On the second irreducibility theorem of I. Schur

Let \(n\) $n$ be a positive integer different from \(8\) $8$ and \(n+1 \neq 2^u\) $n+1\ne 2^u$ for any integer \(u\geq 2\) $u\ge 2$ . Let \(\phi(x)\) $\varphi (x)$ belonging to \(Z[x]\) $Z[x]$ be a monic polynomial which is irreducible modulo all primes less than or equal to \(n+1\) $n+1$ . Let \(a_j(x)\) $a_j(x)$ with \(0\leq j\leq n-1\) $0\le j\le n-1$ belonging to \(Z[x]\) $Z[x]$ be polynomials having degree less than \(\deg\phi(x)\) $deg\varphi (x)$ . Assume that the content of \(a_na_0(x)\) $a_n{a}_0(x)$ is not divisible by any prime less than or equal to \(n+1\) $n+1$ . We prove that the polynomial \(f(x) = a_n\frac{\phi(x)^n}{(n+1)!}+ \sum _{j=0}^{n-1}a_j(x)\frac{\phi(x)^{j}}{(j+1)!}\) $$f(x)=a_n\frac{\varphi {(x)}^n}{(n+1)!}+\sum _{j=0}^{n-1}{a}_j(x)\frac{\varphi {(x)}^j}{(j+1)!}$$ is irreducible over the field \(Q\) $Q$ of rational numbers. This generalises a well-known result of Schur which states that the polynomial \( \sum _{j=0}^{n}a_j\frac{x^{j}}{(j+1)!}\) ${\sum }_{j=0}^n{a}_j\frac{x^j}{(j+1)!}$ with \(a_j \in Z\) $a_j\in Z$ and \(|a_0| = |a_n| = 1\) $|a_0|=|a_n|=1$ is irreducible over \(Q\) $Q$ . For proving our results, we use the notion of \(\phi\) $\varphi$ -Newton polygons and a few results on primes from number theory. We illustrate our result through examples.