# Answer to Question #24897 in Abstract Algebra for Hym@n B@ss

Question #24897

Let R,K be algebras over a commutative ring k such that R is k projective and K ⊇ k.

Show that R ∩ Nil*(R ⊗k K) = Nil*(R).

Show that R ∩ Nil*(R ⊗k K) = Nil*(R).

Expert's answer

Let

Then

*S*=*R**⊗**k K*. Since*R*is*k*-projective,tensoring the inclusion*k → K*by*R*gives an inclusion*R**⊗**k k → R**⊗**k K.*We identify*R**⊗**k k*with*R*, and view it as a subring of*S*=*R**⊗**k K*. We finish by showing that anyprime ideal p of*S*contracts to a prime ideal p0 = p*∩ R*of*R*.Indeed, let*aRb**⊆**p0, where**a, b**∈**R*.Then

*aSb*=*a*(*R**⊗**k K*)*b**⊆**aRb**⊗**k K**⊆*_p0(1*⊗**k K*)*⊆**p**,*so we must have, say,*a**∈**p**∩ R*= p0.Need a fast expert's response?

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