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Does there exist a noetherian domain $A$ and a principal ideal $I = (x)$ in it having an embedded component?

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  • $\begingroup$ In a more geometric term, if $X=\mathrm{Spec}(A)$ is an integral surface with only one non-normal point $p$, then $O_{X,p}$ is not $(S_2)$ (hence has depth 1) by Serre's criterion. Take any non-zero $x \in A$ belonging to the maximal ideal corresponding to $p$, then $A/xA$ has depth $0$ at $p$ and Krull dimension $1$, so $p$ is an embedded point of $A/xA$. $\endgroup$
    – user18119
    Oct 2, 2012 at 19:33

2 Answers 2

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Take the homogeneous coordinate ring of the rational quartic curve, that is, $$R=K[a,b,c,d]/(ad-bc, a^2c-b^3, bd^2-c^3, ac^2-b^2d).$$ (Note that this ring is isomorphic to the affine semigroup ring $K[x^4,x^3y,xy^3,y^4]$ and now we have a connection to the answer from this topic.) Then chose any nonzero noninvertible homogeneous element of $R$ and get the required example.

If you want to study more I can recommend you this paper.

Remark. In the book of B. Hassett, Introduction to Algebraic Geometry, exercise 8.9 provides an example of a principal ideal that has embedded associated primes. His example looks pretty similar to the one given above, but isn't an integral domain. (I wonder whether Hassett missed somehow one generator from the defining ideal of his ring $R$, because for rings which are not integral domains there are much simpler examples.)

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Let $X$ be a locally Noetherian scheme, e.g., we have that $X =$ Spec $A$, where $A$ is a Noetherian ring.

Then, if $X$ is reduced (as a scheme), then $X$ has no embedded points.

In particular, if $A$ is a domain, then $X$ has no embedded points.

The converse isn't true. As you probably know, the scheme Spec $k[u,v]/(u^2,uv)$, where $k$ is a field, admits an embedded point corresponding to the maximal ideal $(u,v)$.

Maybe I'm misunderstanding the question?

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  • $\begingroup$ It's a good thing that you think geometrically, but the question is about the scheme $X=V(x)=Spec(A/(x))$ having embedded components. $\endgroup$ Oct 2, 2012 at 8:53
  • $\begingroup$ That said, I certainly did not downvote your answer. $\endgroup$ Oct 2, 2012 at 10:02

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