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Polynomial Rings Over Fields - Dummit and Foote

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading Dummit and Foote Section 9.5 Polynomial Rings Over Fields II.

I am trying to understand the proof of Proposition 16 which reads as follows:

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Proposition 16. Let F be a field. Let g(x) be a nonconstant monic polynomial of F[x] and let

[TEX] g(x) = {f_1{(x)}}^{n_1} {f_2{(x)}}^{n_2}..... {f_k{(x)}}^{n_k} [/TEX]

be its factorization into irreducibles, where all the [TEX] f_i(x) [/TEX] are distinct.

Then we have the following isomorphism of rings:

[TEX] F[x]/(g(x)) \cong F[x]/{f_1{(x)}}^{n_1} \ \times \ F[x]/{f_2{(x)}}^{n_2} \ \times \ ... ... \ \times \ F[x]/{f_k{(x)}}^{n_k} [/TEX]

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The proof reads as follows:

Proof: This follows from the Chinese Remainder Theorem (Theorem 7.17), since the ideals [TEX] {f_i{(x)}}^{n_i} [/TEX] and [TEX] {f_j{(x)}}^{n_j} [/TEX] are comaximal if [TEX] f_i(x) [/TEX] and[TEX] f_j(x) [/TEX] are distinct (they are relatively prime in the Euclidean Domain F[x], hence the ideal generated by them is F[x]).

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My question: I can follow the reference to the Chinese Remainder Theorem but I cannot follow the argument that establishes the necessary condition that the [TEX] {f_i{(x)}}^{n_i} [/TEX] and [TEX] {f_j{(x)}}^{n_j} [/TEX] are comaximal.

That is, why are[TEX] {f_i{(x)}}^{n_i} [/TEX] and [TEX]{f_j{(x)}}^{n_j}[/TEX] comaximal - how exactly does it follow from [TEX] f_i(x)[/TEX] and [TEX] f_j(x) [/TEX] being relatively prime in the Euclidean Domain F[x]?

Any help would be very much appreciated.

Peter

[This has also been posted on MHF]