Mainly I ask, to check if there are holes in my logic anywhere locally, since these will affect me globally, and perhaps other readers.
If I want to prove that $xy-1\in \Bbb{C}[x,y]$ is irreducible, I believe can proceed as follows:
- Since this is a degree $2$ polynomial, for it to be reducible, it must be the product of two degree $1$ polynomials.
- Then we can take an arbitrary two polynomials of $\Bbb{C}[x,y]$ and we must have that $(ax+by+c)(dx+ey+f)=xy-1$
- We can consider the infinite dimensional $\Bbb{C}$-vectorspace structure of $\Bbb{C}[x,y]$ and consider just the (finite) subset of basis elements $\{1,x,y,xy,x^2,y^2\}$ (so as to match coefficients)
- We can look first at $adx^2=0$ and so $a$ or $d$ is zero. Without loss of generality we take $d=0$. Then we consider $bey^2$ and take without loss of generality $e=0$. Now $(ax+by+c)(f)$ must be degree $1$, since $f\in \Bbb C$. So this polynomial is irreducible.
Edit: Hardmath is totally right. $adx^2$ WLOG $d=0$, then I have $$(ax+by+c)(ey+f)=xy-1$$ and $bey^2=0$ so $b=0$ or $e=0$ and $aexy=xy$ means $a,e\ne 0$ so $b=0$, and $(ax+c)(ey+f)=xy-1$ means $cey=0$ so $c=0$, and $afx=0\implies f=0$, but now we have no scalar, to equal $-1$, contradiction, $xy-1$ is irreducible.