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').



Lci local rings with isolated singularity are irreducible?
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](https://www.jstor.org/stable/2372986?seq=1) or some extensions by [Hochster-Huneke](https://arxiv.org/abs/math/0211172).


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.


