As in the title, I am wondering whether Morita equivalence commutes with tensor products or passing to polynomial rings. More specifically, are any of the following true?
If $A$ is Morita equivalent to $B$, are the rings $A\otimes C$ and $B\otimes C$ Morita equivalent? Here if $A,B$ are rings, the tensor product is over $\mathbb{Z}$, and if $A,B$ are $k$-algebras, it is over $k$.
If $A$ is Morita equivalent to $B$, is $A[x]$ Morita equivalent to $B[x]$?
1 implies 2, and I think the answer is positive for 2. In that case, $A\cong e\mathbb{M}^n(B)e$ for some full idempotent $e$ of the matrix ring $\mathbb{M}^n(B)$, and $e$ stays a full idempotent in the ring $\mathbb{M}^n(B)[x]=\mathbb{M}^n(B[x])$, so we have $A[x]\cong e\mathbb{M}^n(B[x])e$. But is 1 true at all? I don't think it is, but can't come up with a counterexample?
In fact, $1$ is true.
Here's why:
A left $A \otimes_{k} C$-module is the same thing as a left $A$-module $M$ together with a $k$-algebra homomorphism $C \to \operatorname{End}_A(M)$.
Likewise, a left $B \otimes_{k} C$-module is the same thing as a left $B$-module $M$ together with a $k$-algebra homomorphism $C \to \operatorname{End}_B(M)$.
So, if $F:_A\mathrm{Mod} \to _B\mathrm{Mod}$ is a $k$-linear equivalence, then any $k$-algebra homomorphism $C \to \operatorname{End}_A(M)$ includes a $k$-algebra homomorphism $C \to \operatorname{End}_B(F(M))$, which by the above remarks, gives a functor from left $A \otimes_{k} C$-modules to left $B \otimes_{k} C$-modules. This functor is then easily verified to be a $k$-linear equivalence, and so the $k$-algebras $A \otimes_{k} C$ and $B \otimes_{k} C$ are Morita equivalent.
The above applies for any commutative ring $k$, including the special case $k=\mathbb{Z}$.