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What makes a gaussian integer an irreducible integer?


Asked by Aliyah Bernard on Dec 05, 2021 FAQ



As the Gaussian integers form a principal ideal domain they form also a unique factorization domain. This implies that a Gaussian integer is irreducible (that is, it is not the product of two non-units) if and only if it is prime (that is, it generates a prime ideal ). The prime elements of Z[i] are also known as Gaussian primes.
Likewise,
Gaussian primes. This implies that a Gaussian integer is irreducible (that is, it is not the product of two non-units) if and only if it is prime (that is, it generates a prime ideal ). The prime elements of Z [i] are also known as Gaussian primes. An associate of a Gaussian prime is also a Gaussian prime.
Similarly, As for every unique factorization domain, every Gaussian integer may be factored as a product of a unit and Gaussian primes, and this factorization is unique up to the order of the factors, and the replacement of any prime by any of its associates (together with a corresponding change of the unit factor).
In addition,
Principal ideals. Since the ring G of Gaussian integers is a Euclidean domain, G is a principal ideal domain, which means that every ideal of G is principal. Explicitly, an ideal I is a subset of a ring R such that every sum of elements of I and every product of an element of I by an element of R belong to I.
One may also ask,
The only Gaussian integers which are invertible in Z[i] are 1 and i. Proof. It is easy to see 1 and ihave inverses in Z[i]: 1 and 1 are their own inverse and iand iare inverses of each other. For the converse direction, suppose \u000b2Z[i] is invertible, say \u000b\f= 1 for some \f2Z[i]. We want to show \u000b2f 1; ig.