2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Nice answer! Do you know of any ways to prove that the subalgebras of upper triangular mateices cannot be decomposed into local parts? $\endgroup$ – ASKASK Aug 21 '16 at 22:00
  • 1
    $\begingroup$ I do. I suggest you think about how to do it for a while before asking! :) $\endgroup$ – Mariano Suárez-Álvarez Aug 21 '16 at 22:03
1
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.