Are Weyl algebra $A_1$ and it's opposite algebra isomorphic? Let $A$ be a noncommutative ring and $ab=c$ in $A$. $A'$ is it's opposite ring if $ba=c$ in $A'$. If $A$ is a Weyl algebra $A_1$, are $A$ and $A'$ isomorphic?
I have an idea, but I think it's wrong. Let $p, q$ be generators of $A$.
So in $A$ we have $pq=qp+1$.
In $A': qp=qp+1 \implies 0=1$, so there can't be an isomorphism if $0 \neq 1$ in $A$. I think it's a bullshit, but no other ideas... I need your help! It's a task from my research seminar "Geometry and dynamics" for first-year undergraduates in HSE, Moscow.
 A: The Weyl algebra $A$ is given by two generators $p$ and $q$ subject to the single relation $p q - q p = 1$.
The opposite algebra $A^{\mathrm{op}}$ is therefore given by two elements $p'$ and $q'$ subject to the relation $q' p' - p' q' = 1$.
Consequently, we have an isomorphism of algebras $A ≅ A^{\mathrm{op}}$ given on generators by $p \mapsto q'$ and $q \mapsto p'$.
Alternatively, let $$ be the three-dimensional Heisenberg Lie algebra, which given by the vector space basis $p, q, c$ subject to the commutator relation $[p, q] = c$ and $c$ central in $$.
We have $A ≅ \mathrm{U}() / (c - 1)$ where $\mathrm{U}()$ is the universal enveloping algebra of $$.
We have $ ≅ ^{\mathrm{op}}$ via $p \mapsto q$, $q \mapsto p$ and $c \mapsto c$.
This isomorphism of Lie algebras induces isomorphisms of algebras
$$
  A
  ≅ \mathrm{U}() / (c - 1)
  ≅ \mathrm{U}(^{\mathrm{op}}) / (c - 1)
  ≅ \mathrm{U}()^{\mathrm{op}} / (c - 1)
  ≅ ( \mathrm{U}() / (c - 1) )^{\mathrm{op}}
  ≅ A^{\mathrm{op}} \,.
$$
(This is the same isomorphism as in the first argumentation.)
