📂Abstract Algebra

Main Ideals

Definition ¹

A principal ideal is generated by an element $a$ with unity in a commutative ring $R$.

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

Description

Although the notation $\left< a \right> := \left\{ r a \mid r\ \in R \right\}$ is similar to that of a cyclic group, it actually forms a slightly larger structure.

For example, all ideals $n \mathbb{Z} = \left< n \right> = \left\{ \cdots , -2n , -n , 0 , n , 2n , \cdots \right\}$ of $\mathbb{Z}$ are principal ideals.

At first encounter, principal ideals might feel abstract and seemingly useless, but they become valuable when discussing integral domains with good properties. Among the following theorems, especially [2] and [3] act as bridges to PID and UFD, respectively, so it’s recommended to prove them by hand at least once.

Theorems

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

1. All ideals of $F [ x ]$ are principal ideals.
2. If $\left< p(x) \right> \ne \left\{ 0 \right\}$ is a maximal ideal $\iff$ $p(x)$ then $p(x)$ 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 the 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, it equals $g(x) \in F$, and assuming $F$ to be a field, $g(x)$ is a unit in both $F$ and $F [ x ]$. As $g(x)$ is a unit of $F [ x ]$, $N$ is a principal ideal.

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Case 2. $\deg g \ge 1$

Any $f(x) \in N$ can be expressed as $f(x) = g(x) q(x) + r(x)$ according to the division algorithm. Since $N$ is an ideal, $$f(x) - g(x) q(x) = r(x) \in N$$ and the lowest degree polynomial is $g(x)$, then $r(x)=0$ must be true.

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Hence, any $f(x) \in N$ can always be expressed as $f(x) = g(x) q(x)$, making $N = \left< g(x) \right>$ true, and thus $N$ is a principal ideal.

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[2]

$( \implies )$

Let’s assume $p(x)$ is not an irreducible element and is factorized as $p(x) = f(x) g(x)$.

Since $\left< p(x) \right>$ is a maximal ideal of $F [ x ]$, then $\left< p(x) \right> \ne F [ x ]$ and $p(x) \notin F$. A maximal ideal is a prime ideal, so if $\left( f(x) g(x) \right) \in \left< p(x) \right>$, then either $f(x) \in \left< p(x) \right>$ or $g(x) \in \left< p(x) \right>$ must be true. However, the degrees of $f(x)$ and $g(x)$ cannot be less than that of $p(x)$, contradicting our assumption, so $p(x)$ is an irreducible element over $F$.

$( \impliedby )$

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

According to theorem [1], $N$ is a principal ideal of $F [ x ]$, thus for some $g(x) \in F [ x ]$, we can set $N := \left< g(x) \right>$. Since $\left< p(x) \right> \subset N$ from our assumption, for some $q(x) \in F [ x ]$ we have $$p(x) = g(x) q(x)$$ But since $p(x)$ is an irreducible element over $F$, either $g(x)$ or $q(x)$ must be a constant.

• If $g(x)$ is a constant, then $g(x)$ is a unit of $F [ x ]$, hence $N = F [ x ]$.
• If $q(x)$ is a constant, 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 $g(x)$ or $q(x)$ is constant, it contradicts our assumption, thus $\left< p(x) \right>$ is a principal ideal of $F [ x ]$.

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[3]

If $p(x)$ divides $r(x) s(x)$, then $r(x) s(x) \in \left< p(x) \right>$. But as $p(x)$ is an irreducible element over $F$, by theorem [2], $\left< p(x) \right>$ is a maximal ideal and thus a prime ideal.

Thus if $r(x) s(x) \in \left< p(x) \right>$, then either $r(x) \in \left< p(x) \right>$ or $s(x) \in \left< p(x) \right>$, meaning $p(x)$ divides $r(x)$ or $s(x)$.

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