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).

Expert's answer

Let *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.

Then

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