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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).
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.

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