proof of Eisenstein criterion

Let $f(x)\in R[x]$ be a polynomial satisfying Eisenstein’s Criterion with prime $p$.

Suppose that $f(x)=g(x)h(x)$ with $g(x),h(x)\in F[x]$, where $F$ is the field of fractions of $R$. Gauss’ Lemma II there exist $g^{\prime }(x),h^{\prime }(x)\in R[x]$ such that $f(x)=g^{\prime }(x)h^{\prime }(x)$, i.e. any factorization can be converted to a factorization in $R[x]$.

Let $f(x)={\sum }_{i=0}^n{a}_i{x}^i$, $g^{\prime }(x)={\sum }_{j=0}^{\mathrm{\ell }}{b}_j{x}^j$,$h^{\prime }(x)={\sum }_{k=0}^m{c}_k{x}^k$ be the expansions of $f(x),g^{\prime }(x)$, and $h^{\prime }(x)$ respectively.

Let $\phi :R[x]\to R/pR[x]$ be the natural homomorphism from $R[x]$ to $R/pR[x]$. Note that since $p\mid a_i$ for and $p\nmid a_n$, we have $\phi (a_i)=0$ for and $\phi (a_i)=\alpha \ne 0$

$$\phi (f(x))=\phi \left(\sum _{i=0}^n{a}_i{x}^i\right)=\sum _{i=0}^n\phi (a_i)x^i=\phi (a_n)x^n=\alpha x^n$$

Therefore we have $\alpha x^n=\phi (f(x))=\phi (g^{\prime }(x)h^{\prime }(x))=\phi (g^{\prime }(x))\phi (h^{\prime }(x))$ so we must have $\phi (g^{\prime }(x))=\beta x^{{\mathrm{\ell }}^{\prime }}$ and $\phi (h^{\prime }(x)=\gamma x^{m^{\prime }}$ for some $\beta ,\gamma \in R/pR$ and some integers ${\mathrm{\ell }}^{\prime },m^{\prime }$.

Clearly ${\mathrm{\ell }}^{\prime }\le \mathrm{deg}(g^{\prime }(x))=\mathrm{\ell }$ and $m^{\prime }\le \mathrm{deg}(h^{\prime }(x))=m$, and therefore since ${\mathrm{\ell }}^{\prime }{m}^{\prime }=n=\mathrm{\ell }m$, we must have ${\mathrm{\ell }}^{\prime }=\mathrm{\ell }$ and $m^{\prime }=m$.Thus $\phi (g^{\prime }(x))=\beta x^{\mathrm{\ell }}$ and $\phi (h^{\prime }(x))=\gamma x^m$.

If $\mathrm{\ell }>0$, then $\phi (b_i)=0$ for . In particular, $\phi (b_0)=0$, hence $p\mid b_0$. Similarly if $m>0$, then $p\mid c_0$.

Since $f(x)=g^{\prime }(x)h^{\prime }(x)$, by equating coefficients we see that $a_0=b_0{c}_0$.

If $\mathrm{\ell }>0$ and $m>0$, then $p\mid b_0$ and $p\mid c_0$, which implies that $p^2\mid a_0$. But this contradicts our assumptions on $f(x)$, and therefore we must have $\mathrm{\ell }=0$ or $m=0$, that is, we must have a trivial factorization. Therefore $f(x)$ is irreducible.