The ring $F[x]$

In this section, we shall ask and answer the polynomial analog of the question mentioned in the previous section. We state a prove a polynomial analog of Lemma 1.4.2.

Proposition 2.3.3   Let $F$ be a field. Let $I\subset F[x]$ be a closed under addition and subtraction, and under multiplication by any polynomial having integer coefficients ^2.8. Then either

• $I$ is empty,
• $I=\{0\}$.
• there is a polynomial $a(x)$ with coefficients in $F$ such that $I=g(x)F[x]$, i.e., $I$ is the set of all multiples of some polynomial $g(x)$ (which is constructed in the proof below).

The proof below is worthwhile trying to understand well since it contains a basic idea which occurs in other parts of mathematics.

proof: If $I$ is not empty then it must contain 0. If it contains some non-zero element then it must contain a polynomial of degree $>0$. By the well-ordering principle, $I$ contains an element of least degree $g(x)\not= 0$.

Claim: $I=a(x)F[x]$.

If $b\in I$ then the division algorithm says that there is a polynomial $r(x)$ $0\leq {\rm deg}(r(x))<{\rm deg}(g(x))$ such that $b(x)=q(x)g(x)+r(x)$. Since $I$ is closed under addition and subtraction, $r(x)=b(x)-q(x)g(x)$ belongs to $I$. But $g$ is a non-zero element of $I$ of lowest degree, so $r(x)$ must be zero. This implies $b(x)\in g(x)F[x]$. Since $b(x)$ was choosen arbitrarily, this imples $I\subset g(x)F[x]$. The reverse inclusion $g(x)F[x]\subset I$ follows from the assumption that $I$ is closed under addition and subtraction. This proves the claim and the lemma. $\Box$

In other words, any non-empty subset $I$ of polynomials having coefficients in a field $F$ which is closed under addition and polynomial multiplication must be the set of multiples of a fixed polynomial $g(x)$. Using mathematical induction, it is possible to show that an ideal of $F[x]$ is principal. This implies $F[x]$ is a principal ideal domain.

We will study later the ``ring of polynomials modulo $I$'',

$$F[x]/I=\{\overline{f(x)}\ \vert\ f(x)\in F[x]\},$$

where

$$\overline{f(x)}=f(x)+I= \{f(x)+g(x)\ \vert\ g(x)\in I\} =\{f(x)+g(x)p(x)\ \vert\ p(x)\in F[x]\}.$$

This may at first look rather mysterious but it is the polynomial analog of the ring of integers modulo $m$, $\mathbb{Z}/m\mathbb{Z}$. At this point, to understand $F[x]/I$ we need to understand how to factor polynomials better, which we turn to next.

Exercise 2.3.4   Let $I$ be the ideal of $\mathbb{Z}[x]$ generated by $2$ and $x$. Show $I$ is not a principal ideal.

Exercise 2.3.5   Let $I$ be the ideal of $\mathbb{Q}[x]$ generated by $x^2-1$ and $x^3-1$. Show $I$ is a principal ideal and find its generator.