Let $R$ be a ring (not necessary commutative and not necessary contains 1)

In Hungerford, it define $P(R)$ is the prime radical, that is the intersection of all prime ideal of $R$. And, $J(R)$ is the jacobson radical, that is the intersection of the regular maximal left ideals of $R$.

I want to show that if $R$ is left Artinian, $P(R)=J(R).$

Since every element in $P(R)$ is nilpotent, $J(R)$ is nil of $R$, thus is contained in the radical $J(R)$.

For the converse part, that is $J(R)\subset P(R)$,(all I now is if $R$ is left Artinian ring, then $J(R)$ is a nilpotent ideal.) may someone gives some hints for the proof. Thank you!

I see that Hungerford have the same question with the Hint: If $R$ is a left Artinian ring,then the radical $J(R)$ is a nilpotent ideal. Conseqently every nil left or right ideal of $R$ is nilpotent and $J(R)$ is the unique maximal nilpotent left ideal of $R$.

  • $\begingroup$ Assume that $J(R)^{n+1} = J(R)^n ≠ 0$ and let $Q = J(R)^n$. Then there exists left ideals in $P$ such that $IQ≠0$ as $Q^2 = Q ≠ 0$. There is by the Artinian property a left ideal with $IQ ≠0$ and $I \subseteq Q$. There must be some $y \in I$ such that $Qy ≠ 0$ and hence $I = Ry$ is principal by minimality. Choose $x \in Q$ such that $xy = y$. Then $(1-x)y = 0$. But $(1-x)$ is invertible so $y=0$ which is a contradiction $\endgroup$ Commented May 19 at 15:59

1 Answer 1


Picking up from the hint, one can simply observe

Every prime ideal must contain every nilpotent ideal.

Because if $P$ is a prime ideal and $I^n=\{0\}$, you have of course that $I^n\subseteq P$. By primeness, $I\subseteq P$.


$P(R)$ contains every nilpotent ideal.

That being the case, $J(R)\subseteq P(R)$ in a (right or left) Artinian ring.


You must log in to answer this question.

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