Relation how non-reduced and how embedded prime ideal

abstract-algebraalgebraic-geometrycommutative-algebramaximal-and-prime-idealsschemes

Sorry for my bad English.

There is a example of embedded prime ideal as follows; (cited Atiyah-Macdonald $\S$4)

ring $R:=k[x,y]/(x^2,xy)$ and the prime ideal $(x,y)$.

Now $\operatorname{Spec}R$ is $y$-axis s.t. fat at $(0,0)$,
and $(0,0)$ is the only point s.t. the stalk is not reduced ring.

Now $R':=k[x,y]/(x^3,x^2y)$ is not reduced at all points, but $(x,y)$ is the only embedded prime ideal and $(0,0)$ is the only point which has non-reduced order 3. (other points have non-reduced order 2. )

I now define "non-reduced order" as a minimal number $n\ge 1$ s.t. $(\operatorname{nil}A)^n=(0)$.

I feel embedded prime ideal is a point which have bigger non-reduced order than the around points.

Can we construct this theory strictly? or Is there a similarly theory?

Best Answer

I think it's a good, but incorrect, idea and intuition. The problem is that the "non-reduced order" you are defining is too coarse. It measures something like how large the nonreduced structure is at a point. But what it needs to measure is "whether the nonreduced structure is different at that point".

Here is a counterexample. Consider $$R = \frac{k[x, y, z]}{(y^2, z) \cap (x, y, z)^2} = \frac{k[x,y,z]}{(y^2, xz, yz, z^2)}.$$

Now $\mathrm{Spec}(R)$ is the set of points $\{(t, 0, 0)\}$. For $t \ne 0$, the nonreduced order is $2$ because the local ring is isomorphic to a localization of $k[x,y]/(y^2)$. For $t=0$, the local ring has basically the same presentation as above, but observe that the nilradical is $(y, z)$ and $(y, z)^2 = 0$.

So, the order is 2 at all points. But, the primary decomposition has the two pieces given above.

The explanation is that the nonreduced structure has two geometrically incomparable pieces:

  • A "ribbon" of tangent vectors along the entire $(t, 0, 0)$ line.
  • A full 3D tangent space at the origin $(0, 0, 0)$.

Neither piece is contained in the other (otherwise the intersection of ideals would be trivial). But, they are both "of non-reduced order 2" because they only involve tangent vectors.

It's the first fact (incomparability) that matters for primary decomposition: each associated prime ideal has a scheme structure associated to it, that isn't covered by the scheme structures of the other components. The "order" doesn't capture this in full.

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