Irreducible polynomials and prime elements

• I
• darksidemath
In summary, if ##p\in Z## is a prime and ##x^2-3\in \mathbb{Z}_p[x]## is irreducible, then ##\mathbb{Z}_p[\sqrt{3}]\cong \mathbb{Z}[\sqrt{3}]/(p)## is an integral domain, and possibly a field if the ideal generated by ##x^2-3## is maximal. The general case is more complicated and there could exist situations where the ideal generated by an irreducible polynomial is not maximal.
darksidemath
TL;DR Summary
How can I show that p is a prime element of Z[√3]?
let p∈Z a prime how can I show that p is a prime element of Z[√3] if and only if the polynomial x^2−3 is irreducible in Fp[x]? ideas or everything is well accepted :)

If ##p\in R## is prime, then ##R/(p)## is an integral domain.

If ##x^2-3\in \mathbb{Z}_p[x]## is irreducible, then ##\mathbb{Z}_p[x]/(x^2-3)\cong \mathbb{Z}_p[\sqrt{3}]\cong \mathbb{Z}[\sqrt{3}]/(p)## is an integral domain.

darksidemath
fresh_42 said:
If ##p\in R## is prime, then ##R/(p)## is an integral domain.

If ##x^2-3\in \mathbb{Z}_p[x]## is irreducible, then ##\mathbb{Z}_p[x]/(x^2-3)\cong \mathbb{Z}_p[\sqrt{3}]\cong \mathbb{Z}[\sqrt{3}]/(p)## is an integral domain.
I thought the quotient by the ideal generated by an irreducible is a field, not just an integral domain.

It's a field if the ideal is maximal. There are no problems in principal ideal domains, but the general case is more complicated.

Aren't ideals generated by irreducible polynomials maximal?

WWGD said:
Aren't ideals generated by irreducible polynomials maximal?
I'm not sure and have been too lazy to think about it. E.g. we could have a situation ##(p) \subsetneq (p,q) \subsetneq R.##

1. What is an irreducible polynomial?

An irreducible polynomial is a polynomial that cannot be factored into two or more polynomials of smaller degree with coefficients in the same field. In other words, it is a polynomial that cannot be broken down any further.

2. How do you determine if a polynomial is irreducible?

There are a few different methods for determining if a polynomial is irreducible. One method is to check if the polynomial has any roots in its field. If it does not have any roots, then it is irreducible. Another method is to use the Eisenstein criterion, which states that if a polynomial has a prime number as its leading coefficient and all other coefficients are divisible by that prime, then the polynomial is irreducible.

3. What is a prime element in a polynomial ring?

A prime element in a polynomial ring is an irreducible polynomial that cannot be factored into two or more polynomials of smaller degree. Prime elements are important because they are the building blocks of polynomial rings and can be used to define other important concepts, such as prime ideals.

4. How are irreducible polynomials and prime elements related?

Irreducible polynomials and prime elements are closely related because they both cannot be broken down any further. In fact, every irreducible polynomial is also a prime element, but not every prime element is necessarily irreducible. This means that prime elements are a subset of irreducible polynomials.

5. What is the significance of irreducible polynomials and prime elements in mathematics?

Irreducible polynomials and prime elements have many important applications in mathematics, particularly in the field of algebra. They are used in fields such as number theory, algebraic geometry, and coding theory. They also play a crucial role in the construction of field extensions and in the study of algebraic structures.

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