annihilating polynomial

**demode** · June 22nd 2011, 08:30 PM

I'm trying to understand the following worked problem:

So, $\alpha^4 = \alpha^2 . \alpha^2 = (1+x^2)(1+x^2)=1+2x^2+x^4$ . And did they subtract $m(x)=x^4+x^3+x^2+x+1$ from $\alpha^4$ to get $x+x^2+x^3$ because the result had degree 4? Does this mean every time the degree gets larger than or equal to 4, we must subtract m(x) from it (to have everything in mod m(x))?

**chisigma** · June 22nd 2011, 09:07 PM

More exactly is...

$\alpha^{4} = \alpha^{2}\ \alpha^{2}= (1+x^{2})\ (1+x^{2})= 1 + x^{4}= x + x^{2} + x^{3}\ \text{mod}\ m(x)$

Kind regards

$\chi$ $\sigma$

**Deveno** · June 22nd 2011, 11:12 PM

it IS kind of confusing because when the author says " $\alpha = x+1$ " what he really means is:

$\alpha$ is the coset of x+1 in K. and in K, multiples of m(x) are set equal to 0.

so in K, (remembering "x" in K is really x + m(x)), $x^4 = x^3 + x^2 + x + 1$

(more precisely, the coset $x^4 + (m(x))$ is the same coset as $x^3 + x^2 + x + 1$ , since

$x^4 - (x^3 + x^2 + x + 1) = x^4 + (-1)(x^3 + x^2 + x + 1) = x^4 + x^3 + x^2 + x + 1$

$= m(x) \in (m(x)) = 0 + (m(x))$ , the identity of $\mathbb{Z}_2[x]/(m(x))$

with this (unfortunately standard) abuse of notation:

$(\alpha)^4 = (x+1)^4 = x^4 + 1$ (since 4 is a power of 2, the $x^3, x^2$ and $x$ terms are all 0)

$= x^3 + x^2 + x + 1 + 1 = x^3 + x^2 + x$ .

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