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Definition: For a ring with unit $R$, Jacobson Radical is defined as the ideal $J(R) = \{r \in R: rM = 0 \}$ where $M$ is simple.

How do I show the $J(R)$ is nilpotent for an Artinian ring $R$?

A similar questions, but it uses Nakayama, which assumes commutativity. The only lead I got is forming a chain of inclusion $J^n \supset J^{n+1} \supset \dots$ and this chain terminates with $J^{n} = J^{n+k}$ for every $k \geq 1$

Lemon
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    I explicitly said in that solution that I’m using a version of Nakayama’s lemma that is valid for noncommutative rings, and I spell it out. So, you are not reading it correctly... – rschwieb Apr 09 '21 at 01:19
  • But how do I know it is also finitely generated to apply Nakayama? Mine isn't a finite dimensional algebra – Lemon Apr 26 '21 at 02:05
  • Artinian implies Noetherian, so all right ideals are finitely generated. – rschwieb Apr 26 '21 at 02:31
  • Isn't that only when the Artinian Ring is commutative? – Lemon Apr 26 '21 at 02:34
  • Nope. Right Artinian implies right Noetherian in a ring with identity. Always. I guess we have a problem if you want to use the fact we’re working on to prove that theorem, though. – rschwieb Apr 26 '21 at 02:39
  • yeah so I read that theorem and it is by Hopkins, but it uses this result. But that investigation also led me to a proof that doesn't use Nakayama. – Lemon Apr 26 '21 at 03:20
  • ok. You’ve got me thinking I should add a different proof. Or you can! – rschwieb Apr 26 '21 at 11:26

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