- #1
Math Amateur
Gold Member
MHB
- 3,998
- 48
Eisenstein's Criterion is stated in Dummit and Foote as follows: (see attachment)
-----------------------------------------------------------------------------------
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]
------------------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------------
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]
-----------------------------------------------------------------------------------
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]
------------------------------------------------------------------------------------
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.
-----------------------------------------------------------------------------------
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: