Let $Z$ be an irreducible and reduced scheme. Does there exist a reduced complete intersection $Y$ such that $Z$ is an irreducible component of $Y$?


1 Answer 1


As long as $Z$ is projective or quasi-projective over say $k = \mathbb{C}$, this is fine. This can be generalized, but let me keep it simple for now.

The quasi-projective case reduces to the projective case by taking the closure of $Z$ in projective space. Therefore, let's do the projective case, $Z \subseteq X = \mathbb{P}^n_k$. Let $I_Z$ denote the ideal sheaf of $Z$. Consider $$I_Z \otimes O_X(n)$$ for $n \gg 0$. This sheaf is globally generated, and so has no basepoints away from $Z$. For a general section $\gamma \in \Gamma(X, I_Z \otimes O_X(n))$, the hypersurface $H = V(\gamma)$ is therefore smooth away from $Z$ (here I'm using Bertini -- characteristic zero and algebraically closed). Any of these hyperplanes passes through $Z$ by construction.

Now, let $d$ denote the codimension of $Z \subseteq X$. Choose $d$ general hyperplanes $H_1, \dots, H_d$ coming from general global sections $\gamma_1, \dots, \gamma_d \in \Gamma(X, I_Z \otimes O_X(n))$.

The scheme theoretic intersection $W = H_1 \cap \dots \cap H_d$ satisfies what you want.

In fact, $W = Z \cup Y$ where $Y$ is some other irreducible scheme which is smooth away from from $Z$ (the fact that $Y$ is irreducible and smooth away from $W$ comes from Bertini's theorem).

Ok, how do you know that the irreducible component of $W$ corresponding to $Z$ is reduced? This is also pretty easy. This comes from the fact that $Z$ was reduced in the first place. Indeed, your general sections $\gamma_1, \dots, \gamma_d$ generate maximal ideal of the stalk at the generic point of $W$ since $I_Z \otimes O_X(n)$ was globally generated.

Of course, for arbitrary schemes, there's no hope. $Z$ need not embed in a nonsingular scheme at all.

  • $\begingroup$ Thank you very much for the answer, this seems to be more or less what I was looking for! $\endgroup$
    – Blup
    Oct 16, 2012 at 16:02

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