# Formal Cauchy-Riemann equations for formal power series without complex analysis

Consider the ring $\mathbb{C}[[X,Y]]$ and its subring $\mathbb{C}[[X+iY]]$, where $i=\sqrt{-1}$. One can show that $f(X,Y):=u(X,Y)+iv(X,Y)\in \mathbb{C}[[X,Y]]$ lies in $\mathbb{C}[[X+iY]]$ iff $u$ and $v$ satisfy the CR-equations formally. The necessary condition follows by straightforward formal (algebraic) derivation of the homogeneous parts, whereas the "if" part follows from the fact that the homogeneous parts are holomorphic polynomial functions, hence represented by a (homogeneous) polynomial function in a single complex variable (that part uses analysis since we have to look at the polynomials as polynomial functions).

Nevertheless, I cannot shake the feeling that the usage of a complex analysis argument is an overkill since the whole thing seems to play out in the algebraic realm.

Is there perhaps a formal/algebraic argument that entirely circumvents the analysis part of the argument (Kähler differentials?)?

Apologies if my question is inappropriate for MO, I am looking forward to learning about formal differential operators and equations in the algebraic setting.

Thanks!

• Using $z,\overline{z}$ instead of $x,y$ should give this quickly. May 7, 2017 at 0:21
• Ah, change of variables. And the $\overline{z}$ argument should easily disappear. I shall try this approach. Thank you!
– M.G.
May 7, 2017 at 0:25

$f\in C[[x+iy]]$ means $f(x,y)=g(x+iy),$ where $g\in C[[z]]$. Then $$f_x=g'(x+iy),\; f_y=ig'(x+iy),$$ therefore $f_y=if_x$ and this is Cauchy-Riemann: indeed, if $f=u+iv$ then $f_x=u_x+iv_x$ and $f_y=u_y+iv_y$, so $f_y=if_x$ is equivalent to
$u_x=v_y$, $u_y=-v_x$.
To prove the converse, make the linear change of the variables: $z=x+iy,\; \overline{z}=x-iy$. A simple calculation shows that $f_z=(1/2)(f_x-if_y),\; f_{\overline{z}}=(1/2)(f_x+if_y).$ Then the Cauchy-Riemann conditions in the new variables become $f_{\overline{z}}=0$. Assuming that this is satisfied, set $h=f_z$. Then $$h_{\overline{z}}=f_{z\overline{z}}=f_{\overline{z}z}=0,$$ so $h$ does not depend on $\overline{z}$ that is $h\in C[[z]]$. Taking an antiderivative of $h$ in $C[[z]]$ we obtain the $g$ as above.
• Thank you for furnishing out a complete answer, giving closure to the thread. It's a shame that the algebraic aspect is not stressed nearly enough in modern treatments. Two observations for myself: in the converse there is no need to look at the homogeneous polynomials, and $g$ is the derivative of $f$ (rather than $f$ itself as in the first part), so we still have to take a formal anti-derivative in $\mathbb{C}[[z]]$. Moreover, we have implicitly used the fact that $\mathbb{C}$ happens to have characteristic 0 when showing that $g$ does not depend on $\overline{z}$.