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I am reading Dummit and Foote Section 9.4 Irreducibility Criteria. In particular I am struggling to follow the proof of Eisenstein's Criteria (pages 309-310 - see attached).
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
f(x) = x^n + a_{n-1}x^{n-1} + ... ... + a_1x + a_0
be a polynomial in R[x] (here n \ge 1 )
Suppose a_{n-1}, ... ... a_1, a_0 are all elements of P and suppose a_0 is not an element of P^2.
Then f(x) is irreducible in R[x]The proof begins as follows: (see attachment)
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 x^n = \overline{a(x)b(x)} 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 \overline{a(x)} and \overline{b(x)} 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 a_0 of f(x) as the product of these two would be an element of P^2, 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 x^n = \overline{a(x)b(x)} 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 \overline{a(x)} and \overline{b(x)} 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]
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
f(x) = x^n + a_{n-1}x^{n-1} + ... ... + a_1x + a_0
be a polynomial in R[x] (here n \ge 1 )
Suppose a_{n-1}, ... ... a_1, a_0 are all elements of P and suppose a_0 is not an element of P^2.
Then f(x) is irreducible in R[x]The proof begins as follows: (see attachment)
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 x^n = \overline{a(x)b(x)} 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 \overline{a(x)} and \overline{b(x)} 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 a_0 of f(x) as the product of these two would be an element of P^2, 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 x^n = \overline{a(x)b(x)} 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 \overline{a(x)} and \overline{b(x)} 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]