Hi guys I am trying to show the following statement and would appreciate if people can take a look and comment if I am on the right track.
$g(x,y)=y^2-p(x)$ is irreducible if and only if $p(x)$ is not a perfect square.
We are working over complex numbers.
Pf: We will take the contra-positive. Thus we assume $p(x)$ is a perfect square and want to show that $g(x,y)$ is reducible. Clearly we can do $y^2-p(x)=0$ thus $y= \sqrt{p(x)}$ note that $\sqrt{p(x)}$ is a polynomial because p us a perfect square of another polynomial. Thus $g(x,y)=(y+\sqrt{p(x)})(y-\sqrt{p(x)})$.
I am struggling showing that if $g(x,y)$ is reducible then p(x) has to be a perfect square.My attempt so far is let $g$ be reducible this $g(x,y)= k(x,y)m(x,y)$ without lost of generality assme that $m(x,y)= (ax+by)$.
Thus $k(x,y)(ax+by)=g(x,y)=y^2+p(x)$. Now I think I should make a divisibility argument that if you have $(ax+by)$ divide $y^2+p(x)$ we get a contradiction.
Since $g$ has degree $2$ in $y$, if $g$ is reducible then its factors must have degree $1$ in $y$: $$ y^2-p(x) = (A(x)y+a(x))(B(x)y+b(x))=A(x)B(x)y^2+(A(x)b(x)+B(x)a(x))y+a(x)b(x) $$ Therefore, $AB=1$, $Ab+Ba=0$, $ab=-p$.
Now $0=A(Ab+Ba)=A^2b+ABa$ and so $a=-A^2b$, which gives $-p=ab=-A^2b^2$ and so $p=(Ab)^2$.