Eisenstein's criterion

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Let $f(x) \in \mathbb{Z}[x] - \mathbb{Z}$ such that $\forall 0 \le i < \deg(f), p \mid f_i$ and $p \not\mid f_{\deg(f)}$ and $p^2 \not\mid f_0$. Then $f(x)$ is irreducible in $\mathbb{Q}[x]$.

Proof

Assume $f(x)$ is reducible in $\mathbb{Q}[x]$. Therefore, $f(x) = a(x)b(x)$, where $\deg(a) \ge 1$ and $\deg(b) \ge 1$. Since $\mathbb{Q}$ has no zero-divisors, $\deg(f) = \deg(a) + \deg(b)$.

By Gauss' lemma, $f$ also has factors in $\mathbb{Z}[x]$. So we can assume that $a(x), b(x) \in \mathbb{Z}[x]$.

$f_0 = a_0b_0$. Since $p^2 \not\mid f_0$ and $p \mid f_0$, exactly one of $a_0$ and $b_0$ is divisible by $p$. Without loss of generality, $p \mid a_0$ and $p \not\mid b_0$.

Let $k$ be the smallest value such that $p \not\mid a_k$. $k \le \deg(a) < \deg(a) + \deg(b) = \deg(f) \Rightarrow p \mid f_k$.

$f_k = \sum_{j=0}^{k-1} a_jb_{k-j} + a_kb_0$. Since $p \mid a_j \forall j \le k-1$, $p \mid a_kb_0$. Since $p \not\mid b_0$, $p \mid a_k$ by Euclid's lemma. This contradicts the assumption that $k$ is the smallest value such that $p \not\mid a_k$. Therefore, no such smallest value exists and hence $p \mid a_i$ for all $i$.

\[ (\forall i, p \mid a_i) \Rightarrow p \mid a(x) \Rightarrow p \mid f(x) \Rightarrow (\forall i, p \mid f_i) \Rightarrow p \mid f_{\deg(f)} \Rightarrow \bot \]

Therefore $f(x)$ is irreducible.

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Depth: 6
Number of transitive dependencies: 21

Eisenstein's criterion

Dependencies:

Proof

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Info:

Transitive dependencies: