# polynomial ring over a field

Theorem. 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 $$ 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

$${d}^{\prime}\mid d\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}d\mid {d}^{\prime},$$ |

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 coprime^{}.

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

$$\mathrm{gcd}(f,g)=uf+vg$$ |

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 |
---|---|

Canonical name | PolynomialRingOverAField |

Date of creation | 2013-03-22 17:42:55 |

Last modified on | 2013-03-22 17:42:55 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 14 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 13F07 |

Related topic | FieldAdjunction |

Related topic | PolynomialRingOverIntegralDomain |

Related topic | PolynomialRingWhichIsPID |

Defines | coprime |