# Are there examples of an Artinian ring that cannot be broken down into local parts?

I am aware that if you have an Artinian ring $R$ which happens to be commutative and contains unity, then the ring $R$ can be decomposed as: $$R \cong R_1 \times R_2 \times \cdots \times R_n$$

Where each $R_i$ is an artinian local ring. My question is, if we drop the commutativity constraint, must this still hold? Are there any noncommutative left or right Artinian rings that cannot be written as a direct product of local rings?

We say that a ring is connected if it cannot be written as a direct product of non-trivial ideals. There are many examples of connected Artinian rings.

To produce some, just take any finite quiver $Q$ which is connected as a graph, consider the path algebra $\Bbbk Q$ of $Q$ over some field $\Bbbk$, and pick any admissible ideal $I$ in $\Bbbk Q$. The quotient algebra $A=\Bbbk Q/I$ is then connected. You will find a proof of this (and definitions of all the terms I have used) in the book by Assem, Skowroński and Simson.

A concrete example: pick $n\in\mathbb N$ and let $A$ be the subalgebra of the matrix algebra $M_n(\Bbbk)$ over a field consisting of upper triangular matrices.

• Nice answer! Do you know of any ways to prove that the subalgebras of upper triangular mateices cannot be decomposed into local parts? – ASKASK Aug 21 '16 at 22:00
• I do. I suggest you think about how to do it for a while before asking! :) – Mariano Suárez-Álvarez Aug 21 '16 at 22:03

Another observation: since $rad(R)=rad(R_1)\times rad(R_2)\times\ldots \times rad(R_n)$, a decomposition into local rings $R_i$ would imply that $R/rad(R)$ is isomorphic to a finite product of division rings.

So, you can just pick $R$ to be an Artinian ring with $R/rad(R)$ something other than a finite product of division rings.

So, for example, you can take $S=M_2(F)$, and $R=\left\{\begin{bmatrix}a&b \\ 0& a\end{bmatrix}\mid a,b \in S\right\}$, which has $R/rad(R)\cong S\, .$

This distinguishes itself from the ring of upper triangular matrices over a division ring since the quotient by the radical is a product of division rings.

In fact, this ring is also indecomposable (=is not the direct sum of two ideals=has no central idempotents="connected" in the sense of the other solution.)