I have read somewhere that a Hibi ring has a linear resolution if and only if it's regularity is 1. Please refer me the proof of this fact.
Firstly, I need to define the Hibi ring. Let $L$ be a finite distributive lattice and let $P$ be the subposet of $L$ which consists of the join-irreducible elements of $L.$ Then, by a famous theorem of Birkhoff, it follows that $L$ is isomorphic to the lattice $\mathcal{I}(P)$ of the poset ideals of $P.$ Let us assume that $P$ consists of $n$ elements, say $P=\{p_1,\ldots,p_n\}$. Let $A=K[t,x_1,\ldots,x_n]$ be the polynomial ring in $n+1$ variables over a field $K.$ The Hibi ring of $\mathcal{I}(P)$ is the $K$--sublagebra $R[L]$ of $A$ generated over $K$ by the monomials $u_{\alpha}=t\prod_{p_i\in \alpha}x_i,$ with $\alpha\in L.$
$R[L]$ is an algebra with straightening laws on $L$ over $K$. Let $B=K[\{x_\alpha: \alpha \in L\}]$ be the polynomial ring in the indeterminates indexed by the elements of $L.$ Then, the defining ideal $I_L\subset B$ of the toric ring $R[L]$ is generated by the straightening relations, namely $$I_L=(x_\alpha x_\beta-x_{\alpha\cap \beta}x_{\alpha\cup \beta}: \alpha,\beta \in L, \text{ incomparable}).$$ $I_L$ is called the Hibi ideal or the join-meet ideal of $L.$