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Eisenstein criterion (Theorem)
Theorem 1 (Eisenstein criterion)   Let $ f$ be a primitive polynomial over a commutative unique factorization domain $ R$, say
$\displaystyle f(x)=a_0 + a_1x + a_2x^2 + \ldots + a_nx^n \;.$
If $ R$ has an irreducible element $ p$ such that
$\displaystyle p\mid a_m \qquad 0\le m\le n-1$
$\displaystyle p^2 \nmid a_0$
$\displaystyle p \nmid a_n$
then $ f$ is irreducible.
Proof. Suppose
$\displaystyle f=(b_0 + \ldots + b_s x^s)(c_0 + \ldots + c_t x^t)$
where $ s>0$ and $ t>0$. Since $ a_0 = b_0 c_0$, we know that $ p$ divides one but not both of $ b_0$ and $ c_0$; suppose $ p \mid c_0$. By hypothesis, not all the $ c_m$ are divisible by $ p$; let $ k$ be the smallest index such that $ p\nmid c_k$. We have $ a_k = b_0 c_k + b_1 c_{k-1} + \ldots + b_k c_0$. We also have $ p\mid a_k$, and $ p$ divides every summand except one on the right side, which yields a contradiction. QED $ \qedsymbol$



"Eisenstein criterion" is owned by Daume. [ full author list (3) | owner history (3) ]
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See Also: Gauss's lemma II, irreducible polynomial, monic, alternative proof that $\sqrt{2}$ is irrational

Other names:  Eisenstein irreducibility criterion

Attachments:
proof of Eisenstein criterion (Proof) by rspuzio
Eisenstein criterion in terms of divisor theory (Theorem) by pahio
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Cross-references: QED, contradiction, side, right, divisible, hypothesis, divides, irreducible, irreducible element, unique factorization domain, commutative, primitive polynomial
There are 6 references to this entry.

This is version 10 of Eisenstein criterion, born on 2002-02-03, modified 2006-07-31.
Object id is 1724, canonical name is EisensteinCriterion.
Accessed 6513 times total.

Classification:
AMS MSC: 13A05 (Commutative rings and algebras :: General commutative ring theory :: Divisibility)
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