polynomial ring over a field PolynomialRingOverFieldIsEuclideanDomain 2007-12-28 14:34:28 2008-03-06 14:31:21 Theorem Euclidean domain coprime unique factorization % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \theoremstyle{definition} \newtheorem*{thmplain}{Theorem} \textbf{Theorem.}\, The polynomial ring over a field is a Euclidean domain.\\ {\em Proof.}\, Let $K[X]$ be the polynomial ring over a field $K$ in the indeterminate $X$.\, Since $K$ is an integral domain and any polynomial ring over integral domain is an integral domain, the ring $K[X]$ is an integral domain. The degree $\nu(f)$, defined for every $f$ in $K[X]$ except the zero polynomial, satisfies the requirements of a Euclidean valuation in $K[X]$.\, In fact, the degrees of polynomials are non-negative integers.\, If $f$ and $g$ belong to $K[X]$ and the latter of them is not the zero polynomial, then, as is well known, the long division\, $f/g$\, gives two unique polynomials $q$ and $r$ in $K[X]$ such that $$f = qg+r,$$ where\, $\nu(r) < \nu(g)$\, or\, $r$ is the zero polynomial.\, The second property usually required for the Euclidean valuation, is justified by $$\nu(fg) = \nu(f)+\nu(g) \geqq \nu(f).$$ The theorem implies, similarly as in the ring $\mathbb{Z}$ of the integers, that one can perform in $K[X]$ a Euclid's algorithm which yields a greatest common divisor of two polynomials.\, Performing several \PMlinkescapetext{consecutive} Euclid's algorithms one obtains a gcd of many polynomials; such a gcd is always in the same polynomial ring $K[X]$.\\ Let $d$ be a greatest common divisor of certain polynomials.\, Then apparently also $kd$, where $k$ is any non-zero element of $K$, is a gcd of the same polynomials.\, They do not have other gcd's than $kd$, for if $d'$ is an arbitrary gcd of them, then $$d' \mid d \quad \mbox{and} \quad d \mid d',$$ i.e. $d$ and $d'$ are associates in the ring $K[X]$ and thus $d'$ is gotten from $d$ by multiplication by an element of the field $K$.\, So we can write the \textbf{Corollary 1.}\, The greatest common divisor of polynomials in the ring $K[X]$ is unique up to multiplication by a non-zero element of the field $K$. The \PMlinkname{monic}{Monic2} gcd of polynomials is unique.\\ If the monic gcd of two polynomials is 1, they may be called {\em coprime}.\\ Using the Euclid's algorithm as in $\mathbb{Z}$, one can prove the \textbf{Corollary 2.}\, If $f$ and $g$ are two non-zero polynomials in $K[X]$, this ring contains such polynomials $u$ and $v$ that $$\gcd(f,\,g) = uf+vg$$ and especially, if $f$ and $g$ are coprime, then $u$ and $v$ may be chosen such that\, $uf+vg = 1$.\\ \textbf{Corollary 3.}\, If a product of polynomials in $K[X]$ is divisible by an irreducible polynomial of $K[X]$, then at least one \PMlinkname{factor}{Product} of the product is divisible by the irreducible polynomial.\\ \textbf{Corollary 4.}\, A polynomial ring over a field is always a principal ideal domain.\\ \textbf{Corollary 5.}\, The factorisation of a non-zero polynomial, i.e. the \PMlinkescapetext{presentation} of the polynomial as product of irreducible polynomials, is unique up to constant factors in each polynomial ring $K[X]$ over a field $K$ containing the polynomial.\, Especially, $K[X]$ is a UFD.\\ \textbf{Example.}\, The factorisations of the trinomial \,$X^4-X^2-2$\, into monic irreducible prime factors are\\ $(X^2-2)(X^2+1)$\; in\; $\mathbb{Q}[X]$,\\ $(X^2-2)(X+i)(X-i)$\; in\; $\mathbb{Q}(i)[X]$,\\ $(X+\sqrt{2})(X-\sqrt{2})(X^2+1)$\; in\; $\mathbb{Q}(\sqrt{2})[X]$,\\ $(X+\sqrt{2})(X-\sqrt{2})(X+i)(X-i)$\; in\; $\mathbb{Q}(\sqrt{2},\,i)[X]$.