# irreducible polynomial

Let  $f(x)=a_{0}\!+\!a_{1}x\!+\cdots+\!a_{n}x^{n}$  be a polynomial    with complex coefficients $a_{\nu}$ and with the degree (http://planetmath.org/Polynomial)  $n>0$.  If $f(x)$ can not be written as product of two polynomials with positive degrees and with coefficients in the field  $\mathbb{Q}(a_{0},\,a_{1},\,\ldots,\,a_{n})$,  then the polynomial $f(x)$ is said to be .  Otherwise, $f(x)$ is reducible.

Examples.  All linear polynomials are .  The polynomials $x^{2}\!-\!3$, $x^{2}\!+\!1$ and $x^{2}\!-\!i$ are (although they split in linear factors in the fields $\mathbb{Q}(\sqrt{3})$, $\mathbb{Q}(i)$ and $\mathbb{Q}(\frac{1\!+\!i}{\sqrt{2}})$, respectively).  The polynomials $x^{4}\!+\!4$ and $x^{6}\!+\!1$ are not .

The above definition of polynomial is special case of the more general setting where $f(x)$ is a non-constant polynomial in the polynomial ring $K[x]$ of a field $K$; if $f(x)$ is not expressible as product of two polynomials with positive degrees in the ring $K[x]$, then $f(x)$ is (in $K[x]$).

Example.  If $K$ is the Galois field with two elements (0 and 1), then the trinomial $x^{2}\!+\!x\!+\!1$ of $K[x]$ is (because an equation  $x^{2}\!+\!x\!+\!1=(x\!+\!a)(x\!+\!b)$  would imply the two conflicting conditions  $a\!+\!b=1$  and  $ab=1$).

 Title irreducible polynomial Canonical name IrreduciblePolynomial Date of creation 2013-03-22 14:24:22 Last modified on 2013-03-22 14:24:22 Owner pahio (2872) Last modified by pahio (2872) Numerical id 18 Author pahio (2872) Entry type Definition Classification msc 12D10 Synonym prime polynomial Synonym indivisible polynomial Related topic EisensteinCriterion Related topic Irreducible Related topic Monic2 Defines irreducible polynomial Defines reducible