Specific example of Eisenstein's Theorem using R = Z

[Math Amateur]

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

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Proposition 13 (Eisenstein's Criterion) Let P be a prime ideal of the integral domain R and let

[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]

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The beginning of the proof reads as follows:

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

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... .,.. etc. etc.

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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 \(\displaystyle \overline{f(x)} \) 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

[This post is also on MHF]

 

[Opalg]

To discuss this example in terms of Eisenstein's Criterion, you need to point out that neither the hypothesis nor the conclusion of Eisenstein's theorem is satisfied here. The coefficients $9$ and $21$ are both multiples of $3$, but the constant term $9$ is a multiple of $3^2$ (contrary to Eisenstein's Criterion). And the polynomial $f(x)$ is not irreducible because it factorises as $ (x+3) (x^2 +6x + 3)$.

 

[Math Amateur]

To discuss this example in terms of Eisenstein's Criterion, you need to point out that neither the hypothesis nor the conclusion of Eisenstein's theorem is satisfied here. The coefficients $9$ and $21$ are both multiples of $3$, but the constant term $9$ is a multiple of $3^2$ (contrary to Eisenstein's Criterion). And the polynomial $f(x)$ is not irreducible because it factorizes as $ (x+3) (x^2 +6x + 3)$.

Thanks Opalg.

Regarding my specific example, I think that my post was not completely clear in what I was attempting to demonstrate. I was taking a specific example and following the D&F proof on D&F page 310 - see attached - which assumes that f(x) is reducible and then proceeds to reduce f(x) modulo P. So, I took a reducible polynomial that (I thought) followed the Eisenstein rules for coefficients and then was focussed on showing how this led to the equation \(\displaystyle f(x) = \overline{a(x)b(x)} \) when f(x) is reduced modulo P.

Mind you as you point out I was wrong in allowing \(\displaystyle a_0 \in P^2 \). I am not sure it would really alter my exercise in establishing \(\displaystyle f(x) = \overline{a(x)b(x)} \)but I probably should have taken (say) \(\displaystyle f(x) = x^3 + 9x^2 + 24x + 18 = (x +3)(x^2 + 6x + 6) \) and then moved on (in parallel with or following the steps of D&F's proof to show that \(\displaystyle f(x) = \overline{a(x)b(x)} \) - since it is these steps that bother me.

*** Reflecting on the proof, I am confused by the following:

In D&F page 310 (see attached) we find the following:

"Suppose F(x) were reducible, say f(x) = a(x)b(x) in R[x], where a(x) and b(x) are nonconstant polynomials. Reducing the equation modulo P and using the assumptions on the coefficients we obtain the equation \(\displaystyle f(x) = \overline{a(x)b(x)} \) in (R/P)[x] ... ... etc

My confusion is as follows:

P is a prime ideal in R (in my specific example P + (3) is a prime ideal in R = Z)

BUT!

P is not an ideal in R[x] --> so how can we reduce the equation f(x) = a(x)b(x) which is in R[x] by an ideal P which is not even in R[x]?

I would be extremely grateful if someone could clarify this situation for me

Peter