Problem in the "proof" of Eisenstein's criterion on irreducibility.

Solution 1:

Suppose $b$ mod $P$ has nonzero constant term. Then $a$ mod $P$ must be identically zero, since otherwise its lowest nonzero coefficient would multiply by the constant term in $b$ mod $P$ to produce a nonzero term of degree less than $n$ in the product $ab$ mod $P$, since $P$ is prime. (It wouldn't be cancelled by anything else since all lower degrees of $a$ mod $P$ are $0$.) But that means $a$ is in $P$, and yet the original polynomial is not in $P$, so this can't happen.

Solution 2:

Since $R/P$ is a domain, $R/P[x]$ is a domain. In this ring you have $$x^n = \overline{a(x)} \cdot \overline{b(x)}$$ and $\overline{a(x)},\overline{b(x)}$ have degree $<n$.

Call $$\overline{a(x)} = \sum_{k \le i \le K} \overline{a_i} x^i \\ \overline{b(x)}=\sum_{h \le j \le H} \overline{b_j} x^j $$ Then $$x^n = \left( \sum_{k \le i \le K} \overline{a_i} x^i\right) \cdot \left( \sum_{h \le j \le H} \overline{b_j} x^j\right) = \sum_{m=0}^n \left(\sum_{i+j=m} \overline{a_i b_j} \right) x^m$$