Is there a finite distributive lattice that is not isomorphic to the lattice of ideals of a finite ring?
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asked Feb 26, 2018 at 11:48
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2$\begingroup$ Do you consider the lattice of (1) right ideals, (2) two-sided ideals of a possibly non-commutative ring, or the lattice of (3) ideals of a commutative ring? Do rings have an identity element? $\endgroup$– Luc GuyotFeb 27, 2018 at 17:39
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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, 2018 at 18:50
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$\begingroup$ Ideal generally means two-sided 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 SteinbergFeb 27, 2018 at 20:29
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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 non-isomorphic 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 SteinbergFeb 28, 2018 at 22:33
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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, 2018 at 2:48
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1 Answer
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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.
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$\begingroup$ @ToddTrimble No, he certainly didn't and I shamefully missed that point, thanks. I have now extended my answer. $\endgroup$ Feb 27, 2018 at 16:44
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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 $(p\otimes q)r=p\otimes qr$. $\endgroup$ Feb 28, 2018 at 21:35
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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. $\endgroup$ Mar 1, 2018 at 0:48
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1$\begingroup$ @LucGuyot: "isn't the Jacobson radical both a maximal and minimal two-sided 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. $\endgroup$ Mar 1, 2018 at 1:23