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All rings to be considered are associative, but need not have an identity element.

A ring R is said to be a Jacobson ring if, in every homomorphic image of R, the Jacobson radical coincides with the prime radical. Equivalently every prime ideal of R is an intersection of primitive ideals.

I know that:

The Jacobson radical (J(R)) of R is the intersection of all primitive ideals.

The Prime radical ($Nil_*(R)$) of R is the intersection of all prime ideals.

A/I is a primitive ideal of R/I iff A is a primitive ideal of R containing I.

A/I is a prime ideal of R/I iff A is a prime ideal of R containing I.

for the inclusion $\Leftarrow)$ Let I be an ideal of R. $$Nil_*(R/I) = \bigcap A_i/I$$ where each $A_i/I$ is prime in $R/I$, so $A_i$ is prime in R, and by hypothesis, $A_i = \cap B_k$ ($B_k$ primitive ideals), so $B_k/I$ is primitive. But then

$$Nil_*(R/I) = \bigcap B_k/I = J(R/I)$$ Is this part correct?

for the inclusion $\Rightarrow)$ I have no clue. Any help would be great.

Thanks in advance.

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1 Answer 1

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Every primitive ideal is prime. Always, then, you have that the intersection of all primes is contained in the intersection of all primitives.

Suppose first the prime and Jacobson radicals coincide in every quotient. For a particular prime $P$, the prime and Jacobson radicals of $R/P$ coincide. But the prime radical is zero, and so also must be the Jacobson radical. It must be an intersection of primitive ideals of $R/P$, which pull up to a collection of primitive ideals of $R$ which intersect to $P$.

Now suppose every prime ideal of $R$ is an intersection of primitive ideals. By ideal correspondence, everything we are about to say holds in all nontrivial quotient rings. It follows that each prime ideal contains the intersection of all primitive ideals. Again it follows that the intersection of all prime ideals contains the intersection of all primitive ideals. From the first paragraph I wrote above, we already know the reverse containment always holds. So we have equality of the Jacobson and prime radicals. (Again, this reasoning holds in all nontrivial quotient rings.)

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  • $\begingroup$ From $0=\bigcap A_i/P$, $A_i/P$ all primitive ideals in R/P (P prime), we get some indexes $i_k$ such that $A_{i_k} \subset P\ \forall i_k$? $\endgroup$
    – P.G
    May 29, 2018 at 22:08
  • $\begingroup$ @P.G I’m not sure what your question is, but the first equality you wrote implies the intersection of the $A_i$ is $P$ $\endgroup$
    – rschwieb
    May 29, 2018 at 23:42
  • $\begingroup$ oh!! ok, thank you! $\endgroup$
    – P.G
    May 29, 2018 at 23:46
  • $\begingroup$ I was thinking about this definition. If we suppose every prime ideal of R is an intersection of primitive ideals, with this that you said we only show that the radicals coincide in homomorphic image of the form $R/I$. But if we have an epimorphism $R \to S$, how to show that both radicals coincide in S? $\endgroup$
    – P.G
    Jun 9, 2018 at 21:53
  • $\begingroup$ @P.G I think you're not using the full power of the correspondence. The primitive ideals of $R/I$ correspond to the primitive ideals of $R$ containing $I$. $\endgroup$
    – rschwieb
    Jun 9, 2018 at 22:07

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