## 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.