Triviality of $\mathrm{Ann}(\mathfrak m)$ This question is regarding the first paragraph of the proof of Proposition 2.4 from this paper.
QUESTION: Is it true that if $(0)$ is irreducible, then $\mathrm{Soc}(R)=\mathrm{Ann}(\mathfrak m)=(0:_R\mathfrak m)$ is non-zero for a general commutative local ring $(R,\mathfrak m)$ which is not a domain?
The proof provided there in the first paragraph starts by assuming that $\dim(\mathrm{Soc}(R))\geq 2$ and gets a contradiction. But it could still be zero, in particular this proof does not work for arbitrary (non-noetherian) commutative ring.
 A: It is obviously not $0$ because an Artinian ring always has nonzero socle. 
A commutative ring with socle zero cannot have any minimal ideals.
A: If $R$ is a local integral domain which is not a field, then $(0)$ is irreducible and its socle is obviously $0$. 
More general, we have

If $R$ is an irreducible Noetherian ring then every element of $R$ is a non-zerodivisor or nilpotent.

Suppose $a\in R$ is not nilpotent, and let $\mathrm{Ann}_R(a^n)$ maximal among the ideals of this form. Then $\mathrm{Ann}_R(a^{2n})=\mathrm{Ann}_R(a^n)$. Suppose $ab=0$, and $b\ne0$. We have $Ra^n\cap Rb\ne0$ (here one uses that $R$ is irreducible). Let $c\ne0$, $c\in Ra^n\cap Rb$. Write $c=ra^n$. Since $c\in Rb$ and $a^nb=0$ we get $ca^n=0$, so $ra^{2n}=0$. Then $ra^{n}=0$, that is, $c=0$, a contradiction.
We conclude: 

If $R$ is an irreducible Noetherian local ring then $\mathrm{Soc}(R)=0$ unless $R$ is Artinian. 

If $R$ is an irreducible Noetherian local ring, and $\mathrm{Soc}(R)\ne0$ we get that all elements of $\mathfrak m$ are zerodivisors. By the result above all elements of $\mathfrak m$ are necessarily nilpotent, so $\mathfrak m$ is nilpotent. This shows that $R$ is Artinian. 
