The Principal Ideal Domain of Polynomials over a Field

# The Principal Ideal Domain of Polynomials over a Field

Recall from the Principal Ideals and Principal Ideal Domains (PIDs) page that if $(R, +, *)$ is a ring then an ideal of the form $I = aR = \{ ar : r \in R \}$ is called a principal ideal, and if every ideal in $R$ is a principal ideal, then $R$ is said to be a principal ideal domain.

We will now look at an important example of a principal ideal domain.

Theorem 1: Let $(F, +, *)$ be a field. Then the ring of polynomials $F[x]$ is a principal ideal domain. |

**Proof:**Let $I$ be an ideal of $F[x]$. There are two cases to consider.

**Case 1:**If $I = \{ 0 \}$ then trivially, $I$ is a principal ideal since $I = 0F[x]$.

**Case 2:**If $I \neq \{ 0 \}$ then there exists a nonzero polynomial $f \in I$ with minimal degree. Let $g \in I$. By the division algorithm there exists polynomials $q, r \in R[x]$ such that:

\begin{align} \quad g(x) = f(x)q(x) + r(x) \end{align}

- where $r(x) = 0$ or $0 \leq \deg r \leq \deg f$. We rewrite the equation above as:

\begin{align} \quad r(x) = g(x) - f(x)q(x) \end{align}

- Since $f \in I$ and $q \in R[x]$ we have by definition of $I$ being an ideal that $fq \in I$. Since $g \in I$ we have that $(g - fq) \in I$. So $r \in I$. But by minimality of the degree of $f$ in $I$ we must have that $r(x) = 0$. Therefore:

\begin{align} \quad g(x) = f(x)q(x) \end{align}

- Since $g \in I$ was arbitrary we have that $I = f(x)F[x]$. So $I$ is a principal ideal.

- Hence every ideal in $F[x]$ is a principal ideal. So $F[x]$ is a principal ideal domain. $\blacksquare$