Proposition 4

Let X be a closed subset of V, and $H_1, ..., H_m$ hypersurfaces in V. Put $X' = X \cap H_1 \cap ... \cap H_m$.

a) For every closed subset Y of V contained in X', one has

b) One has $\operatorname{codim}(Z, X) \leq m$ for every irreducible component Z of X'. If X' is nonempty, one has $\operatorname{codim}(X', X) \leq m$.

c) If Z is an irreducible closed subset of V contained in X such that $\operatorname{codim}(Z, X) \leq m$, there exist hypersurfaces $H'_1, ..., H'_m$ such that Z is an irreducible component of $X \cap H'_1 \cap ... \cap H'_m$.

Let α be an ideal of A and $x_1, ..., x_m$ elements of A such that $X = V(\alpha)$ and $H_i = V(x_i)$ for $1 \leq i \leq m$. Let Z be an irreducible closed subset of V contained in X; there exists a prime ideal p of A containing α and such that $Z = V(p)$.

Suppose first that Z is contained in X' and denote by $\xi_i$ the image of $x_i$ in the noetherian local ring $B = A_p/\alpha A_p$. By Proposition 7, b) of § 1, No. 3, one has

By Proposition 2 of No. 1, one therefore has

(11)

If Z is an irreducible component of X', one has $\operatorname{codim}(Z, X') = 0$, whence $\operatorname{codim}(Z, X) \leq m$; this proves b). One proves a) by taking in both members of (11) the greatest lower bound over the set of irreducible components Z of Y.

1 Recall that $V(x)$ consists of the prime ideals of A containing x.

Conversely, suppose that one has codim(Z, X) $\leq m$, that is to say dim(B) $\leq m$. Since every element of $A_p$ is the product of an invertible element of $A_p$ by the image of an element of A, the scholium of No. 2 proves the existence of elements $x'_1, ..., x'_m$ of A whose images in B generate an ideal of definition of B. Put $H'_i = V(x'_i)$ for $1 \leq i \leq m$. It is clear that Z is an irreducible component of $X \cap H'_1 \cap ... \cap H'_m$.