This question concerns reduced scheme structure on locally complete intersection, and I guess the answer is related to the number of generators of a radical ideal.

I am confused about the following proposition in Hartshorne's Algebraic Geometry, Chapter 2:

Proposition 8.23 Let $Y$ be a locally complete intersection subscheme of a nonsingular variety $X$ over $k$. Then:

(a) $Y$ is Cohen-Macaulay;

(b) $Y$ is normal if and only if it is regular in codimension 1.

In the proof of (b), the author says that "regular in codimension 1" is just (R1) condition and then uses his Theorem 8.22A, but I think he forgets to treat the case of codimension 0. In order to use (8.22A), one needs the fact that any local ring $\mathcal{O}_{Y,y}$ of $Y$ of dimension 0 is regular.

It is trivially true that a dimension 0 reduced local ring is regular (and hence is a field). In contrast, if $\mathcal{O}_{Y,y}$ is not reduced, then it might not be regular. For example, $k[x]/(x^2)$ is a dimension 0 non-reduced non-regular local ring.

So I guess that in (8.23), the author assumed $Y$ to be reduced.

Now my question comes: if $Y$ is not reduced at first, and let $Y^\prime$ be $Y$ with the reduced closed subscheme structure from $X$, is it true that $Y^\prime$ is still a locally complete intersection?

Locally, the question is: Let $A$ be a finitely generated local ring over a field $k$, and $I=\langle x_1,\dots,x_r\rangle$ be an ideal generated by $x_1,\dots,x_r$ in $A$ (the $x_i$'s in fact form a regular sequence in the situation of locally complete intersection). Then is $\sqrt{I}$ also generated by $r$ elements?

I am sorry if this question turns out to be easy.

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    $\begingroup$ (R1) should imply (R0), right? $\endgroup$ Jan 3, 2018 at 2:17
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    $\begingroup$ If I did not misread, (R1) means regular in codimension $\leq 1$, which includes (R0), while the point in my question is that regular in codimensnion 1 implies regular in codimension 0. $\endgroup$
    – Yeah
    Jan 3, 2018 at 2:37

1 Answer 1


Localization of a regular local ring is still regular. So, if $Y$ is regular in codimension one, then it is also regular in codimension zero, in particular reduced. So, Hartshorne is correct. Your question has a negative answer. For a standard example, take the curve given by $(t^3, t^4, t^5)\subset\mathbb{A}^3, t\in\mathbb{C}$. Then, it is not a local complete intersection, but it is the reduced of a complete intersection (of quadric and a cubic).

  • $\begingroup$ Thank you Mohan, I just did not realize that regular in codimension $k$ implies regular in codimension $l$ for all $l<k$. $\endgroup$
    – Yeah
    Jan 3, 2018 at 2:31

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