📂Abstract Algebra

Principal Ideal

Definition¹

For an element $a$ of a commutative ring $R$ with unity, the $\left< a \right>$ generated by $a$ is called the principal ideal generated by $a$.

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• The identity element $1$ for multiplication is called the unity.

Explanation

The notation $\left< a \right> := \left\{ r a \mid r\ \in R \right\}$ is the same as that of a cyclic group, but it actually forms a slightly larger structure than a cyclic group.

For example, every ideal $n \mathbb{Z} = \left< n \right> = \left\{ \cdots , -2n , -n , 0 , n , 2n , \cdots \right\}$ of $\mathbb{Z}$ is a principal ideal.

When encountering principal ideals for the first time, one cannot help but feel like grasping at clouds. They are of no immediate use, but they become valuable later when discussing integral domains with various good properties. Among the following theorems, [2] and [3] in particular serve as stepping stones toward PIDs and UFDs respectively, so it is recommended to prove them by hand at least once.

Theorem

For a field $F$, let $p(x), r(x), s(x) \in F [ x ]$.

1. Every ideal of $F [ x ]$ is a principal ideal.
2. $\left< p(x) \right> \ne \left\{ 0 \right\}$ is a maximal ideal $\iff$ $p(x)$ is an irreducible element over $F$
3. If an irreducible element $p(x)$ over $F$ divides $r(x) s(x)$, then $p(x)$ divides $r(x)$ or $s(x)$.

Proof

[1]

Consider the polynomial $g(x)$ of lowest degree in an ideal $N \ne \left\{ 0 \right\}$ of $F [ x ]$.

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Case 1. $\deg g = 0$

Since $g(x)$ is a constant function, $g(x) \in F$, and since $F$ was assumed to be a field, $g(x)$ is a unit of $F$ and at the same time a unit of $F [ x ]$. Since $g(x)$ is a unit of $F [ x ]$, $N = F [ x ] = \left< 1 \right>$, and therefore $N$ is a principal ideal.

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Case 2. $\deg g \ge 1$

Any $f(x) \in N$ can be expressed as $f(x) = g(x) q(x) + r(x)$ by the division algorithm. Since $N$ is an ideal, $$ f(x) - g(x) q(x) = r(x) \in N $$ but since the polynomial of lowest degree is $g(x)$, it must be that $r(x)=0$.

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In other words, any $f(x) \in N$ can always be expressed as $f(x) = g(x) q(x)$, so $N = \left< g(x) \right>$, and therefore $N$ is a principal ideal.

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[2]

$( \implies )$

Assume that $p(x)$ is not an irreducible element, so that it factors as $p(x) = f(x) g(x)$.

Since $\left< p(x) \right>$ is a maximal ideal of $F [ x ]$, we have $\left< p(x) \right> \ne F [ x ]$ and $p(x) \notin F$. Since a maximal ideal is a prime ideal, if $\left( f(x) g(x) \right) \in \left< p(x) \right>$, then $f(x) \in \left< p(x) \right>$ or $g(x) \in \left< p(x) \right>$ must hold. However, the degrees of $f(x)$ and $g(x)$ cannot be smaller than the degree of $p(x)$, which contradicts the assumption, so $p(x)$ is an irreducible element over $F$.

$( \impliedby )$

Assume that $\left< p(x) \right>$ is not a maximal ideal, so that there exists an ideal $N$ satisfying $\left< p(x) \right> \subsetneq N \subsetneq F [ x ]$.

By theorem [1], $N$ is a principal ideal of $F [ x ]$, so we can set $N := \left< g(x) \right>$ for some $g(x) \in F [ x ]$. Since $\left< p(x) \right> \subset N$ by assumption, for some $q(x) \in F [ x ]$ it can be expressed as $$ p(x) = g(x) q(x) $$ However, since $p(x)$ is an irreducible element over $F$, one of $g(x)$ or $q(x)$ must be a constant.

• If $g(x)$ is a constant, then $g(x)$ is a unit of $F [ x ]$, so $$ N = F [ x ] $$
• If $q(x)$ is a constant, then for some $c \in F [ x ]$ we have $\displaystyle g(x) = {{1} \over {c}} p(x)$, so $$ N = \left< g(x) \right> = \left< p(x) \right> $$

Whether $g(x)$ is a constant or $q(x)$ is a constant, it contradicts the assumption, so $\left< p(x) \right>$ is a maximal ideal of $F [ x ]$.

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[3]

If $p(x)$ divides $r(x) s(x)$, then $r(x) s(x) \in \left< p(x) \right>$. However, since $p(x)$ is an irreducible element over $F$, by theorem [2], $\left< p(x) \right>$ is a maximal ideal and therefore a prime ideal. That is, if $r(x) s(x) \in \left< p(x) \right>$, then $r(x) \in \left< p(x) \right>$ or $s(x) \in \left< p(x) \right>$, which means that $p(x)$ divides $r(x)$ or $s(x)$.

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