Ring Theory: Eisenstein's Criterion & Z[x]/pZ[x]

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The discussion centers on the quotient ring Z[x]/pZ[x], where Z[x] represents polynomials with integer coefficients and pZ[x] denotes polynomials divisible by a prime number p. The user, Mathguy, seeks clarification on Eisenstein's criterion and whether the reducibility of polynomials applies to general integers or is specific to primes. DonAntonio clarifies that the equality involving the quotient ring holds true in general for any ring, but emphasizes that the context of Eisenstein's criterion specifically requires p to be a prime ideal. Ultimately, Mathguy acknowledges understanding the argument used to prove Eisenstein's criterion.

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Mathguy15
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Hello,

I wanted to know something regarding the quotient ring Z[x]/pZ[x], where Z[x] is the set of all polynomials with integer coefficients and pZ[x], for a prime number p, is the set of all polynomials with integer coefficients divisible by p. I'm currently working through some notes on Applying Basic Abstract Algebra to Problems of Number Theory, and I don't understand the proof of Eisenstein's criterion. My confusion hinges upon this issue. The author of the notes says that if p(x) is a reducible polynomial with integer coefficients, then p(x)+Z[x]={q(x)+Z[x]}{r(x)+Z[x]}, where q(x) and r(x) are elements of Z[x] such that q(x)r(x)=p(x). Now, I know this seems really basic, but I have a question: Doesn't this hold for general integers rather than prime numbers? Must p be prime?

Thanks,
Mathguy
 
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Mathguy15 said:
Hello,

I wanted to know something regarding the quotient ring Z[x]/pZ[x], where Z[x] is the set of all polynomials with integer coefficients and pZ[x], for a prime number p, is the set of all polynomials with integer coefficients divisible by p. I'm currently working through some notes on Applying Basic Abstract Algebra to Problems of Number, and I don't understand the proof of Eisenstein's criterion. My confusion hinges upon this issue. The author of the notes says that if p(x) is a reducible polynomial with integer coefficients, then p(x)+Z[x]={q(x)+Z[x]}{r(x)+Z[x]}, where q(x) and r(x) are elements of Z[x] such that q(x)r(x)=p(x). Now, I know this seems really basic, but I have a question: Doesn't this hold for general integers rather than prime numbers? Must p be prime?

Thanks,
Mathguy



Where in what you quoted is the word "prime" written? If \,\,f(x)\in\mathbb Z[x]\,\, is reducible then \,\,\exists q(x)\,,\,r(x)\in\mathbb Z[x]\,\, s.t.

f(x)=q(x)r(x) . Why you add to each term in this equation the \,\,+\mathbb Z[x]\,\, is beyond my comprehension, but I can try to guess that

what the book/you is/are really trying to convey is that from the last equality it follows at once that f(x)+p\mathbb Z[x]=\left(q(x)+p\mathbb Z[x]\right)\left(r(x)+p\mathbb Z[x]\right) which is an equality in the quotient ring \,\,\mathbb Z[x]/p\mathbb Z[x]\cong \left(\mathbb Z/p\mathbb Z\right)[x] .


DonAntonio
 
DonAntonio said:
Where in what you quoted is the word "prime" written? If \,\,f(x)\in\mathbb Z[x]\,\, is reducible then \,\,\exists q(x)\,,\,r(x)\in\mathbb Z[x]\,\, s.t.

f(x)=q(x)r(x) . Why you add to each term in this equation the \,\,+\mathbb Z[x]\,\, is beyond my comprehension, but I can try to guess that

what the book/you is/are really trying to convey is that from the last equality it follows at once that f(x)+p\mathbb Z[x]=\left(q(x)+p\mathbb Z[x]\right)\left(r(x)+p\mathbb Z[x]\right) which is an equality in the quotient ring \,\,\mathbb Z[x]/p\mathbb Z[x]\cong \left(\mathbb Z/p\mathbb Z\right)[x] .


DonAntonio

Sorry, I meant to add pZ[x] to each term. Regarding the equality you stated, doesn't that hold more generally, or is it special to primes because pZ[x] is a prime ideal?

Thanks,
mathguy
 
Mathguy15 said:
Sorry, I meant to add pZ[x] to each term. Regarding the equality you stated, doesn't that hold more generally, or is it special to primes because pZ[x] is a prime ideal?

Thanks,
mathguy



If you meant the equality \,\,f(x)=q(x)r(x)\,\, then this is the meaning of "the element f(x) is reducible" in any ring , or when

it is given that the element q(x) divides the element f(x)...and this much is true for any kind of elements in a ring, not only polynomials.

DonAntonio
 
DonAntonio said:
If you meant the equality \,\,f(x)=q(x)r(x)\,\, then this is the meaning of "the element f(x) is reducible" in any ring , or when

it is given that the element q(x) divides the element f(x)...and this much is true for any kind of elements in a ring, not only polynomials.

DonAntonio

Oh, sorry again, I'm being awfully slow today. I meant the equality f(x)+pZ[x]={q(x)+pZ[x]}{r(x)+pZ[x]}. Doesn't this hold more generally for integers rather prime numbers?
 
Mathguy15 said:
Oh, sorry again, I'm being awfully slow today. I meant the equality f(x)+pZ[x]={q(x)+pZ[x]}{r(x)+pZ[x]}. Doesn't this hold more generally for integers rather prime numbers?



Oh, this is only the definition of product in the quotient ring \,\,\mathbb{Z}[x]/p\mathbb{Z}[x]\,\, ...

I don't understand your question about integers and primes,

DonAntonio
 
Oh, well, I understand the argument used to prove Eisenstein's criterion now. My original question is now irrelevant.

Thanks for the help, though.
 

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