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- Jun 22, 2012

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Eisenstein's Criterion is stated in Dummit and Foote as follows: (see attachment)

__Let P be a prime ideal of the integral domain R and let__

**Proposition 13 (Eisenstein's Criterion)**[TEX] f(x) = x^n + a_{n-1}x^{n-1} + ... ... + a_1x + a_0 [/TEX]

be a polynomial in R[x] (here [TEX] n \ge 1[/TEX] )

Suppose [TEX] a_{n-1}, ... ... a_1, a_0 [/TEX] are all elements of P and suppose [TEX] a_0 [/TEX] is not an element of [TEX] P^2 [/TEX].

Then f(x) is irreducible in R[x]

The proof begins as follows: (see attachment)

__Suppose f(x) were reducible, say f(x) = a(x)b(x) in R[x] where a(x) and b(x) are nonconstant polynomials.__

**Proof:**Reducing the equation modulo P and using the assumptions on the coefficients of f(x) we obtain the equation [TEX] x^n = \overline{a(x)b(x)}[/TEX] in (R/P)[x] where the bar denotes polynomials with coefficients reduced modulo P. Since P is a prime ideal, R/P is an integral domain, and it follows that [TEX] \overline{a(x)} [/TEX] and [TEX] \overline{b(x)} [/TEX] have zero constant term i.e. the constant terms of both a(x) and b(x) are elements of P. But then the constant term [TEX]a_0 [/TEX] of f(x) as the product of these two would be an element of [TEX] P^2 [/TEX], a contradiction.

**Questions/Issues**(1) I cannot see exactly how the following follows:

"Reducing the equation modulo P and using the assumptions on the coefficients of f(x) we obtain the equation [TEX] x^n = \overline{a(x)b(x)}[/TEX] in (R/P)[x]"

I am struggling to see just exactly how this follows: can someone please clarify this?

(2) I am somewhat unsure of the following:

"t follows that [TEX] \overline{a(x)} [/TEX] and [TEX] \overline{b(x)} [/TEX] have zero constant term"

This probably is a consequence of the mechanics of the reduction modulo P process but can someone please clarify exactly how it follows?

Peter

[NOTE: This has also been posted on MHF]