## Tag `07EN`

Chapter 15: More on Algebra > Section 15.73: Pseudo-coherent and perfect ring maps

Lemma 15.73.6. A local complete intersection homomorphism is perfect.

Proof.Let $A \to B$ he a local complete intersection homomorphism. By Definition 15.30.2 this means that $B = A[x_1, \ldots, x_n]/I$ where $I$ is a Koszul ideal in $A[x_1, \ldots, x_n]$. By Lemmas 15.73.2 and 15.67.3 it suffices to show that $I$ is a perfect module over $A[x_1, \ldots, x_n]$. By Lemma 15.67.11 this is a local question. Hence we may assume that $I$ is generated by a Koszul-regular sequence (by Definition 15.29.1). Of course this means that $I$ has a finite free resolution and we win. $\square$

The code snippet corresponding to this tag is a part of the file `more-algebra.tex` and is located in lines 18513–18516 (see updates for more information).

```
\begin{lemma}
\label{lemma-lci-perfect}
A local complete intersection homomorphism is perfect.
\end{lemma}
\begin{proof}
Let $A \to B$ he a local complete intersection homomorphism.
By Definition \ref{definition-local-complete-intersection} this
means that $B = A[x_1, \ldots, x_n]/I$ where $I$ is a Koszul ideal
in $A[x_1, \ldots, x_n]$.
By Lemmas \ref{lemma-perfect-ring-map} and \ref{lemma-perfect-module}
it suffices to show that $I$ is a perfect module over $A[x_1, \ldots, x_n]$.
By Lemma \ref{lemma-glue-perfect} this is a local question. Hence we
may assume that $I$ is generated by a Koszul-regular sequence (by
Definition \ref{definition-regular-ideal}).
Of course this means that $I$ has a finite free resolution and we win.
\end{proof}
```

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