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 $$R$$.

• 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 Commented May 19 at 15:59

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.