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Could anyone tell me why $g(x, y) = x^3 - y^2$ is irreducible in $\mathbb C[[x]][y]$ while $f(x, y) = x^3 - x^2 + y^2$ is not?

Yoshinobu Osawa
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    What is C{x}? {} –  Jan 12 '13 at 06:20
  • Since you accepted an answer assuming that $\mathbb C{x}$ is $\mathbb C[[x]]$, I think this is what you meant, although as far as I know $\mathbb C{x}$ denotes the ring of convergent power series; see here. –  Jan 12 '13 at 14:25

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I supposing you mean the ring $\mathbf C[[x]][y]$ of polynomials in$~y$ over power series in$~x$. A non-trivial decomposition of those polynomials, made monic in $y$, would require finding in $\mathbf C[[x]]$ a square root of the negated constant term (coefficient of $y^0$) in both cases. Now $x^3$ cannot have a square root because it has odd valuation. However $x^2-x^3$ has valuation $2$, and a square root would be $x$ times $\sqrt{1-x}$, where the latter square root exists in $\mathbf C[[x]]$ by Newton's binomial formula $$ (1-x)^\frac12 = \sum_{k=0}^\infty\binom{1/2}k(-x)^k. $$

Marc van Leeuwen
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  • Does everything you said also apply to $\mathbb{C}[[x]][[y]] = \mathbb{C}[[x, y]]$? Also wen you talk about valuations, what valuation function are you using? And why does $x$ having odd valuation imply it can't have a square root? – Anakhand Dec 03 '23 at 18:30
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    @Anakhand Irreducible monic polynomials in some ring $D[Y]$ over an integral domain $D$ will still be irreducible in $D[[Y]]$, and this applies here with $D=\mathbf{C}[[X]$, so yes, the question of reducibility is the same in $\mathbf{C}[[X,Y]]$. The valuation in $\mathbf{C}[[X]$ is the standard discrete valuation: the degree of the lowest non-zero term in the power series. – Marc van Leeuwen Dec 05 '23 at 09:37
  • Thank you. Could you provide a reference for "irreducible monic polynomial is also irreducible as power series"? – Anakhand Dec 05 '23 at 20:57
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    It seems this is a counterexample to what you stated. Perhaps I misunderstood, or it holds in under some extra assumptions on $D$? – Anakhand Dec 06 '23 at 21:22
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Assuming C{x} means the power series ring in $x$ over the complex numbers,

$$f(x,y) = y^2 + x^3 - x^2 = y^2 - (x^2 - x^3) = y^2 - x^2(1-x) = (y - x \sqrt{1-x})(y + x \sqrt{1-x})$$ and $\sqrt{1-x}$ can be expanded as a power series is $x$. Therefore $f$ is reducible.

Ted
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