Talk:Additive polynomial

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

$\{w_{1},...,w_{m}\}\subset {\bar {k}}$

instead of

$\{w_{1},...,w_{m}\}\subset k$?

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)

Non-absolutely additive example dispute

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)

I agree with you. However, I also want to dispute the clarity of the definition. The article first says that an additive polynomial is such that $P(a+b)=P(a)+P(b)$ as polynomials in $a$ and $b$. What does the underlined part mean? Isn't that just evaluating $P(x)$ at both values? Then, the article goes on to say that this (which looks like the merely additive version) is equivalent to assume that this equality holds for all a and b in some infinite field containing k, such as its algebraic closure (which looks like the absolutely additive version). Wisapi (talk) 14:45, 6 January 2016 (UTC)

Yes, $P(a+b)=P(a)+P(b)$ is evaluating $P(x)$ at both values. The clarification in the underlined part has to do with the difference between functions and polynomials. In finite characteristic you can have equality of functions without equality of polynomials - this makes the equals sign "=" ambiguous as to whether you mean as functions or polynomials. For instance, consider the field $Z_{2}=\{0,1\}$ and the functions $f(x)=x^{2}+x,g(x)=0$. As functions over $Z_{2}$, $f=g$ because $f(x)=g(x)$ for all $x\in Z_{2}$. However, as polynomials $f\neq g$ Jbeyerl (talk) 18:15, 6 January 2016 (UTC)