Exercise 7.5.1

Proof. By hypothesis, $P\in F[x]$ divides in $F[x,y]$ the polynomial

so $Q(x,y)=P(x)S(x,y),S\in F[x,y]$. The evaluation $y\mapsto 0$ gives

thus $P\mid a_0$.

By induction, we suppose that $P\mid a_i,0\le i<k$, where $k\le m$.

Then $P$ divides $a_k(x)y^k+\cdots a_m(x)y^m=y^k(a_k(x)+\cdots +a_m(x)y^{m-k})$.

In the UFD $F[x,y]$, every irreducible factor of $y^k$ is associate to $y$, and $y$ doesn’t divide $P(x)$. Therefore $P(x)$ and $y^k$ are relatively prime, so $P$ divides $a_k(x)+\cdots +a_m(x)y^{m-k}$. The same evaluation $y\mapsto 0$ gives then $p\mid a_k$, so the induction is done. Consequently

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