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Specific example of Eisenstein's Theorem using R = Z

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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]
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
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)$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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