Let $R$ be a noetherian local ring; I say it has isolated singularity if its spectrum is regular outside the closed point. Such rings certainly don't need to be irreducible, for example the localisation of $k[x,y]/(x,y)$ at $(x,y)$. However, in dimension 2 I have to work a little harder to build an example; the first that springs to mind is 'the union of two planes through the origin in $\mathbb A^4$', which I write more formally as the localisation of $$k[w,x,y,z]/((w, x)(y,z))$$ at $(w, x,y,z)$. However, this is a rather horrible ring (in particular not lci). In fact, I've just about convinced myself that this cannot happen for hyper surfaces:

Claim: Let $R$ be a regular local ring of dimension at least 3, and let $r \in R$ be non-zero and such that $R/(r)$ has isolated singularity. Then $R/(r)$ is irreducible.

(Idea of proof: otherwise we factor $r = r_1r_2$, let's imagine distinct irreducibles for simplicity; then the intersection of $V(r_1)$ and $V(r_2)$ should have codimension at most 1 in either of them).

This motivates my question:

Let $R$ be a complete intersection local ring of dimension at least 2 with isolated singularity. Does it follow that $R$ is irreducible?

If I had to guess my money would be somewhat on the 'no' side, but if it were true I could shorten a proof, so seems worth asking. Thanks in advance for any thoughts!

(p.s. by 'irreducible' I mean 'every zero-divisor is nilpotent').


2 Answers 2


Claim: A noetherian local normal ring $R$ is a domain.

Proof: recall that normal is equivalent to $(S_2)$ and $(R_1)$. Since $R$ is $(S_2)$, $Spec(R)$ it is connected in codimension one (removing any subset of codimension at least $2$ leaves it connected). There is a pleasant characterization using the connectedness of "dual graph": we can arrange the minimal primes of $R$, $p_1,...,p_n$ so that the height of $p_i+p_{i+1}$ is one. You can find the details in Hartshorne's original paper or some extensions by Hochster-Huneke.

So if $n>1$, we can choose a height one prime $P$ that contains $p_1,p_2$. This implies that $R_P$ has at least 2 minimal primes. But since $R_P$ is regular local (here is where we use $(R_1)$), it is a domain, contradiction. Thus $R$ has unique minimal prime and since the localization at that prime is also regular, it is a domain.

If your ring is a complete intersection then it is Cohen-Macaulay so certainly $(S_2)$. It has dimension $\geq 2$ and isolated singularity, so it is $(R_1)$. So it must be a domain.


In fact it seems you only need the $S_2$ condition. The following is proposition 3.1.12 of the book Joins and Intersections by Flenner, O'Carroll and Vogel:

Let $(R,m)$ be a catenary local ring satisfying Serre's condition $S_2$, then $\mathrm{Spec}(R)$ is equidimensional and connected in dimension $\mathrm{dim}R-1$ (meaning that any open subset of $\mathrm{Spec}(R)$ whose complementary has dimension strictly less than $\mathrm{dim}(R)-1$ is connected).

In your situation, if $\mathrm{Spec} R$ was not irreducible then each component would meet in codimension 1, hence the singular locus would be of dimension bigger or equal to $1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.