0

I am trying to show that $y-x^2 \in \Bbb C[x,y]$ is irreducible by striving for a contradiction as follows.

Suppose that $y-x^2$ factors as $y-x^2=f(x,y)g(x,y)$ for $f,g \in \Bbb C[x,y]$. Then we can consider $y-x^2 \in \Bbb C[y][x]$ and we'll have that $y-x^2=(x+f_0(y))(g_0(y)-x)=f_0(y)g_0(y)-x^2+x(g_0(y) -f_0(y))$

so $g_0(y)-f_0(y)=0$ and $f_0(y)g_0(y)=y$. From here we'll get that $(g_0(y))^2= y$, but I'm not sure how I can get to a contradiction from here?

Jonas
  • 29
  • 1
    Just a degree argument? Anyway, one can use Eisenstein - see this post. – Dietrich Burde Apr 11 '22 at 16:32
  • I would use Eisenstein, but I don't know how to use it with polynomials of multiple variables. And by degree argument do you mean with respect to what? – Jonas Apr 11 '22 at 16:36
  • Did you read the answers at my link? One answer is exactly about Eisenstein for polynomials in several variables, even with a link. And degree with respect to $y$. – Dietrich Burde Apr 11 '22 at 16:37
  • It was a bit confusing as they were considering some invertible linear transformations also. I think I can just argue that $\deg( (g_0(y))^2) \ne \deg(y)$? – Jonas Apr 11 '22 at 17:08

1 Answers1

0

Suppose $$x^2-y=(ax+by+c)(a'x+b'y+c')\\=aa'x^2+(ab'+a'b)xy+(ac'+a'c)x+(bc'+b'c)y+bb'y^2+cc'$$

First we have $cc'=0$, assume without loss of generality that $c=0$. We then have $ac'=0$. $a\neq 0$, so $c'=0$. However, this is also impossible as $bc'+b'c\neq 0$.

cansomeonehelpmeout
  • 14,020
  • 3
  • 29
  • 55