Lemma 10.135.4. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q$ be a prime of $S$. Choose any presentation $S = k[x_1, \ldots , x_ n]/I$. Let $\mathfrak q'$ be the prime of $k[x_1, \ldots , x_ n]$ corresponding to $\mathfrak q$. Set $c = \text{height}(\mathfrak q') - \text{height}(\mathfrak q)$, in other words $\dim _{\mathfrak q}(S) = n - c$ (see Lemma 10.116.4). The following are equivalent

1. There exists a $g \in S$, $g \not\in \mathfrak q$ such that $S_ g$ is a global complete intersection over $k$.

2. The ideal $I_{\mathfrak q'} \subset k[x_1, \ldots , x_ n]_{\mathfrak q'}$ can be generated by $c$ elements.

3. The conormal module $(I/I^2)_{\mathfrak q}$ can be generated by $c$ elements over $S_{\mathfrak q}$.

4. The conormal module $(I/I^2)_{\mathfrak q}$ is a free $S_{\mathfrak q}$-module of rank $c$.

5. The ideal $I_{\mathfrak q'}$ can be generated by a regular sequence in the regular local ring $k[x_1, \ldots , x_ n]_{\mathfrak q'}$.

In this case any $c$ elements of $I_{\mathfrak q'}$ which generate $I_{\mathfrak q'}/\mathfrak q'I_{\mathfrak q'}$ form a regular sequence in the local ring $k[x_1, \ldots , x_ n]_{\mathfrak q'}$.

Proof. Set $R = k[x_1, \ldots , x_ n]_{\mathfrak q'}$. This is a Cohen-Macaulay local ring of dimension $\text{height}(\mathfrak q')$, see for example Lemma 10.135.3. Moreover, $\overline{R} = R/IR = R/I_{\mathfrak q'} = S_{\mathfrak q}$ is a quotient of dimension $\text{height}(\mathfrak q)$. Let $f_1, \ldots , f_ c \in I_{\mathfrak q'}$ be elements which generate $(I/I^2)_{\mathfrak q}$. By Lemma 10.20.1 we see that $f_1, \ldots , f_ c$ generate $I_{\mathfrak q'}$. Since the dimensions work out, we conclude by Proposition 10.103.4 that $f_1, \ldots , f_ c$ is a regular sequence in $R$. By Lemma 10.69.2 we see that $(I/I^2)_{\mathfrak q}$ is free. These arguments show that (2), (3), (4) are equivalent and that they imply the last statement of the lemma, and therefore they imply (5).

If (5) holds, say $I_{\mathfrak q'}$ is generated by a regular sequence of length $e$, then $\text{height}(\mathfrak q) = \dim (S_{\mathfrak q}) = \dim (k[x_1, \ldots , x_ n]_{\mathfrak q'}) - e = \text{height}(\mathfrak q') - e$ by dimension theory, see Section 10.60. We conclude that $e = c$. Thus (5) implies (2).

We continue with the notation introduced in the first paragraph. For each $f_ i$ we may find $d_ i \in k[x_1, \ldots , x_ n]$, $d_ i \not\in \mathfrak q'$ such that $f_ i' = d_ i f_ i \in k[x_1, \ldots , x_ n]$. Then it is still true that $I_{\mathfrak q'} = (f_1', \ldots , f_ c')R$. Hence there exists a $g' \in k[x_1, \ldots , x_ n]$, $g' \not\in \mathfrak q'$ such that $I_{g'} = (f_1', \ldots , f_ c')$. Moreover, pick $g'' \in k[x_1, \ldots , x_ n]$, $g'' \not\in \mathfrak q'$ such that $\dim (S_{g''}) = \dim _{\mathfrak q} \mathop{\mathrm{Spec}}(S)$. By Lemma 10.116.4 this dimension is equal to $n - c$. Finally, set $g$ equal to the image of $g'g''$ in $S$. Then we see that

$S_ g \cong k[x_1, \ldots , x_ n, x_{n + 1}] / (f_1', \ldots , f_ c', x_{n + 1}g'g'' - 1)$

and by our choice of $g''$ this ring has dimension $n - c$. Therefore it is a global complete intersection. Thus each of (2), (3), and (4) implies (1).

Assume (1). Let $S_ g \cong k[y_1, \ldots , y_ m]/(f_1, \ldots , f_ t)$ be a presentation of $S_ g$ as a global complete intersection. Write $J = (f_1, \ldots , f_ t)$. Let $\mathfrak q'' \subset k[y_1, \ldots , y_ m]$ be the prime corresponding to $\mathfrak qS_ g$. Note that $t = m - \dim (S_ g) = \text{height}(\mathfrak q'') - \text{height}(\mathfrak q)$, see Lemma 10.116.4 for the last equality. As seen in the proof of Lemma 10.135.3 (and also above) the elements $f_1, \ldots , f_ t$ form a regular sequence in the local ring $k[y_1, \ldots , y_ m]_{\mathfrak q''}$. By Lemma 10.69.2 we see that $(J/J^2)_{\mathfrak q}$ is free of rank $t$. By Lemma 10.134.16 we have

$J/J^2 \oplus S_ g^ n \cong (I/I^2)_ g \oplus S_ g^ m$

Thus $(I/I^2)_{\mathfrak q}$ is free of rank $t + n - m = m - \dim (S_ g) + n - m = n - \dim (S_ g) = \text{height}(\mathfrak q') - \text{height}(\mathfrak q) = c$. Thus we obtain (4). $\square$

Comment #716 by Keenan Kidwell on

In the last paragraph, in the first difference of heights, $\mathfrak{q}$ and $\mathfrak{q}^{\prime\prime}$ should be switched, and in the second difference of heights, $\mathfrak{q}$ and $\mathfrak{q}^\prime$ should be switched.

Comment #2914 by Dario Weißmann on

Could you add a line why $\text{dim}_{\mathfrak{q'}}(k[x_1,\dots,x_n])=n$? It follows from (the proof of) 10.133.2, but I didn't see it right away. Or does this appear as an example or remark somewhere?

Comment #2943 by on

@#2914: I'm sorry, but I don't know which equality you mean.

Comment #2944 by Dario Weißmann on

It's implicitly in the statment of the lemma. Lemma 10.115.4 gives us $\text{dim}_{\mathfrak{q}}(S)=\text{dim}_{\mathfrak{q}'}(k[x_1,\dots,x_n])-c$. But it is stated as '=n-c'. This is used in the proof $(2),(3),(4) \Rightarrow (1)$.

Comment #2945 by on

@#2944: The lemma cited in the statement tells us that the difference $c$ in the heights is the codimension of $\Spec(S)$ inside the spectrum of the polynomial ring. OK, perhaps you are referring to the fact that the spectrum of the polynomial algebra in $n$ variables has dimension $n$ and that the same remains true for any nonempty open subset: namely, you can center your coordinate axes anywhere you want (OK, I understand that this does not work if the field is finite -- but the picture is clear). In fact, in general the dimension does not change when passing to nonempty opens of irreducible spectra of type algebras over fields; this follows from Lemma 10.116.1, or Lemma 10.114.4, or Lemma 10.114.5. I have tried several times in courses taught and in the Stacks project to explain succintly, precisely, and completely all the facts algebraic geometers need on the dimension theory of finite type algebras over fields. For example see Section 33.20 for an attempt in the setting of geometry. But I feel I have failed each time. So maybe somebody else should perhaps write a bit more about this and then we can refer the reader to that?

Comment #2946 by Dario Weißmann on

Yes, I was referring to that. I'm happy with either of those references. A section like that would be very nice, however I'm not great with intuition. Thank you for the picture (and the references)!

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