UIC Master’s Exam- Spring 2007 A2

Problem Statement: (a) State the division algorithm for \mathbb{Q}[x] (b) Show that \mathbb{Q}[x] is a principle ideal domain.

Solutions:

(a) The Division Algorithm: If f(x), q(x)\in\mathbb{Q}[x] with \deg q(x)\leq\deg f(x) then there exist q(x), r(x)\in\mathbb{Q}[x] such that f(x)=g(x)q(x)+r(x) where \deg r(x)<\deg g(x).

(b) Proof: Let I be an ideal of \mathbb{Q}[x]. If I is the trivial ideal then it is principal. So suppose that I is not the trivial ideal. Then there is some non-zero polynomial f\in I such that f has the least degree of any polynomial in I.

Let g\in I, then by the division algorithm we may write

g(x)=f(x)q(x)+r(x)

where q(x), r(x)\in\mathbb{Q}[x] and \deg r(x)<\deg f(x). Now, since I is an ideal and f, g,q\in I it follows that g(x)-f(x)q(x)\in I, so r\in I. But \deg r(x)<\deg f(x) which is the least possible degree in I, so it must be that r(x)=0. And so we have shown that there is a q(x)\in I for every g(x)\in I such that g(x)=f(x)q(x), thus, I=(f(x)) and so \mathbb{Q}[x] is a Principal Ideal Domain.

\Box

Reflection: They key in this proof is that we chose our f to have the least possible degree. This makes the polynomial act kind of like a gcd does for a list of numbers.

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