# “Weierstrass preparation” of $\mathbb{C}[[X,Y]]$

In Lang's book "Algebra", theorem 9.2, it said that suppose $f\in \mathbb{C}[[X,Y]]$, then by some conditions imposed to $f$, $f$ can be written as a product of a polynomial $g\in \mathbb{C}[[X]][Y]$ and a unit $u$ in $\mathbb{C}[[X,Y]]$.

It suggests the following claim is not true in general.

Let $f\in \mathbb{C}[[X,Y]]$, then there exists a $g\in \mathbb{C}[X,Y]$ and a unit $u\in \mathbb{C}[[X,Y]]$ such that $f=gu$.

I would like to find a counter-example.

Thanks.

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It is known that there are transcendental power series $h(X)\in \mathbb C[[X]]$ over $\mathbb C[X]$. Note that $Xh(X)$ is also transcendental. Let $$f(X,Y)=Y-Xh(X)\in\mathbb C[[X,Y]].$$ Suppose $f=gu$ with $g$ polynomial and $u$ invertible. Consider the ring homomorphism $\phi: \mathbb C[[X,Y]]\to \mathbb C[[X]]$ which maps $X$ to $X$ and $Y$ to $Xh(X)$. Applying this homomorphism to $f=gu$, we get $$0 = g(X, Xh(X))\phi(u), \quad \phi(u)\in \mathbb C[[X]]^*.$$ So $g(X, Xh(X))=0$. As $Xh(X)$ is transcendental over $\mathbb C[X]$, and $g(X,Y)\in \mathbb C[X][Y]$, this implies that $g(X,Y)=0$. Hence $f(X,Y)=0$, absurd.
Wonderful! How stupid me. A typo, $d Xh(X)$ should be $Xh(X)$. –  wxu May 18 '12 at 23:57