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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!

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

1 Answer 1

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$f\in C[[x+iy]]$ 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.

Remark. This is how old textbooks like Whittaker-Watson explain the Cauchy-Riemann conditions. Pure algebra.

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    $\begingroup$ 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 $\overline{z}$. $\endgroup$
    – M.G.
    Commented May 7, 2017 at 9:48
  • $\begingroup$ @July: Thanks for your comments, I edited to make the notation more consistent. Many old textbooks take an "algebraic approach" proving as much as possible with power series, without integrals. With this approach, analytic functions are defined as sums of power series. This has also advantage of generality (think of p-adic analytic functions!) The last text that uses this approach that I know is Cartan. $\endgroup$
    – Alexandre Eremenko
    Commented May 7, 2017 at 10:00
  • $\begingroup$ See also mathoverflow.net/questions/116896/… where this elementary approach is used. $\endgroup$
    – Alexandre Eremenko
    Commented May 7, 2017 at 10:02

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