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 $f(x)=\sum_{i=0}^{n}a_{i}x^{i}$, $g^{\prime}(x)=\sum_{j=0}^{\ell}b_{j}x^{j}$, $h^{\prime}(x)=\sum_{k=0}^{m}c_{k}x^{k}$ be the expansions of $f(x),g^{\prime}(x)$, and $h^{\prime}(x)$ respectively.

Let $\varphi:R[x]\rightarrow R/pR[x]$ be the natural homomorphism from $R[x]$ to $R/pR[x]$. Note that since $p\mid a_{i}$ for $i and $p\nmid a_{n}$, we have $\varphi(a_{i})=0$ for $i and $\varphi(a_{i})=\alpha\neq 0$

 $\varphi(f(x))=\varphi\left(\sum_{i=0}^{n}a_{i}x^{i}\right)=\sum_{i=0}^{n}% \varphi(a_{i})x^{i}=\varphi(a_{n})x^{n}=\alpha x^{n}$

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

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

If $\ell>0$, then $\varphi(b_{i})=0$ for $i<\ell$. In particular, $\varphi(b_{0})=0$, 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 $\ell>0$ 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 $\ell=0$ or $m=0$, that is, we must have a trivial factorization. Therefore $f(x)$ is irreducible.

Title proof of Eisenstein criterion ProofOfEisensteinCriterion 2013-03-22 12:42:11 2013-03-22 12:42:11 rspuzio (6075) rspuzio (6075) 11 rspuzio (6075) Proof msc 11C08 msc 13F15 GausssLemmaII