Is there a finite distributive lattice that is *not* isomorphic to the lattice of ideals of a finite ring?



Finite distributive lattices as lattice of ideals of a finite ring
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](https://en.wikipedia.org/wiki/Hasse_diagram) is a *diamond with
a tail*). Then $L$ is a [distributive
lattice](https://en.wikipedia.org/wiki/Lattice_(order)) 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"](https://shareok.org/bitstream/handle/11244/3379/7229919.PDF?sequence=1&isAllowed=y).



> 
> 
> > 
> > *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](https://en.wikipedia.org/wiki/Arithmetical_ring), i.e., if
> >  the localization $R\_{\mathfrak{m}}$ of $R$ at $\mathfrak{m}$ is a
> >  [uniserial ring](https://en.wikipedia.org/wiki/Serial_module) 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](https://en.wikipedia.org/wiki/Local_ring), 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.


