A non-commutative analog of: two polynomials are algebraically dependent iff their Jacobian is zero Let $f,g \in \mathbb{C}[x,y]$.
There is a well-known result, that can be found for example 
here, pages 19-20, that says the following:

$f,g$ are algebraically dependent over $\mathbb{C}$ if and only if their Jacobian $Jac(f,g):=f_xg_y-f_yg_x$ is zero. 

Actually, this result is valid for $f_1,\ldots,f_n \in \mathbb{C}[x_1,\ldots,x_n]$, any $n \in \mathbb{N}$.
Now, let $f,g \in A_1(\mathbb{C})$, where $A_1(\mathbb{C})$
is the first Weyl algebra over $\mathbb{C}$, namely, the $\mathbb{C}$-algebra
generated by $x,y$ such that $[y,x]=yx-xy=1$.
I wonder if there exists an analog result to the above in $A_1(\mathbb{C})$,
namely:

$f,g \in A_1(\mathbb{C})$ are 'algebraically dependent' over $\mathbb{C}$ if and only if $[f,g]=0$.

One has to be careful because:
(1) One has to define algebraic dependence over $\mathbb{C}$ of two non-commuting elements $f$ and $g$.
Should it be $\sum \lambda_{ij}f^ig^j=0$, with $\lambda_{ij} \in \mathbb{C}$ not all zero,
or $\sum \lambda_{ij}f^ig^j + \sum \mu_{ij}g^if^j=0$, with $\lambda_{ij},
\mu_{ij} \in \mathbb{C}$ not all zero?.
(Perhaps the first definition should be called 'one-sided algebraic dependence',
while the second definition should be called 'two-sided algebraic dependence').
(2) Perhaps this question is relevant. The example there (of Dixmier) is of $U,V \in A_1$, $[U,V]=0$, and $U^3-V^2+1=0$, so it does not contradict my plausible analog result, since those $U$ and $V$ are algebraically dependent.
((3) I am not sure if this is relevant, but in the above mentioned reference, on page 11, the Gelfand-Kirillov dimension is mentioned with connection to transcendence degree; for $\mathbb{C}[x,y]$ those notions coincide. Is the fact that the Gelfand-Kirillov dimension of $A_1(\mathbb{C})$ is two relevant to my question?).
Thank you very much for any help!
 A: The reverse implication is true in a considerably more general setting (Burchnall-Chaundy theory). Namely, for any pair $(U,V)$ of commuting meromorphic coefficient differential operators in one variable of order at least one, there is a two-variable polynomial $P(z,w)$ such that $P(U,V)=0$ (the polynomial evaluation is unambiguous because $U$ and $V$ commute). There is an enormous body of literature on this topic related to integrable systems.  
The proposed forward implication does not appear interesting or meaningful to me. (However, see a very interesting formulation proposed by David Speyer in the comments.) For example, the defining relation $yx-xy-1=0$ between $x$ and $y$ may be viewed as a form of "noncommutative algebraic dependence", yet of course $x$ and $y$ do not commute. And the proposed "one-sided algebraic dependence" need not have good formal properties, such as symmetry or transitivity. At best, one can attempt to extract such properties from commutativity.
A: The first Weyl algebra over $C$ is isomorphic to the algebra of polynomials $C[x_1,x_2]$ equipped with a new multiplication, as follows:
define a linear operator $L$ on $C[x_1,x_2,y_1,y_2]$ to be the composite of partial differentiation w.r.t. $x_2$ and of partial differentiation w.r.t. $y_1$. Then $exp(L)$ defines an associative binary operation on $C[x_1,x_2]$ taking $f(x_1,x_2)$, $g(x_1,x_2)$ to the image of $exp(L)f(x_1,x_2)g(y_1,y_2)$ under the map taking the $y$s to $x$s. Sorry my lack of LaTeX skill prevents me writing this more elegantly. In fact, replacing $L$ by $tL$ gives you a $t$-parameter family of associative multiplications. By considering partial differentiation w.r.t. $t$ you can reduce the question of algebraic dependence in the Weyl algebra to that over the polynomial ring.
