Gaussian Integers?

Question

Gaussian Integers?

Answer 1

No, not really, but the ring of Gaussian integers (that is, numbers of the form a + ib, with a, b integers and i² = -1) do share many properties with the ring of ordinary integers: you may define Gauss primes, prove unique factorization results, both are Euclidean Domains, their algebraic closure are the Gaussian rationals Q[i], etc. (One thing they do not share is order: i being a member block the existence of a total order in Z[i]).



Incidentally, Z[i] is but one of a number of ways of extending the integers to the complex domain, and the theory behind these extensions has non-trivial connections with Galois Theory and deep results in Algebraic Number theory.



See the link below.

Answer 2

Briefly, no, the integers are a proper subset of the Gaussian integers. The adjective Gaussian here serves to label this as an extension of the concept of an integer to a larger domain, not to constrain it to those integers which are Gaussian.



Of course, algebraic number theorists work with such extensions so frequently that "integer", unless otherwise specified, will usually be taken to refer to an algebraic integer, and if you want to indicate that you are talking about the ordinary integers you learned about in elementary school, you have to specify that it is a "rational integer." In that context, therefore, Gaussian integers are (algebraic) integers since a+bi is a root of the monic polynomial z² - 2az + (a² + b²).



Outside of algebraic number theory, though, integer is taken to refer only to elements of the set {...-3, -2, -1, 0, 1, 2, 3...}, and so most gaussian integers are not integers.

Answer 3

They are a subset of algebraic integers.

Algebraic integers are quantities that

satisfy a MONIC polynomial equation with

integer coefficients.

They are not rational integers, which are your

"actual" intetgers, but quadratic integers.