Undergrad Irreducible polynomials and prime elements

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To show that a prime p in Z is a prime element of Z[√3] if and only if the polynomial x^2−3 is irreducible in Fp[x], one must consider the properties of integral domains and fields. If x^2-3 is irreducible in Z_p[x], then the quotient ring Z_p[x]/(x^2-3) forms an integral domain, implying that Z[√3]/(p) is also an integral domain. The discussion raises questions about whether ideals generated by irreducible polynomials are maximal, noting that while this is true in principal ideal domains, the general case can be more complex. Clarification is sought on the relationship between irreducibility and maximal ideals in the context of polynomial rings. Understanding these concepts is crucial for establishing the connection between prime elements and irreducible polynomials.
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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 :)
 
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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.
 
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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.##
 
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