Dear QiL, I'm a bit confused by your last sentence. Isn't $\mathbb Z[1/2] = \mathbb Z[x]/(2x - 1)$ already finite type over $\mathbb Z$? Regards, $\endgroup$ – Matt E Mar 24 '13 at 23:10
  • $\begingroup$ Dear @MattE: you are right. I wanted to say $\mathbb Z[1/2]$ is never finite over a subring isomorphic to a polynomial ring over $\mathbb Z$, so the Noether Normalization Lemma can't be translated verbatim for rings. This example is cheap because $\mathrm{Spec}(\mathbb Z[1/2])$ is not surjective over $\mathrm{Spec}(\mathbb Z)$, but it is easy to find an example with surjectivity. $\endgroup$ – user18119 Mar 25 '13 at 8:59
  • $\begingroup$ The existence of a quasi-finite dominant morphism $X\to\mathbb A^d_R$ still holds if $R$ is a ring of integers or a ring of regular functions of a regular affine curve over a finite field, but this requires some extra works. $\endgroup$ – user18119 Mar 25 '13 at 9:05

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