Ideals in the Set of Polynomials over a Field
 Table of Contents

# Ideals in the Set of Polynomials over a Field

Recall that if $(R, +, \cdot)$ is a ring then $I \subseteq R$ is an ideal of $R$ if:

• (1) $(I, +)$ is a subgroup of $(R, +)$ (That is, $I$ is closed under $+$ and contains the additive identity $0$).
• (2) For all $x \in I$ and for all $r \in R$ we have that $xr, rx \in I$.

We now prove a remarkable result. Given any field $(F, +, \cdot)$, any ideal $I$ of $F$ can be generated by a single element. In other words, if $I$ is an ideal of $F$ then there exists a polynomial $d(x) \in I$ with $d(x) \neq 0$ of minimal degree such that:

(1)
\begin{align} \quad I = \{ q(x)d(x) : q \in F[x] \} \end{align}

Any ideal that can be generated by a single element is called a principal ideal. Thus every ideal if $F[x]$ is a principal ideal.

 Theorem 1: Let $(F, +, \cdot)$ be a field and let $I \subseteq F[x]$ satisfy the following properties: 1) There exists an polynomial $f \in I$ such that $f(x) \neq 0$. 2) If $f, g \in I$ then $(f + g) \in I$. 3) If $f \in I$ and $g \in F[x]$ then $(f \cdot g) \in I$. Then there exists a polynomial $d \in I$ with $d(x) \neq 0$ and $d$ being of minimal degree in $I$ such that $I = \{ q(x)d(x) : q\in F[x] \}$.

The element $d \in I$ of minimal degree stated above exists because the set of degrees of the polynomials in $I$ is nonempty by (1) and is a collection of natural numbers. By the well-ordering principle, any subset of the natural numbers has a least element. In other words, there will exist such a polynomial $d$ of minimal degree in $I$.

• Proof: Let $I \subseteq F[x]$ that satisfies (1)-(3) and let $d \in I$ with $d(x) \neq 0$ be of minimal degree in $I$.
• Consider the set $\{ q(x)d(x) : q \in F[x] \}$ and let $f \in \{ q(x)d(x) : q \in F[x] \}$. Then for some $q \in F[x]$ we have that:
(2)
\begin{align} \quad f(x) = q(x)d(x) \end{align}
• Since $d \in I$ and $q \in F[x]$ by (3) this implies that $f \in I$. Therefore:
(3)
\begin{align} \quad \{ q(x)d(x) : q \in F[x] \} \subseteq I \quad (*) \end{align}
• Now let $f \in I$. Since $d(x) \neq 0$, by the division algorithm there exists unique polynomials $q, r \in F[x]$ such that:
(4)
\begin{align} \quad f(x) = q(x)d(x) + r(x) \end{align}
• Where either $r = 0$ or $\deg (r) < \deg (d)$.
• Now we write the equality above as:
(5)
\begin{align} \quad r(x) = f(x) - q(x)d(x) \end{align}
• Since $q \in F[x]$ and $d \in I$ we have by (3) that $qd \in I$. And since $f \in I$ and $qd \in I$ by (2) we have that $f - qd \in I$. So $r \in I$. Since $d$ has minimal degree in $I$ we cannot have $\deg(r) < \deg(d)$ unless $r(x) = 0$. So $f(x) = q(x)d(x)$ which shows that $f \in \{ q(x)d(x) : q \in F[x] \}$. Thus:
(6)
\begin{align} \quad \{ q(x)d(x) : q \in F[x] \} \supseteq I \quad (**) \end{align}
• From $(*)$ and $(**)$ we obtain:
(7)
\begin{align} \quad I = \{ q(x)d(x) : q \in F[x] \} \quad \blacksquare \end{align}
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