Polynomial $f(X) g(Y)- g(X) f(Y)$ cannot be divided by a non-constant $h(X) \in F[X]$

Question

Let $F$ be a field and $f(X), g(X) \in F[X]$ two coprime non constant polynomials. Now the question is how to show that the symmetric polynomial $P(X,Y):= f(X) g(Y)- g(X) f(Y)$ can not be divided by a non constant polynomial $h(X) \in F[X] \setminus F$ in one variable $X$?

By this question $P(X,Y)$ need not to be irreducible, so we can probably find divisors in $F[X,Y]$. Now I think that it's not possible to find a divisor in $F[X]$ but I'm not convinced I can show it.

Is there standard approach to show it? I used following method about which I quite non convinced if this could lead me to a correct way to prove it: We know that for an arbitrary ring $A$ and ideal $I$ of $A$ there exist a reduction modulo $I$ isomorphism $A[Y]/I \cdot A[Y] \cong (A/I)[Y]$. For any $f = \sum^n a_i Y^i$ and $a \in A$ this implies that following statements are equivalent:

$I$ "divides" $f$ in the sense $f \in I \cdot A[Y]$ iff $a_i \in I$ for every $i $.

Now assume that $A$ is a $F$-algebra (in uor case $A=F[X]$) and non constant $f(Y),g(Y) \in F[Y]$ are $F$-linear independent. Let $a, a_f, a_g \in A$.

Is then $a | (a_f f(Y) + a_g g(Y))$ equivalent to $a | a_f$ & $ a| a_g$? I try think about it as an application of reduction modulo isomorphism to different basis instead the standard basis $1, Y, Y^2,...$. Does it make sense?

Otherwise, is there a more conceptional way to prove that $f(X) g(Y)- g(X) f(Y)$ can not be divided by a non constant polynomial $h(X) \in F[X] \backslash F$.

Answer 1

Let $f,g\in F[t]$ be non-constant and suppose $p(x,y):=f(x)g(y)-g(x)f(y)\in F[x,y]$ is divisible by $h(x)$ for some $h\in F[t]$. We will show that $f$ and $g$ share a common irreducible factor, which will show the desired result. Now, $p(x,y)$ is divisible by every factor of $h(x)$, so we may assume that $h$ is irreducible and monic. Embed $F[x,y]$ in $\overline{F}[x,y]$, where $\overline{F}$ is the algebraic closure of $F$, and let $\alpha\in\overline{F}$ be a root of $h$. (Note that, by construction, $h$ is the minimal polynomial of $\alpha$ over $F$.) Then we have $p(\alpha,y)=0$, so $f(\alpha)g(y)=g(\alpha)f(y)$. In particular, since $g(y)$ and $f(y)$ are both non-zero, $f(\alpha)=0$ if and only if $g(\alpha)=0$.

We thus have two cases; first suppose that $f(\alpha)$ and $g(\alpha)$ are both non-zero. Then we have $f(y)=\lambda g(y)$, where $\lambda=f(\alpha)\big/g(\alpha)\in\overline{F}$ is non-zero. In particular, since both $f(y)$ and $g(y)$ lie in $F[y]$, we must have $\lambda\in F^\times$, so $f(t)$ and $g(t)$ are associates and hence share all irreducible factors.

On the other hand, suppose that $f(\alpha)$ and $g(\alpha)$ are both zero. We claim that $h(t)$ is a common factor of both $f(t)$ and $g(t)$. To see this, note that, by the division algorithm, there are $q,r\in F[t]$ such that $f=qh+r$, where $\deg r<\deg h$. Then, embedding $F[t]$ in $\overline{F}[t]$, we have $$r(\alpha)=f(\alpha)-q(\alpha)h(\alpha)=0.$$ Since $h$ is the minimal polynomial of $\alpha$ over $F$, and $\deg r<\deg h$, this means $r$ must be the zero polynomial, whence $f=qh$ in $F[t]$. Exactly the same argument shows that $g$ is divisible by $h$ in $F[t]$, so $f$ and $g$ share a common irreducible factor, as desired.

Answer 2

For the sake of completeness I would like to add two alternative proofs which recently came in my mind:

1. Assume $h(X) \in F[X] \setminus F$ divides $P(X,Y)$. Let us apply the division algorithm twice and obtain $f(X)= a_f(X)h(X)+r_f(X)$ and $g(X)= a_g(X)h(X)+r_g(X)$. Then we get $P(X,Y) = h(X)(a_f(X)g(Y)) - h(X) (a_g(X)f(Y)) + (r_f(X)g(Y)-r_g(X)f(Y))$.
   Thus $h(X)$ divides $r_f(X)g(Y)-r_g(X)f(Y)$. Comparing the degrees in $X$ on both sides we conclude $r_f(X)g(Y)-r_g(X)f(Y)=0$. But $F[X,Y]$ is a UFD, so $r_f, r_g=0$ and we are done, and we obtain a contradiction to coprime assumption.

2. Since $f(X)= \sum^n f_i X^i$ and $g(X)= \sum^m g_i X^i$ are coprime they are of course $F$-linear independent and we can find two indices $k \neq l$ with the property that the two vectors $(f_k, g_k), (f_l, g_l) \in F^2$ are linearly independent. Now if $h(X)$ divides $P(X,Y)$, this implies by reduction modulo theorem that for every $i$ the $h(X)$ divides the linear combinations $g_if(X) + f_ig(X)$ as we as their own linear combinations. As by construction $g_kf(X)+ f_k g(X)$ and $g_lf(X)+ f_l g(X)$ are linear independent and divisible by $h(X)$, the procedure solving linear systems by building to diagonal shows that $h(X)$ divides also $f(X)$ and $g(X)$, and this again violates coprimeness of $f$ and $g$.