Homogeneous polynomials Let $f$ be a homogeneous polynomial of degree $k$ in $K[x,y]$ where $K$ is a field, let $g$ be a homogeneous polynomial of degree $j$ in $K[x,y]$ such that $f(x,y)/x^{k} = g(x,y)/y^{j}$. Why this implies that $f(x,y)=x^k$ and $g(x,y)=y^j$?
 A: This is not quite true. The conclusion should instead by that $f=cx^k$ and $g=cy^k$ for some $c\in \mathbb k$ (I use $\mathbb k$ to denote the field to avoid confusion with the degree $k$).
Proof: Note that the degrees of $f$ and $g$ have to be the same since the degrees of $x^k$ and $y^k$ are the same. Furthermore, $f$ must be divisible by $x^k$ and $g$ by $y^k$ since $f(x,y)/x^k=g(x,y)/y^k\in k(y)[x]$ and $g(x,y)/y^k=f(x,y)/x^k\in k(x)[y]$, thus $f=f'x^k$ and $g=g'y^k$ for some homogeneous $f',g'\in k[x,y]$. But the degrees of $f$ and $g$ are both $k$, so the degrees of $f'$ and $g'$ must be $0$. They must also be equal, as $f'=f/x^k=g/y^k=g'$, so $f(x,y)=cx^k$ and $g(x,y)=cy^k$ for some $c\in \mathbb k$.
A: Have a closer look at $f(x,y)/x^k$. Since $f$ is homogeneous, the quotient is a polynomial in $y/x$. Similarly $g/y^k$ is a polynomial in $x/y$. Since they are equal, they have to be constant.
Remark: In fact they do not actually have to be equal to $x^k$ or $y^k$, but to a scalar multiple. I assume that in your (algebraic geometrical ?) setting you consider the polynomials up to a unit anyway.
