|
|
|
Revision difference : irreducible polynomial |
| Version 14 |
Version 13 |
|
Let\, $f(x) = a_0\!+\!a_1x\!+\cdots+\!a_nx^n$\, be a polynomial with complex coefficients $a_{\nu}$ and with the \PMlinkname{degree}{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 \PMlinkescapetext{{\em irreducible}}.\, Otherwise, $f(x)$ is {\em reducible}.
|
Let\, $f(x) = a_0\!+\!a_1x\!+\cdots+\!a_nx^n$\, be a polynomial with complex coefficients $a_{\nu}$ and with the \PMlinkname{degree}{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 \PMlinkescapetext{{\em irreducible}}.}.
|
|
|
| \textbf{Examples.}\, All linear polynomials are \PMlinkescapetext{irreducible}.\, The polynomials $x^2\!-\!3$, $x^2\!+\!1$ and $x^2\!-\!i$ are \PMlinkescapetext{irreducible} (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 \PMlinkescapetext{irreducible}. |
\textbf{Examples.}\, All linear polynomials are \PMlinkescapetext{irreducible}.\, The polynomials $x^2\!-\!3$, $x^2\!+\!1$ and $x^2\!-\!i$ are \PMlinkescapetext{irreducible} (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 \PMlinkescapetext{irreducible}. |
|
|
| The above definition of \PMlinkescapetext{irreducible} 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 \PMlinkescapetext{{\em irreducible}} (in $K[x]$). |
The above definition of \PMlinkescapetext{irreducible} 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 \PMlinkescapetext{{\em irreducible}} (in $K[x]$). |
|
|
| \textbf{Example.}\, If $K$ is the Galois field with two elements (0 and 1), then the trinomial $x^2\!+\!x\!+\!1$ of $K[x]$ is \PMlinkescapetext{irreducible} (because an equation\, $x^2\!+\!x\!+\!1 = (x\!+\!a)(x\!+\!b)$\, would imply the two conflicting conditions\, $a\!+\!b = 1$\, and\, $ab = 1$). |
\textbf{Example.}\, If $K$ is the Galois field with two elements (0 and 1), then the trinomial $x^2\!+\!x\!+\!1$ of $K[x]$ is \PMlinkescapetext{irreducible} (because an equation\, $x^2\!+\!x\!+\!1 = (x\!+\!a)(x\!+\!b)$\, would imply the two conflicting conditions\, $a\!+\!b = 1$\, and\, $ab = 1$). |
|
|
|
|
