Lemma 10.135.7. Let $k$ be a field. Let $S$ be a local $k$-algebra essentially of finite type over $k$. The following are equivalent:

1. $S$ is a complete intersection over $k$,

2. for any surjection $R \to S$ with $R$ a regular local ring essentially of finite presentation over $k$ the ideal $\mathop{\mathrm{Ker}}(R \to S)$ can be generated by a regular sequence,

3. for some surjection $R \to S$ with $R$ a regular local ring essentially of finite presentation over $k$ the ideal $\mathop{\mathrm{Ker}}(R \to S)$ can be generated by $\dim (R) - \dim (S)$ elements,

4. there exists a global complete intersection $A$ over $k$ and a prime $\mathfrak a$ of $A$ such that $S \cong A_{\mathfrak a}$, and

5. there exists a local complete intersection $A$ over $k$ and a prime $\mathfrak a$ of $A$ such that $S \cong A_{\mathfrak a}$.

Proof. It is clear that (2) implies (1) and (1) implies (3). It is also clear that (4) implies (5). Let us show that (3) implies (4). Thus we assume there exists a surjection $R \to S$ with $R$ a regular local ring essentially of finite presentation over $k$ such that the ideal $\mathop{\mathrm{Ker}}(R \to S)$ can be generated by $\dim (R) - \dim (S)$ elements. We may write $R = (k[x_1, \ldots , x_ n]/J)_{\mathfrak q}$ for some $J \subset k[x_1, \ldots , x_ n]$ and some prime $\mathfrak q \subset k[x_1, \ldots , x_ n]$ with $J \subset \mathfrak q$. Let $I \subset k[x_1, \ldots , x_ n]$ be the kernel of the map $k[x_1, \ldots , x_ n] \to S$ so that $S \cong (k[x_1, \ldots , x_ n]/I)_{\mathfrak q}$. By assumption $(I/J)_{\mathfrak q}$ is generated by $\dim (R) - \dim (S)$ elements. We conclude that $I_{\mathfrak q}$ can be generated by $\dim (k[x_1, \ldots , x_ n]_{\mathfrak q}) - \dim (S)$ elements by Lemma 10.135.6. From Lemma 10.135.4 we see that for some $g \in k[x_1, \ldots , x_ n]$, $g \not\in \mathfrak q$ the algebra $(k[x_1, \ldots , x_ n]/I)_ g$ is a global complete intersection and $S$ is isomorphic to a local ring of it.

To finish the proof of the lemma we have to show that (5) implies (2). Assume (5) and let $\pi : R \to S$ be a surjection with $R$ a regular local $k$-algebra essentially of finite type over $k$. By assumption we have $S = A_{\mathfrak a}$ for some local complete intersection $A$ over $k$. Choose a presentation $R = (k[y_1, \ldots , y_ m]/J)_{\mathfrak q}$ with $J \subset \mathfrak q \subset k[y_1, \ldots , y_ m]$. We may and do assume that $J$ is the kernel of the map $k[y_1, \ldots , y_ m] \to R$. Let $I \subset k[y_1, \ldots , y_ m]$ be the kernel of the map $k[y_1, \ldots , y_ m] \to S = A_{\mathfrak a}$. Then $J \subset I$ and $(I/J)_{\mathfrak q}$ is the kernel of the surjection $\pi : R \to S$. So $S = (k[y_1, \ldots , y_ m]/I)_{\mathfrak q}$.

By Lemma 10.126.7 we see that there exist $g \in A$, $g \not\in \mathfrak a$ and $g' \in k[y_1, \ldots , y_ m]$, $g' \not\in \mathfrak q$ such that $A_ g \cong (k[y_1, \ldots , y_ m]/I)_{g'}$. After replacing $A$ by $A_ g$ and $k[y_1, \ldots , y_ m]$ by $k[y_1, \ldots , y_{m + 1}]$ we may assume that $A \cong k[y_1, \ldots , y_ m]/I$. Consider the surjective maps of local rings

$k[y_1, \ldots , y_ m]_{\mathfrak q} \to R \to S.$

We have to show that the kernel of $R \to S$ is generated by a regular sequence. By Lemma 10.135.4 we know that $k[y_1, \ldots , y_ m]_{\mathfrak q} \to A_{\mathfrak a} = S$ has this property (as $A$ is a local complete intersection over $k$). We win by Lemma 10.135.6. $\square$

Comment #2359 by Simon Pepin Lehalleur on

Suggested slogan: On a local algebra essentially of finite type over a field, all reasonable (algebraic, geometric, local, global,...) notions of complete intersection coincide.

There are also:

• 2 comment(s) on Section 10.135: Local complete intersections

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).