2
$\begingroup$

I am studying for a qualifying exam, but I'm at a loss for what theorems or methods to use to solve this problem:

Let $R$ be an Artinian ring with identity and $I$ a minimal ideal in $R$. Show that $I$ is a direct sum of isomorphic simple $R$-modules.

I know that $I$ has no proper sub-ideals because it is minimal, but it could have left or right sub-ideals.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

The most direct way I can think of is

  1. Let $L$ be a non-trivial simple left submodule of $I$, and $r\in R$, show that either $Lr=0$ or $Lr\cong L$. (Consider that right multiplication is a left module homomorphism.)
  2. Use (1) to observe that for any two ring elements $r,s\in R$, either $Lr=Ls$ (literally equals, not isomorphic to) or $Lr\cap Ls=0$.
  3. Prove that $LR=\{l\cdot r~|~l\in L,r\in R\}$ is equal to $I$ (this is where minimality is used.)
  4. Use the fact that Artinian $\Rightarrow$ Noetherian to deduce that there is a finite colletion of elements $r_1,r_2,\dots ,r_n\in R$ such that $$Lr_1\oplus Lr_2\oplus\dots \oplus Lr_n = I$$

Another method would be to use the Artin-Wedderburn theorem. The result follows immediately once you prove that for $j\in J(R)$ and $i\in I$, $ji=0$, where $J(R)$ denotes the Jacobson Radical of $R$.

$\endgroup$
1
  • $\begingroup$ Thanks for being so clear! $\endgroup$ Dec 9, 2013 at 17:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .