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 letProposition 13 (Eisenstein's Criterion)

$\displaystyle f(x) = x^n + a_{n-1}x^{n-1} + ... ... + a_1x + a_0 $

be a polynomial in R[x] (here $\displaystyle n \ge 1$ )

Suppose $\displaystyle a_{n-1}, ... ... a_1, a_0 $ are all elements of P and suppose $\displaystyle a_0 $ is not an element of $\displaystyle P^2 $.

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

------------------------------------------------------------------------------------

The beginning of the proof reads as follows:

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 $\displaystyle x^n = \overline{a(x)b(x)}$ in (R/P)[x] where the bar denotes polynomials with coefficients reduced modulo P... .,.. etc. etc.

-----------------------------------------------------------------------------------

I will now take a specific example with R= Z as the integral domain concerned and P = (3) as the prime ideal in Z.

Also take (for example) $\displaystyle f(x) = x^3 + 9x^2 + 21x + 9 = (x+3) (x^2 +6x + 3) $

Now as the proof requires, reduce f(x) mod P

Now using D&F Proposition 2 (see attached) - namely $\displaystyle R[x]/(I) \cong R/I)[x] $ we have

$\displaystyle Z[x]/(3) \cong (Z/(3))[x]$

and so we to obtain $\displaystyle \overline{f(x)} $ we simply reduce the coefficients of f(x) by mod 3

Since $\displaystyle 9, 21 \in \overline{0} $

we have $\displaystyle \overline{f(x)} = \overline{x^3} $

The coset TEX] \overline{f(x)} [/TEX] would include elements such as $\displaystyle x^3 + 3, x^3 + 6x^2 + 24x - 3, ... ... $ and so on.

Can someone please confirm my working in this particular case of the Eisenstein proof is correct?

Peter