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

2$\begingroup$ Do you consider the lattice of (1) right ideals, (2) twosided ideals of a possibly noncommutative ring, or the lattice of (3) ideals of a commutative ring? Do rings have an identity element? $\endgroup$ – Luc Guyot Feb 27 '18 at 17:39

2$\begingroup$ There are however results that imply that a finite distributive lattice is isomorphic to the lattice of ideals of a ring (not necessarily finite). See for example tandfonline.com/doi/abs/10.1080/00927878008822518 $\endgroup$ – Todd Trimble♦ Feb 27 '18 at 18:50

$\begingroup$ Ideal generally means twosided ideal. Since finite rings are artinian and Morita equivalent rings have isomorphic lattices of ideals you can assume the quotient by the radical is commutative (as finite division algebras are commutative) $\endgroup$ – Benjamin Steinberg Feb 27 '18 at 20:29

1$\begingroup$ @LucGuyot, it is a standard fact that any Artinian ring R is Morita equivalent to a basic Artinian ring. An Artinian ring R is basic if $R/J(R)$ is a direct product of fields. The way you do this is let $P_1,\ldots, P_s$ be a complete list of nonisomorphic projective indecomposable modules and Let $P=P_1\oplus \cdots \oplus P_S$. Then $P$ is a projective generator so $End(P)$ is Morita equivalent to $R$. It is not difficult to check that $End(P)$ is basic. You can find this in standard books on Artin algebras like Auslander et al. $\endgroup$ – Benjamin Steinberg Feb 28 '18 at 22:33

2$\begingroup$ Dominic, this is a frustrating thread because you have failed to respond to the questions/issues addressed to you in the comment thread under the question. In addition, you have accepted an answer where is it unclear (in view of discussion by Keith Kearnes) that it really answered the intended question. $\endgroup$ – Todd Trimble♦ Mar 1 '18 at 2:48
The answer is yes, there is a finite distributive lattice which is not isomorphic to the lattice
 of right ideals in a noncommutative 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 wellknown facts regarding finite rings with identity, see e.g., Chapter I of Richard Wirt's PhD thesis, "Finite noncommutative 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 nonzero $K$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 twosided ideals in a finite noncommutative ring.

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$\begingroup$ @ToddTrimble No, he certainly didn't and I shamefully missed that point, thanks. I have now extended my answer. $\endgroup$ – Luc Guyot Feb 27 '18 at 16:44

1$\begingroup$ @LucGuyot: In fact, L IS the ideal lattice of a finite ring. Let $K$ be a finite field that is a quadratic extension of its prime field. Let $R$ be the ring of all $2\times 2$ matrices $\begin{bmatrix} a&m\\0&a\end{bmatrix}$ where $a\in K$, $m\in K\otimes_{\mathbb Z} K$, left/right actions of $K$ on $K\otimes_{\mathbb Z} K$ are determined by $r(p\otimes q)=rp\otimes q$ and $(p\otimes q)r=p\otimes qr$. $\endgroup$ – Keith Kearnes Feb 28 '18 at 21:35

1$\begingroup$ @LucGuyot: The problem is that if K is an arbitrary field, then Mn(K) ⊗Z Mn(K)^op need not be a matrix ring over a field. This fails already when n=1. $\endgroup$ – Keith Kearnes Mar 1 '18 at 0:48

1$\begingroup$ @LucGuyot: "isn't the Jacobson radical both a maximal and minimal twosided ideal?" It is maximal, but not minimal. I took $K$ to be a quadratic extension of its prime field so that the $(K,K)$bimodule $K\otimes_Z K$ would not be simple, hence the radical of the ring would not be minimal. $\endgroup$ – Keith Kearnes Mar 1 '18 at 1:23