Factorization on polynomial rings

Overview

• Also important facts when to do with Prime and maximal ideals, leads to Extension fields

Relevant theorems:

• (Theorem) Chinese remainder
• (Theorem) Long division with remainder
• (Theorem) Unique factorization in polynomial rings

Related notes:

• Polynomial roots
• Irreducible polynomials
• Factorization on integral domains
• Unique factorization domains

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General definitions

The following definitions and proofs build up to a proof of (Theorem) Unique factorization in polynomial rings, a.k.a. the fundamental theorem of arithmetic.

Greatest common divisor of two polynomials

Greatest common divisor

If $f,g\in F[x]$ are polynomials with coefficients in a field $F$ where $f,g$ are not both $0$, then a greatest common divisor of $f,g$ is a polynomial $h\in F[x]$ such that:

• (i) The polynomial $h$ divides both $f,g$, i.e., $h|f$ and $h|g$;
• If $k\in F[x]$ is any other polynomial that satisfies the same, i.e., $k|f$ and $k|g$, we have $k|h$ as well.

Modern Algebra II 2.4

Let $f,g\in F[x]$, not both $0$.

• (i) If $p$ is a greatest common divisor of $f,g$, then so is $cp$ for every $c\in F^{\ast }$, the group of nonzero constant polynomials.
• (ii) If $p,p’$ are two greatest common divisors of $f,g$, then there exists a $c\in F^{\ast }$ such that $p=cp’$.
• (iii) A greatest common divisor $p$ of $f,g$ exists and is of the form $p=rf+sg$ for some $r,s\in F[x]$.

Proof from Modern Algebra II.

• (i) Clear from the definition.
• (ii) If $p,p^{\prime }$ are two greatest common divisors of $f,g$, then by definition we have $p|p’$ and $p’|p$. Since $F[x]$ is an integral domain, the elements are associates, meaning there exists some unit $c\in F^{\ast }$ such that $p=cp’$
• (iii) Consider $(f,g)=(f)+(g)=\{rf+sg:r,s\in F[x]\}.$ This is the ideal sum of $(f),(g)$, hence an ideal in its own right, and certainly $f,g\in (f,g)$. Then since every ideal in $F[x]$ is principal, there exists $p\in F[x]$ such that $(f,g)=(p)$. In particular $p=rf+sg$ for some $r,s\in F[x]$ and $p|f$ and $p|g$. Finally if $q|f$ and $q|g$, then check that $q$ divides every expression of the form $rf+sg$ to conclude that $q|p$, and hence $p$ satisfies the definition of a greatest common divisor. $\square$

Relatively prime polynomials

Relatively prime

Two nonzero polynomials $f,g\in F[x]$ are relatively prime if $1$ is a greatest common divisor of $f,g$. Equivalently, there exist $r,s\in F[x]$ such that $1=rf+sg$.

Modern Algebra II 2.7: Relatively prime elements do not divide each other

If $f,g\in F[x]$ are relatively prime and $f|gh$ for some $h\in F[x]$, then $f|h$.

Proof from Modern Algebra II.    Suppose $r,s\in F[x]$ such that $1=rf+sg$, so $h=(rf+sg)h=rfh+sgh.$ Clearly $f|rfh$ for the left term in the sum. For the right term, we have by hypothesis $f|gh$, hence $f|sgh$. It follows that $f|rfh+sgh=h$. $\square$