# polynomial ring over a field

The polynomial ring over a field is a Euclidean domain.

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 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 $d^{\prime}$ is an arbitrary gcd of them, then

 $d^{\prime}\mid d\quad\mbox{and}\quad d\mid d^{\prime},$

i.e. $d$ and $d^{\prime}$ are associates in the ring $K[X]$ and thus $d^{\prime}$ is gotten from $d$ by multiplication by an element of the field $K$.  So we can write the

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 monic (http://planetmath.org/Monic2) gcd of polynomials is unique.

If the monic gcd of two polynomials is 1, they may be called .

Using the Euclid’s algorithm as in $\mathbb{Z}$, one can prove the

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  $uf+vg=1$.

Corollary 3.  If a product of polynomials in $K[X]$ is divisible by an irreducible polynomial of $K[X]$, then at least one factor (http://planetmath.org/Product) of the product is divisible by the irreducible polynomial.

Corollary 4.  A polynomial ring over a field is always a principal ideal domain.

Corollary 5.  The factorisation of a non-zero polynomial, i.e. the 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.

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  $\mathbb{Q}(\sqrt{2},\,i)[X]$.

Title polynomial ring over a field PolynomialRingOverAField 2013-03-22 17:42:55 2013-03-22 17:42:55 pahio (2872) pahio (2872) 14 pahio (2872) Theorem msc 13F07 FieldAdjunction PolynomialRingOverIntegralDomain PolynomialRingWhichIsPID coprime