# Is being reduced a generic property of schemes?

1. (Naive formulation:) Let $X$ be an (irreducible) affine variety (over an algebraically closed field $k$) and $I$ be an ideal of the coordinate ring $R$ of $X$. Assume $Y = V(I)$ is equidimensional. The question is, can we detect if $Y$ is reduced as a subscheme on generic points of $Y$? More precisely, assume there is a (proper) closed subvariety $Z$ of $Y$ such that for all $y \in Y \setminus Z$, the localization of $I$ at the maximal ideal of $y$ is a radical ideal. Then is it true that $I$ is a radical ideal?

The answer to the naive question is false, even in the simplest case that $I$ is principal, e.g. take $X = Spec~ k[x,xy,y^2]$ and $I$ to be the ideal generated by $x$. Then $I$ is radical on $X\setminus\{O\}$, where $O$ is the "origin". A perhaps more natural set of counter-example can be constructed as follows: Take a subvariety $Y'$ of $X$ which is not a complete intersection, take a minimal collection of generators $g_1, \ldots, g_k$ of the ideal of $Y'$, and set $I$ to be the ideal generated by $g_1, \ldots, g_{k-1}, g_k^2$. So the `real question', which is perhaps too broad, is:

1. Are there 'natural' conditions on $X$ and $I$ under which the reducedness of $V(I)$ can be detected at generic (closed) points on $Y$?

A 'natural' situation that I have in mind (and that disallows both counter-examples above) is the following:

1. Is the answer to Question 2 affirmative when $X$ is normal and $I$ is generated by a regular sequence, i.e. $Y$ is a set-theoretic complete intersection? (The answer is affirmative in the simplest case, i.e. when $I$ is principal.)

I don't think normal is quite enough, but something similar should work. So,

• normal is equivalent to $S_2$ and $R_1$
• reduced is equivalent to $S_1$ and $R_0$

If $Y$ is a regular hypersurface in something normal and it is not entirely singular which follows from you generically reduced assumption, then $Y$ is reduced. (This is just restating what you are saying in #3).

A simple sufficient condition is to assume that $X$ is at least $S_{t+1}$ where $t=\mathrm{codim}_X Y$. That way $Y$ will be $S_1$ and your generically reduced assumption implies that it is $R_0$.

Here is an example that the question in #3 only holds if the codimension of $Y$ is $1$:

Let $X_0 = \mathrm{Spec}\ k[x,y,z,t,w]/(xz,xt,yz,yt)$. This is a threefold in $\mathbb A^5$. Near any point with any of $x,y,z,t$ non-zero, this is locally isomorphic to $\mathbb A^3$, hence smooth. When $x=y=z=t=0$, that's a line in $\mathbb A^5$, so $X_0$ is smooth away from that line, so it is $R_1$. We will see below that it is also $S_2$. In case you would like your $X$ be integral, then do this: This $X_0$ is really just two copies of $\mathbb A^3$ intersecting in a line. So, take two skew lines in $\mathbb A^3$ and glue them together using the local structure of $X_0$ near its singular line. That way you get an irreducible $X$ which is locally like $X_0$.

Next let $Z = Z(w)\subset X_0$. This is the union of two planes meeting in a single point, the famous example of something not normal, not Cohen-Macaulay, etc.. It is easily seen to be $S_1$: For instance $x+z$ is a regular element, but modding out by that we get $Y=\mathrm{Spec} k[x,y,z,t]/(x^2,xy,xt,yt)$ which is two lines intersecting in a fat point, which is $R_0$, but not reduced at the fat point and hence non-reduced and cannot be $S_1$.

Now in order to work on the irreducible $X$, take the $Z$ and $Y$ to be what you get while gluing the two lines of $\mathbb A^3$ together. Both $Z$ and $Y$ are linear away from the origin, so we can do the same gluing on them as on $X$.

So, naming the new "glued" objects $X$, $Y$, and $Z$, we have that

• $Y$ is $S_0$ and $R_0$, in particular non-reduced,
• $Z$ is $S_1$ and $R_1$, in particular reduced, but not normal
• $X$ is $S_2$ and $R_1$, in particular normal.

So, $Y$ is the complete intersection of a regular sequence, $w, x+z$ in the normal $X$, it is generically reduced, but not everywhere reduced.

I think you can adapt this example to show that if $X$ is not $S_{t+1}$, then there is a codimension $t$ complete intersection which is generically reduced but not everywhere reduced.

Otherwise there is still taking general hypersurface sections, those retain even normality.

• Dear Sándor, not sure I get the example: your $Y$ is nowhere reduced, so it does not satisfy the assumption of being generically reduced. – auniket Jan 19 '15 at 23:40
• You're right. My original answer was different, then I forgot half of the assumptions and made an edit (actually several) and ended up with this. I have a new example which looks much better now. Cheers! – Sándor Kovács Jan 20 '15 at 2:34

There's an exercise in [Eisenbud] that says that reducedness is R0 + S1, i.e. generic reducedness plus Serre's condition S1, which says that there are no embedded primes. This is analogous to normality being R1+S2, where R1 is regularity in codimension 1.

So the cheap way to get what you want is to take a Cartier divisor in a normal scheme. That'll be S1. Then check generic reducedness on each component.

A more expensive way is for your scheme to be Cohen-Macaulay, e.g. a complete intersection inside something regular. That'll be S1 too, so again, check generic reducedness.

I used this latter trick to show that the space of pairs (X,Y) of matrices, such that XY, YX are upper triangular, is reduced.