Finite distributive lattices as lattice of ideals of a finite ring Is there a finite distributive lattice that is not isomorphic to the lattice of ideals of a finite ring?
 A: The answer is yes, there is a finite distributive lattice which is not isomorphic to the lattice


*

*of right ideals in a non-commutative ring with identity,

*of ideals in a commutative ring with identity.


Let $L = \left\{ \{0, 1, 2\}, \{1, 2\}, \{1\}, \{2\}, \emptyset
\right\}$ be partially ordered by inclusion (the Hasse
diagram is a diamond with
a tail). Then $L$ is a distributive
lattice with join the
union of subsets and meet the intersection of subsets.
Rings are supposed unital, but not necessarily commutative.


Claim 1. The lattice $L$ is not isomorphic to the lattice of
    right ideals of a finite ring.


I will make implicit use of well-known facts regarding finite rings with
identity, see e.g., Chapter I of Richard Wirt's PhD thesis, "Finite
non-commutative local
rings".


Proof of Claim 1. Let $R$ be a finite ring whose lattice of right
    ideals is isomorphic to $L$. Then $R$ is a local ring whose radical
    $\mathfrak{m}$ is the direct sum of two minimal right ideals
    $\mathfrak{a}$ and $\mathfrak{b}$. As $\mathfrak{a}\mathfrak{m}$ is
    either null or equal to $\mathfrak{a}$ and $\mathfrak{m}$ is nilpotent,
    we deduce that $\mathfrak{a} \mathfrak{m} = 0$. Likewise, $\mathfrak{b}
\mathfrak{m} = 0$. As a result, $\mathfrak{m}^2 = 0$. Therefore
    $\mathfrak{m}$ is a vector space of dimension at least $2$ over $K =
R/\mathfrak{m}$. Hence $\mathfrak{m}$ contains at least a third
    non-zero $K$-subspace, a contradiction.


In the commutative setting, the same example is valid without further
finiteness assumption.


Claim 2. The lattice $L$ is not isomorphic to the ideal lattice
    of a commutative ring with identity.
Proof of Claim 2. The ideal lattice of a commutative ring $R$ with
    identity is distributive if and only if $R$ is
    arithmetic, i.e., if
    the localization $R_{\mathfrak{m}}$ of $R$ at $\mathfrak{m}$ is a
    uniserial ring for every
    maximal ideal $\mathfrak{m}$ of $R$. If $R$ is a commutative ring with
    identity whose ideal lattice is $L$, then $R$ is
    local, but not uniserial, a
    contradiction.


Addendum. As observed by Keith Kearnes in the comments below, the lattice $L$ is isomorphic to the lattice of two-sided ideals in a finite non-commutative ring.
