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I am studying this article. In proving Theorem 2.14—

Let $R$ be an $\alpha$-ring and $S=R[x,\alpha]$. Then $S$ is $\alpha$-Jacobson if and only if $R$ is $\alpha$-Jacobson

—the author says that:

. . . we can factor out $(P \cap R)S $, which is contained in $P$, and assume that $P \cap R = 0$

(where $P$ is an $\alpha$-prime ideal of $S$).

I don't understand what this “factor out” means. Any help would be great.

gen-ℤ ready to perish
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P.G
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  • See this answer. – Bill Dubuque Oct 24 '18 at 01:36
  • @BillDubuque, I read your answer. I didn't understand the following : "For example, in many ring theoretic problems involving an ideal I, one can reduce to the case I=P prime, then reduce to R/P, thus reducing to the case when the ring is a domain." How can we reduce to the case when R is a domain? – P.G Oct 24 '18 at 03:33

1 Answers1

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It means to consider the quotient. When you do $P/(P\cap R)$, now $P\cap R$ is the zero of the quotient.

Martin Argerami
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  • I edited the question, its $(P \cap R)S$ rather than $P \cap R$. I read what Bill commented but I still don't see why we can only considerer the case $P \cap R = 0$. In another article, the author justifies by the isormophism $S/(P \cap R)S \cong R/(P \cap R) [x,\alpha]$, but I still don't see it. Can you explain more, please? – P.G Oct 24 '18 at 02:04