Polynomial Rings Over Fields - Dummit and Foote

[Math Amateur]

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

$$g(x) = {f_1{(x)}}^{n_1} {f_2{(x)}}^{n_2}... {f_k{(x)}}^{n_k}$$

be its factorization into irreducibles, where all the $$f_i(x)$$ are distinct.

Then we have the following isomorphism of rings:

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

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The proof reads as follows:

Proof: This follows from the Chinese Remainder Theorem (Theorem 7.17), since the ideals $${f_i{(x)}}^{n_i}$$ and $${f_j{(x)}}^{n_j}$$ are comaximal if $$f_i(x)$$ and$$f_j(x)$$ 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 $${f_i{(x)}}^{n_i}$$ and $${f_j{(x)}}^{n_j}$$ are comaximal.

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

Any help would be very much appreciated.

Peter

[This has also been posted on MHF]

 

[jvicens]

Dear Peter,

Thank you for your question about Proposition 16 in Dummit and Foote's Section 9.5 on Polynomial Rings Over Fields II. I am happy to help clarify the argument for you.

To understand why the ideals {f_i(x)}^{n_i} and {f_j(x)}^{n_j} are comaximal, we need to first understand the concept of "relatively prime" in the context of Euclidean domains.

In a Euclidean domain, two elements a and b are said to be relatively prime if their greatest common divisor is 1. In other words, there is no element that divides both a and b except for 1. This is equivalent to saying that a and b have no common factors (except for units) in their factorizations.

Now, returning to the proof of Proposition 16, we are given that the polynomials f_i(x) and f_j(x) are distinct, meaning they are different irreducible factors of g(x). This implies that they have no common factors in their factorizations. Additionally, since g(x) is a monic polynomial, its factorization into irreducibles is unique. This means that the factorization of g(x) into f_i(x)^{n_i} and f_j(x)^{n_j} is also unique.

Therefore, since f_i(x) and f_j(x) have no common factors, their powers f_i(x)^{n_i} and f_j(x)^{n_j} also have no common factors. This implies that the ideals they generate, {f_i(x)}^{n_i} and {f_j(x)}^{n_j}, are also relatively prime. And since they are relatively prime, they are comaximal by definition.

I hope this helps clarify the argument for you. Please let me know if you have any further questions.