Showing that a polynomial is irreducible It seems intuitively clear that $xy-1$ in $\mathbb C[x,y]$ is irreducible. But I can't prove it rigorously. Could anyone show me how to prove it?
 A: Assume it is reducible, the degrees of the factors must be one.
So you are looking for two polynomials $P=ax+by+c$ and $P'=a'x+b'y+c'$ with $PP'=xy-1$. (we can take $P$ unitary in $x$).
Solving the constraints gives you $0=aa'=bb'=ac'+ca'=bc'+cb'$, $cc'=-1$ and $ab'+ba'=1$.
Without loss of generality, we obtain $a=0$. To avoid $P$ constant, we then must have $b'=0$.
We get $ca'=bc'=0$ but also still $cc'=-1$. This means $a'=b=0$, and both $P$ and $P'$ are constant, contradiction.
A: If $xy - 1$ is reducible, it must be the product of $(ax+by-c)$ and $(a'x+b'y-c')$ for some constants $a',b',c'$.  
Since there are no $x^2$ or $y^2$ terms, we can assume WLOG that $a' = b = 0.$  So, $xy - 1 = (ax - c)(b'y - c')$.
This would mean that for some $k = c/a$, $ky - 1 = 0$ for all $y$.  This is not the case.
A: Suppose $xy-1=f(x,y)g(x,y)$, where $f,g\in\mathbb{C}[x,y]$. Then $f(1,1)=0$ or $g(1,1)=0$. Moreover $f(x,0)g(x,0)=-1$ and $f(0,y)g(0,y)=-1$. Thus both $f$ and $g$ are constant polynomials with respect to $x$ and $y$. But this is a contradiction.
