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Field: Algebra

## The fundamental theorem of additive polynomials

Let $P(x)$ be a polynomial with coefficients in k, and $\{w_1,...,w_m\}\subset k$ be the set of its roots. Assuming that $P(x)$ is separable, then P(x) is additive if and only if $\{w_1,...,w_m\}$ form a group.

They are a subgroup in respect to addition or multiplication?

Should it be

$\{w_1,...,w_m\}\subset \bar{k}$

$\{w_1,...,w_m\}\subset k$?

## Mathworld

Hmm, the structure of this article shows remarkable similarity to the Mathworld article. linas 05:03, 10 Jun 2005 (UTC)

I noticed that a while ago too.
By the way, could you check my edits for correctness? Back then both of us were green and we fought like hell. :) Oleg Alexandrov 05:14, 10 Jun 2005 (UTC)

Is the example under additive versus absolutely additive correct? That is, I think $x^q-x$ is absolutely additive. $q$ is the order of the field, and if finite it must then be $q=p^n$ for some $n\in\mathbb{N}$. But $p$ is the characteristic, so that $x^q-x=x^\left(p^n\right)-x$ is a linear combination of $\tau_p^n=x^\left(p^n\right)$ and $\tau^0=x$ and thus absolutely additive. GromXXVII (talk) 12:31, 5 May 2008 (UTC)