# 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!

• Since $x$ and $y$ are noncommuting formal variables, shouldn't your general noncommutative monomials rather be of the form $x^{0,1}y^{0,1}x^{0,1}y^{0,1}...$, where $0,1$ are the possible exponents? In other words, you should also have things like $fgf^2g^3fgf$. – M.G. Nov 29 '17 at 21:34
• @July, thank you for your comment. I think you are right. So the definition of algebraic dependence should be slightly more complicated than what I suggested. – user237522 Nov 29 '17 at 21:37

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.
• DavidSpeyer, nice idea, thank you. If I am not wrong, your statement is true, see ysharifi.wordpress.com/category/…, Algebras of GK dimension one, Theorem 2: Let $k$ be an algebraically closed field and let $A$ be a $k$-algebra. If $A$ is a domain and the GK dimension of $A$ is $\leq 1$, then $A$ is commutative". Perhaps it will be interesting to show the inverse statement: A commutative subalgebra of $A_1(\mathbb{C})$ has GK dimension at most one (perhaps this is a known result?). – user237522 Nov 30 '17 at 9:03
• @user237522 Victor's answer is 95% of the way to the inverse statement: He shows that, if $U$ and $V$ commute, then the ring generated by $U$ and $V$ has Krull dimension $1$. It follows from that the same is true for any finitely generated commutative subring. I bet you don't have to work much further to do all subalgebras whatsoever. – David E Speyer Dec 1 '17 at 1:37
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.