Let $h(t,x)$ and $f ( t, x)$ be nonzero elements of $\mathbb C[t,x]$. Suppose that $h$ is not divisible by any polynomial of the form $t — \alpha$. If $h$ divides $f$ in $\mathcal F[x]$, then $h$ divides $f$ in $\mathbb C[t,x]$. (This proposition is in "Algebra" by Artin book).
The proof seemed to be easy but when I tried to embed the intuition into logic, I got stuck.
Proof:
we have, $f = h q$ , where $q \in \mathcal F[x]$, it is a polynomial in x whose coefficients are rational functions in $t$. We multiply both sides of the equation $f = h q$ by a monic polynomial in $t$ to clear denominators in these coefficients. This gives us an equation of the form $u ( t) f ( t, x) = h (t, x) q_1(t, x )$.
Let $\alpha $ be a root of $u(t)$, then $t-\alpha$ divides RHS as well, $t — \alpha$ does not divide $h$, so $h(\alpha,x)$ is not zero. Since the polynomial ring $\mathbb C[x]$ is an integral domain, $q_1 (\alpha, x) = 0$, and the lemma shows that $t-\alpha$ divides $q_1(t, x)$. We cancel $t-\alpha$ from $u$ and $q_1$.
Problem: (All the implications below are mine, thus can be wrong)
we have, $q \in \mathcal F[x]$ where $\mathcal F = \mathbb C (t)=$rational field of $\mathbb C [t]$.
So, $q=\frac{a_n(t)}{b_n(t)}x^n+\frac{a_{n-1}(t)}{b_{n-1}(t)}x^{n-1}+\frac{a_{1}(t)}{b_{1}(t)}x+\frac{a_{0}(t)}{b_{0}(t)}$
$ q_1 \in \mathbb{C}[t,x]$, Let $q_1=u(t)*q=c_{n}(t)x^n+c_{n-1}(t)x^{n-1}+\cdots+c_{1}(t)x+c_{0}(t)$, where, $u(t)=lcm(b_{n}(t),b_{n-1}(t),\cdots, b_{0}(t))$ and $c_j(t)=\frac{u(t)}{b_j(t)}a_{j}(t)$
$(t-\alpha)|q_1 \Rightarrow (t-\alpha)| c_{j}(t) \; \forall \; j={1,\cdots,n} $. let $(t-\alpha)$ has multiplicity $k$ in $u(t)$ which is due to some $b_{p}(t)$. Now, $c_{p}(t)$ will not have any power of $(t-\alpha)$ which would mean $(t-\alpha)$ does not divide $c_j(t) \; \forall j$, which is a contradiction to what I have assumed initially. What am I doing wrong here?
I asked this question on mathoverflow as well. Here is the link.