Let \( R = \mathbb{Z}[\sqrt{-n}] \) where \( n \) is a square-free integer greater than 3 and \( -n \equiv 2, 3 \pmod{4} \).

(a) Prove that \( 2, \sqrt{-n} \), and \( 1 + \sqrt{-n} \) are irreducibles in \( R \).

(b) Prove that \( R \) is not a UFD. Conclude that the quadratic integer ring \( \mathcal{O} \) is not a UFD for \( D \equiv 2, 3 \pmod{4}, D < -3 \) (so also not an ED and not a PID).

Hint: Show that either \( \sqrt{-n} \) or \( 1 + \sqrt{-n} \) is not prime.

Question

12-10-2023
Mathematics

High School

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Let \( R = \mathbb{Z}[\sqrt{-n}] \) where \( n \) is a square-free integer greater than 3 and \( -n \equiv 2, 3 \pmod{4} \).

(a) Prove that \( 2, \sqrt{-n} \), and \( 1 + \sqrt{-n} \) are irreducibles in \( R \).

(b) Prove that \( R \) is not a UFD. Conclude that the quadratic integer ring \( \mathcal{O} \) is not a UFD for \( D \equiv 2, 3 \pmod{4}, D < -3 \) (so also not an ED and not a PID).

Hint: Show that either \( \sqrt{-n} \) or \( 1 + \sqrt{-n} \) is not prime.

Answer :

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anuragpalchoudhury19 anuragpalchoudhury19 · Answer 1 · 2023-09-28T16:35:58-04:00

Final answer:

To show the irreducibility of 2, \(\sqrt{-n}\), and 1+\(\sqrt{-n}\) in the ring Z[\(\sqrt{-n}\)], we must prove that these elements have no non-unit factors. For R to not be a UFD, we must demonstrate that either \(\sqrt{-n}\) or 1 + \(\sqrt{-n}\) is not prime. The failure of R being a UFD implies the quadratic integer ring O is not a UFD, ED, or PID for D = 2,3 mod 4 when D < -3.

Explanation:

To show that elements 2, \(\sqrt{-n}\), and 1+\(\sqrt{-n}\) are irreducible in the ring R = Z[\(\sqrt{-n}\)], we examine if they have non-unit factors within the ring. For a number to be irreducible, it cannot be factored into the product of two non-units of the ring. In the given ring R, it is necessary to show that these elements cannot be factored, or if factored, the factors are units or associated with the numbers themselves.

To prove that R is not a Unique Factorization Domain (UFD), we can demonstrate that either \(\sqrt{-n}\) or 1 + \(\sqrt{-n}\) is not prime. Since one of these elements is not prime, they cannot be unique in their factorization, and as such, R fails to be a UFD. The failure of R to be a UFD leads to the conclusion that the quadratic integer ring O is not a UFD for D = 2,3 mod 4, D < -3; hence it is neither a Euclidean Domain (ED) nor a Principal Ideal Domain (PID).