# Lci local rings with isolated singularity are irreducible?

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').

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 $$1$$.