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Let $A\rightarrow B$ be a homomorphism of commutative rings. Then $B\otimes_A A[x]\cong B[x]$ as $B$-algebras. How can one demonstrate this nicely, i.e. using universal properties alone and the Yoneda lemma?

What I did so far for this holy purpose: by the canonical definition of $B[x]$ we have $hom_{B-Alg}(B[x],-)=hom_{Set}(\left\lbrace x\right\rbrace, U-)=id_{Set}$ all = are functorial isomorphisms.

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1 Answer 1

up vote 11 down vote accepted

The point is that ${-} \otimes_A B$ is the left adjoint of the forgetful functor from commutative $B$-algebras to commutative $A$-algebras. Thus, $$\mathbf{CAlg}_B (A [x] \otimes_A B, C) \cong \mathbf{CAlg}_A (A [x], C) \cong U(C) \cong \mathbf{CAlg}_B (B [x], C)$$ and therefore $A [x] \otimes_A B \cong B [x]$.

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nice, thanks! the forgetful functor is simply the restriction of scalars functor, right? also we don't that the map is an inclusion, innit? – user88576 Feb 13 '14 at 12:56
Yes, that's right. – Zhen Lin Feb 13 '14 at 13:29

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