📂Abstract Algebra

Irreducible Elements of Polynomial Functions

Definition¹

When a non-constant $f(x) \in F [ x ]$ cannot be expressed as a product $f(x) = g(x) h(x)$ of some $g(x) , h(x) \in F [ x ]$ of lower degree than $f(x)$, $f(x)$ is called an irreducible element over $F$.

Explanation

For example, consider $\mathbb{Q} [x ]$. Here $x^2 - 2$ is an irreducible element over $\mathbb{Q}$, but $x^2 - 2$ in $\mathbb{R} [ x ]$ can be factored over $\mathbb{R}$ as $$ (x + \sqrt{2} ) ( x - \sqrt{2} ) $$ Also, $x^2 + 1$ is an irreducible element over $\mathbb{R}$, but in $\mathbb{C}$ it can be factored as $$ (x + i ) ( x - i ) $$

Just as in topology, where what one takes as the whole set matters when speaking of openness and closedness, one must pay attention to where something is irreducible. Of course, by the definition, a polynomial function of $F [ x ]$ only needs to be considered over $F$, but for an extension $E$ of $F$ we would have $f(x) \in E[x]$.

By considering this irreducibility, we can think of the true completion of the factor theorem.

Theorem

• [1]: If an irreducible element $p(x) \in F [ x ]$ divides $r_{1} (x) \cdots r_{n} (x) \in F [ x ]$, then $p(x)$ must divide one of $r_{1} (x) , \cdots , r_{n} (x) $.
• [2]: If $F$ is a field, every non-constant $f(x) \in F [ x ]$ can be factored into a product of irreducible elements, and this factorization is unique.

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• Uniqueness here does not take into account the order or multiplication by units. For instance, differences such as $$ x^2-1 = (x+1)(x-1) = [-(x+1)][-(x-1)] = (x-1)(x+1) $$ are simply not worried about.

Significance

This theorem can be seen as an extension of the fundamental theorem of arithmetic from number theory to abstract algebra. If the factor theorem proved the existence of factorizations of polynomial functions, the fact above guarantees their uniqueness.