3

Suppose $R$ is a ring with identity and $R$ is of prime power order. Also suppose jacobson radical of $R$ is not trivial. Under which condition there exist a subring $S$ such that $$R=J(R)\oplus S\ ?$$

user26857
  • 52,094
Hamid
  • 31
  • If $R$ contains a field (which will be finite, hence perfect), then this follows from Wedderburn's Principal Theorem. Hrm. Actually: do you mean that the ring is of prime power order or that the identity element is of prime power order? – Mariano Suárez-Álvarez Nov 06 '14 at 19:42
  • Well, the first thing to observe is that $J(S)=0$. Now take the Jacobson radical of both sides of your equality and use the fact that your ring is artinian to make a conclusion. – David Hill Nov 06 '14 at 19:45
  • Dear @DavidHill : That would certainly be true if $S$ was an ideal and the product ring structure was componentwise, but in that case $J(R)={0}$. If the componentwise product wasn't intended, then it seems like it would be harder to determine what $J(S)$ is. Regards – rschwieb Nov 10 '14 at 21:04
  • What does $\oplus$ mean here? Abelian group sum? Product of rings? Or as modules over some ring? – rschwieb Nov 10 '14 at 21:16

1 Answers1

1

It remains to be seen what $\oplus$ means here.

If, by this notation, you mean a subring $S$ such that $R=J(R)\oplus S$ where the ring operations $(j+s)(j'+s')=jj'+ss'$, then $S$ has to be an ideal of $R$.

If that's the case, then the answer is "When $J(R)=\{0\}$" because the $J(R)$ is never a nontrivial direct summand of $R_R$.

This is because the Jacobson radical is a superfluous ideal. If you try to add $J(R)+I$ for any proper ideal, $I$ is contained in some maximal ideal $M$, which $J(R)$ is also contained in. So, you see you can't ever get all of $R$ unless $I=R$.

rschwieb
  • 153,510