Lci local rings with isolated singularity are irreducible?

Question

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').

Answer 1

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.

Answer 2

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 $1$.