Talk:Additive polynomial

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 Field: Algebra

The fundamental theorem of additive polynomials[edit]

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}

instead of

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


Mathworld[edit]

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)

Non-absolutely additive example dispute[edit]

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)