Scalar extension in polynomial ring Let $A$ be a commutative ring, $x$ an indeterminate over $A$, $B$ be a commutative, associative and unital $A$-algebra.

It's $B\otimes_AA[x]\cong B[x]$ a polynomial algebra over $B$ in $x$?

The answer is yes if:


*

*$A\to B$ is a ring localization as shown here.

*$A\to B$ is
surjective as shown here.


My attempt for the general case.
Let $B\to B\otimes_AA[x]$ be the canonical ring homomorphism.
Let $C$ be a $B$-algebra and $c\in C$.
There exists one and only one $A$-algebra homomorphism
\begin{align}
&A[x]\to C&
&p\mapsto p(c)
\end{align}
The composition $B\otimes_AA[x]\to B\otimes_A C$ with the canonical $B\otimes_AC\to C$ gives the ring homomorphism
\begin{align}
&\varphi:B\otimes_AA[x]\to C&
&b\otimes p\mapsto bp(c)
\end{align}
which makes the following diagram commutative

and sends $1\otimes x\mapsto c$.
Conversely, let $\varphi:B\otimes_AA[x]\to C$ be a ring homomorphism making the diagram commutative and sending $1\otimes x\mapsto c$.
By composing with the canonical ring homomorphism $A[x]\to B\otimes_AA[x]$ we get $\varphi(1\otimes p)=p(c)$, while by composing with the canonical $B\to B\otimes_AA[x]$ we get $\varphi(b\otimes 1)=b1_C$.
Consequently,
\begin{align}
\varphi(b\otimes p)
&=\varphi(b\otimes 1)\varphi(1\otimes p)\\
&=bp(c)
\end{align}
thus proving uniqueness.

Other proof by abstract nonsense are appreciated.
 A: Since $B$ is an $A$-algebra, you have a ring homomorphism $f:A\to B$, and this induces a functor $f^*:B-\mathbf{Alg}\to A-\mathbf{Alg}$, the restriction of scalars, that takes a $B$-algebra $K$ to $K$, with the same abelian group structure but scalar multiplication defined by
$$a\cdot_A k=f(a)\cdot_B k$$
for all $k\in K$.
If we denote $U_A:A-\mathbf{Alg}\to \mathbf{Set}$ and $U_B:B-\mathbf{Alg}\to \mathbf{Set}$ the forgetful functors, then we have $U_A\circ f^*=U_B$.
Now the forgetful functors have left adjoints $F_A\dashv U_A$, $F_B\dashv U_B$, which are precisely the "polynomial algebra" functors; and moreover the functor $(B\otimes_A\_) : L\mapsto B\otimes_AL$ is the left adjoint to $f^*$. Then the composition of two left adjoints is left adjoint to the composition of their right adjoints, i.e. $(B\otimes_A \_)\circ F_A \dashv U_A\circ f^*=U_B$; and since adjoints are unique up to isomorphism this implies $(B\otimes_A \_)\circ F_A\simeq F_B$. In particular,
$$B\otimes_A A[x]=B\otimes F_A(\{\ast\})\simeq F_B(\{\ast\})=B[x].$$
