Eisenstein's criterion

Revision as of 14:47, 14 August 2018 by Naman12 (talk | contribs) (Proof and Extended Eisentein's Criterion)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Let $a_0, a_1, ... ,a_n$ be integers. Then, Eisenstein's Criterion states that the polynomial $a_nx^n+a_{n-1}x^{n-1}+ ... + a_1x+a_0$ cannot be factored into the product of two non-constant polynomials if:

$1) p$ is a prime which divides each of $a_0,a_1,a_2,...,a_{n-1}$

$2) a_n$ is not divisible by $p$

$3) a_0$ is not divisible by $p^2$

Proof

Assume $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+ a_1x+a_0$ and $f(x)=g(x)h(x)$ for non-constant polynomials $g(x)=g_rx^r+g_{r-1}x^{r-1}+\cdots+ g_1x+g_0$ and $h=h_sx^s+h_{s-1}x^{s-1}+\cdots+ h_1x+h_0$. Since $a_0$ has only one factor of $p$, we know that $p|g_0$ or $p|h_0$. WLOG, assume $p|g_0$. Then, we know that $a_1 \equiv g_0h_1+g_1h_0 \equiv g_1h_0 \equiv 0 \pmod{p}$. This means $p|g_1$. Similarily, we see, since $r<n$, $p|g_i$ for all $0\leq i \leq r$. This means that $p|g(x)$, so $p|f(x)$. However, we know that $p\nmid a_n$, a contradiction. Therefore, $f(x)$ is irreducible.

Extended Eisenstein's Criterion

Let $a_0, a_1, ... ,a_n$ be integers. Then, Eisenstein's Criterion states that the polynomial $a_nx^n+a_{n-1}x^{n-1}+ ... + a_1x+a_0$ has an irreducible factor of degree more than $k$ if:

$1) p$ is a prime which divides each of $a_0,a_1,a_2,...,a_{k}$

$2) a_{k+1}$ is not divisible by $p$

$3) a_0$ is not divisible by $p^2$

Proof

Let $f(x)=g(x)h(x)$, where $g(x)=g_rx^r+g_{r-1}x^{r-1}+\cdots+ g_1x+g_0$ and $h=h_sx^s+h_{s-1}x^{s-1}+\cdots+ h_1x+h_0$. Since $a_0$ has only one factor of $p$, we know that $p|g_0$ or $p|h_0$. WLOG, assume $p|g_0$. Then, we know that $a_1 \equiv g_0h_1+g_1h_0 \equiv g_1h_0 \equiv 0 \pmod{p}$. This means $p|g_1$. Similarily, we see, if $r\leq k$, $p|g_i$ for all $0\leq i \leq r$. This means that $p|g(x)$, so $p|f(x)$. However, we know that $p\nmid a_{k+1}$, a contradiction. Therefore, $r\geq k+1$.


This article is a stub. Help us out by expanding it.