Let $A$ be a subring of a commutative unital ring $B$.
Can you tell me if my proof of the following claim is correct?
Claim: $B \otimes_A A[X] \cong B[X]$
Proof:
It's enough to show that $B[X]$ satisfies the universal property of $B \otimes_A A[X]$, that is if $N$ is any $R$-module and $b^\prime: B \times A[X] \to N$ any bilinear map then there exists a unique linear map $l: B[X] \to N$ such that $l \circ b = b^\prime$.
Define $b: B \times A[X] \to B[X]$ as $(r,p(x)) \mapsto rp(x)$. This is bilinear. Now define $l: B[X] \to N$ as $l: p(x) \mapsto b^\prime((1,p(x)))$. It remains to be shown that this $l$ is unique and linear. Linearity directly follows from the bilinearity of $b^\prime$. So let $l^\prime: B[X] \to N$ be another linear map such that $l^\prime \circ b = b^\prime$. Then $ l(p(x)) = b^\prime ((1,p(x)) = l^\prime \circ b ((1,p(x)) = l^\prime(p(x))$ hence $l$ is unique and $B \otimes_A A[X] \cong B[X]$ follows.
Thanks!