Question #24896

If R ⊆ Z(S), show that R ∩ Nil*(S) = Nil*(R).

Expert's answer

Under the assumption *R **⊆** Z*(*S*), any prime ideal p of *S *contractsto a prime p0 : = p *∩ R *of *R*. Indeed, if *a, b **∈** R *are such that *ab **∈*p0, then *aSb *= *abS **⊆*p0*S **⊆*p*, *so we have, say, *a **∈*p *∩ R *=p0. Therefore, any *r **∈*Nil**R *lies in all primes p of*S*, and so *r **∈*Nil**S*. This gives *R ∩ *Nil*(*S*)= Nil*(*R*).

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