5

I'm studing the ring of formal series with complex coefficients $\mathbb{C}[[x]]$. I proved that the polynomial $y^2-x^3-x^2$ is irreducible in $\mathbb{C}[x,y]$ but reducible in $\mathbb{C}[[x,y]]$. In fact we can factorize $y^2-x^2(x+1)$. So if we take $u=x\sqrt{x+1}=x(1+1/2x^2-1/8x^3+\cdots)$ we can write $(y-u)(y+u)$.

But sometimes these polynomials can't be factorized in $\mathbb{C}[[x,y]]$, because geometrically are cusps.

Is there a method to classify the polynomials that correspond to ''irreducible series''? If not, could you give some examples?

Community
  • 1
ArthurStuart
  • 4,932
  • For an irreducible $p(x,y)$ you can always find a uniformizing parameter $t$, which means $x=t^n$, $y=\sum_{k=m}^\infty a_k t^k=:s(t)$, with $a_m\neq0$ and $(m,n)=1$. By excluding $t$ you'll get $p(x,y)=0$; namely $p(x,y)=\prod_{\zeta^n=1}(y-s(\zeta t))$. BTW there is a method of factorization to irreducibles, due to Newton, namely the Newton polygon. – user8268 Sep 26 '13 at 18:50
  • @user8268 I didn't undertand. Is it simil to Weiestrass preparation theorem? Can tou give me some reference? – ArthurStuart Sep 26 '13 at 18:53
  • It is more like a local coordinate on a Riemann surface (given by $p(x,y)=0$). I should abstain from writing careless comments (i.e. sorry, no references). – user8268 Sep 26 '13 at 19:27
  • Yes... I think that they are cusps – ArthurStuart Sep 28 '13 at 05:32
  • @YACP For example $p(x,y)=y^2-x^3$ – ArthurStuart Sep 28 '13 at 06:34

1 Answers1

-1

I’m late, I know, but any way, there is something like a “criteria” on formal power series.

Let $k$ be an algebraically closed field. If $f(X,Y)\in k[[X,Y]]$ is irreducible then its leading form of $f$ is a power of a linear form.

The proof of this fact involves Weierstrass Preparation Theorem and Hensel’s Lemma.

Edit:

For example: $f(X,Y)=f_r(X,Y)+ f_{r+1}(X,Y)+ f_{r+2}(X,Y)+\ldots$, where $f_i$ is a homogeneous polynomial of degree $i$. If $f$ is irreducible then $f_r(X,Y)=(aX+bY)^r$

Claudiopc
  • 1