Irreducibility of plane curves

peccavi_2006 · **Posted:** Fri, 27 Apr 2012 19:27:08 UTC

So I was looking through past exam questions with a friend and we came across this one:

"Show that $f(x,y) = x^{4} - xy^{2} + y^{3}$ is irreducible over $\mathbb{C}^{2}$ ."

My first thought was that it's obviously true because because I can't factor anything out, but I have no idea how to prove it rigorously enough for exam-standard.

I thought something along the lines of
Assuming it can be written as the product of a line and a lower dimensional curve. Then
$f(x,y) = (y - \lambda x)(y^{2} + axy + x^{3})$
but this leads to a contradiction.
Would I also need to consider the case where I can pull out some sort of conic as well?

outermeasure · **Posted:** Fri, 27 Apr 2012 19:38:41 UTC

peccavi_2006 wrote:

So I was looking through past exam questions with a friend and we came across this one:

"Show that $f(x,y) = x^{4} - xy^{2} + y^{3}$ is irreducible over $\mathbb{C}^{2}$ ."

My first thought was that it's obviously true because because I can't factor anything out, but I have no idea how to prove it rigorously enough for exam-standard.

I thought something along the lines of
Assuming it can be written as the product of a line and a lower dimensional curve. Then
$f(x,y) = (y - \lambda x)(y^{2} + axy + x^{3})$
but this leads to a contradiction.
Would I also need to consider the case where I can pull out some sort of conic as well?

You need more than just $y-\lambda x$ .

Recall $\mathbb{C}[x,y]$ is a UFD.

Suppose f(x,y)=g(x,y)h(x,y)

be a (nontrivial) factorisation. Then taking degrees in y, you have $3=\deg_y f=\deg_y g+\deg_y h$ . So we may assume WLOG $\deg_y g=1$ (since the presence of the y^3 term means you cannot have $\deg_y g=0$ --- in exam you need to expand that a little bit more).

Hence we may assume g(x,y)=y-p(x)

, where p(x) divides x^4, so is one of $\lambda,\lambda x,\lambda x^2,\lambda x^3,\lambda x^4$ . Now examine each one in turn.

peccavi_2006 · **Posted:** Fri, 27 Apr 2012 19:44:18 UTC

I assume $\mathrm{deg}_{y}(g) \neq 0$ because we can't have that $f(x,y) = (h(x,y))^{2}$ , right? Hence there must be some g component in there.

outermeasure · **Posted:** Fri, 27 Apr 2012 19:57:49 UTC

peccavi_2006 wrote:

I assume $\mathrm{deg}_{y}(g) \neq 0$ because we can't have that $f(x,y) = (h(x,y))^{2}$ , right? Hence there must be some g component in there.

No. $\deg_y g=0$ means g is a polynomial in x alone. Thus when we write f as a polynomial in y with coefficients in $\mathbb{C}[x]$ , every coefficient must be divisible by g. That means $g\mid 1$ (1=coefficient of y^3), which gives trivial factorisation.

peccavi_2006 · **Posted:** Fri, 27 Apr 2012 20:28:07 UTC

ofc yeah - thanks outermeasure.

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Irreducibility of plane curves

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