Consider locally Noetherian schemes over an algebraically closed base field $k$. Given a scheme $X$, the scheme $X_{red}$ is the smallest closed subscheme of $X$ with the same underline topological space. In the sense that for every other closed subscheme $Z$ of $X$ with this property there are closed embeddings $X_{red}\to Z\to X$.

Passing form $X$ to $X_{red}$ we have shaved off every non-reduced structure of every associated point. That is, as it is well known $$Ass(X_{red})=\{\mbox{generic points of irreducible components of }X\}$$

I am interested into the closed subschemes $Z$ of $X$ with $X_{red}\to Z\to X$. Since it is a local matter we may assume $X$ irreducible, Noetherian and affine for some ring $R$. Since the underline topological spaces $X$, $Z$ and $X_{red}$ are equal I guess that we have just to look at the associated points.

First I thought that such a $Z$ just would be an scheme with $\{p_0\}=Ass(X_{red})\subseteq Ass(Z)\subseteq Ass(X)$. But the example $X$ a double line, $Z$ a line with and embedded point and $X_{red}$ a simple line shows that more interesting things may occur. Algebraically the example is $R=k[x,y]/(x^2)\to k[x,y]/(x^2,xy)\to k[x,y]/(x)$, $Ass(X)=\{(x)=(0:x)\}$, $Ass(Z)=\{(x,y)=(xy:x),(x)=(xy:y)\}$, $Ass(X_{red})=\{(x)=(x:x))\}$.

In this example is clear that the extra associated point of $Z$ comes from collapsing one line. So, $Ass(Z)$ may be bigger than $Ass(X)$ but, at least apparently, we should be able to determine the possible geometries of such schemes $Z$ from the geometry of $X$. Is that make sense or I am too innocent about the geometry of schemes?

Translating to algebra, the nilradical $p_0$ of $R$ is the unique minimal associated prime ideal of $R$. The ideals determining such schemes $Z$ are ideals $I$ of $R$ for which $(0)\subseteq I\subseteq p_0$.

Let $p_i=(0:r_i)$ be the associated prime ideals of $R$. For every $i$ the element $r_i\in p_0$. Since for every $s\in p_i$ $sr_i=0\in p_0$, if $r_i\not\in p_0$ then $p_i\subseteq p_0$. So, is it true that if $r_i\not\in I$ then $p_i\in Ass(Z)$?

The converse case, a prime ideal $q=(I:f)\in Ass(Z)\setminus Ass(X)$, looks even harder. I have not made any progress.

Just recall that in the previous example there are a family of ideals $I_a=(x(y-a))$ such that $(x)=(I_a:f_a)\in Ass(Z_a)\setminus Ass(X)$, we are just collapsing the doubled line to the point $(x,y-a)$ of $X$. In this case geometricly it is clear that this schemes $Z_a$ are the only ones (besides $X$ and $X_{red}$) such that there are closed embeddings $X_{red}\to Z_a\to X$. How do we translate this to the algebra setting? Could I expect to prove that if for a scheme $W$ there are closed embeddings $X_{red}\to W\to X$, then $W=Z_a$ for some $a$?

  • $\begingroup$ Characterize- can you tell me what you have in mind for this word? $\endgroup$
    – Mohan
    Commented May 11, 2018 at 1:02
  • $\begingroup$ Dear @Mohan, let $p\in Ass(X)$ with $p=(0:r)$, is there some condition on $r$, $p$ and $I$ (or over some other objects) implying $p\not\in Ass(Z)$? and implying $p\in Ass(Z)$? And vice versa for a given $q\in Ass(Z)$ with $q=(I:s)$? Actually, I would like to understand the schemes $Z$ between $X$ and $X_{red}$, for example when $Z$ has more associated points, from where they come? In the example, we have a double line and we collapse one of them into an embedded point, so it looks like the geometry of $X$ encodes which kind of associated points there may (dis)appear on $Z$. $\endgroup$
    – MonLau
    Commented May 11, 2018 at 6:06


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