# 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 $$t=\mathrm{codim}_X Y$$, where $$Y$$ is a complete intersection in $$X$$. 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. Jan 19, 2015 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! Jan 20, 2015 at 2:34
• @SándorKovács I'm sorry for reopening this post after so much time, but I am confused about something you said: if I understood correctly you claimed that if $Y$ is a subvariety of $X$, and $X$ is at least $S_{t+1}$ with $t=codim_XY$, then $Y$ is $S_1$. But let's take $X=\mathbb A^2$ and $Y=V(x^2, xy)$, i.e. the vertical line with a double origin: $X$ should be $S_k$ for any $k$ (since it's Cohen–Macaulay), but $Y$ is not $S_1$, since it has an embedded prime...where am I wrong?
– Utf
21 hours ago
• Dear @Utf, the point of this answer was to give an example that the assumption in #3 in the original question was still not enough for the desired conclusion. In other words, the $Y$ here is supposed to be a complete intersection in $X$. I added a clarification to that end. I hope it makes more sense now. 14 hours ago
• Dear @SándorKovács, thanks a lot for the clarification. I had just posted a somehow similar question, when I read your comment and thought that my question could maybe be already answered by that...but unfortunately that's just not the case, apparently.
– Utf
9 hours ago

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.

• Hi Allen, I'm interested in your last result: the space of pairs (X,Y) of matrices, such that XY, YX are upper triangular, is reduced. Could you be precise where can I find it? Feb 18, 2022 at 16:20
• It's here arxiv.org/abs/math/0306275 Feb 19, 2022 at 18:55
• Thanks a lot! I will read your beautiful results. Feb 19, 2022 at 21:12