The Elegant Way To Generate The Complex Number Field

* A note on notation: Notation is very important in mathematics, witness the notation differences between Newton's and Leibnitz's calculus. Who won out? I've tried to avoid confusion by trying to be as careful as I can with the relevant notation. In particular, [.....] stands for equivalence class or coset, and <.....> for an ideal of a ring. However, in the act of factoring, the ideal takes on the role of zero equivalence class -- the additive identity element -- in the quotient ring thus generated. So, it can be a little confusing. *

* It's almost like we're going in the back door to develop F[x] into a field. Find a polynomial that has no zeros or roots in the field F, use it to generate a maximal (principle) ideal, then use this to factor the ring of polynomials thereby constructing a field of polynomials, and by so accomplishing, ensure that all polynomials over the extended field have zeros or solutions. The existence of multiplicative inverses insures the absence of zero divisors, an important feature when factoring. *

* Outline of Proof: Given: F[x] is a commutative ring with unity [the multiplicative identity]. We need to show F[x]/(p(x)), for p(x) irreducible, is a field, and in order to do that we need to show that nonzero elements are invertible, that is to say, have inverses. The proof of inverses for non-zero polynomials is too lengthy to include here. It can be found on the page on Quotient Rings below. In the proof, as it turns out, if p(x) is not irreducible, then F[x]/(p(x)) has zero divisors, in which case it fails to even be an integral domain.*

Book Review of An Imaginary Tale: The Story of the √-1 | by Paul J.Nahin
Review by Brian E. Blank | PDF File

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