Fake Proofs in Mathematics

Let $$f(x)=\sum_{r=0}^m a_rx^r, \qquad g(x)=\sum_{s=0}^n b_sx^s,$$ and suppose that $f(x)g(x)\in P[x]$.

Assume that neither $f$ nor $g$ belongs to $P[x]$. Then there are indices $i$ and $j$ such that $$a_i\notin P, \qquad b_j\notin P.$$ The coefficient of $x^{i+j}$ in $fg$ is $$a_ib_j+ \sum_{\substack{r+s=i+j\\(r,s)\neq(i,j)}}a_rb_s.$$ Since $fg\in P[x]$, this coefficient belongs to $P$. Because $P$ is an ideal, subtracting the remaining terms shows that $$a_ib_j\in P.$$ Since $P$ is prime, either $a_i\in P$ or $b_j\in P$, contradicting their choice. Therefore $f\in P[x]$ or $g\in P[x]$, so $P[x]$ is prime.