# Answer to Question #24896 in Abstract Algebra for Hym@n B@ss

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*).Need a fast expert's response?

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