Additive polynomial

In mathematics, the additive polynomials are an important topic in classical algebraic number theory.

Let ${\displaystyle k}$ be a field of prime characteristic ${\displaystyle k}$. A polynomial ${\displaystyle P(x)}$ with coefficients in ${\displaystyle k}$ is called an additive polynomial, or a Frobenius polynomial, if

$${\displaystyle P(a+b)=P(a)+P(b)}$$

as polynomials in ${\displaystyle a}$ and ${\displaystyle b}$. It is equivalent to assume that this equality holds for all ${\displaystyle a}$ and ${\displaystyle b}$ in some infinite field containing ${\displaystyle k}$, such as its algebraic closure.

Occasionally absolutely additive is used for the condition above, and additive is used for the weaker condition that ${\displaystyle P(a+b)=P(a)+P(b)}$ for all ${\displaystyle a}$ and ${\displaystyle b}$ in the field.¹ For infinite fields the conditions are equivalent,² but for finite fields they are not, and the weaker condition is the "wrong" as it does not behave well. For example, over a field of order ${\displaystyle q}$ any multiple ${\displaystyle P}$ of ${\displaystyle x^{q}-x}$ will satisfy ${\displaystyle P(a+b)=P(a)+P(b)}$ for all ${\displaystyle a}$ and ${\displaystyle b}$ in the field, but will usually not be (absolutely) additive.

The polynomial ${\displaystyle x^{p}}$ is additive.¹ Indeed, for any ${\displaystyle a}$ and ${\displaystyle b}$ in the algebraic closure of ${\displaystyle k}$ one has by the binomial theorem

$${\displaystyle (a+b)^{p}=\sum _{n=0}^{p}{p \choose n}a^{n}b^{p-n}.}$$

Since ${\displaystyle p}$ is prime, for all ${\displaystyle n=1,\dots ,p-1}$ the binomial coefficient ${\displaystyle {\tbinom {p}{n}}}$ is divisible by ${\displaystyle p}$, which implies that

$${\displaystyle (a+b)^{p}\equiv a^{p}+b^{p}\mod p}$$

as polynomials in ${\displaystyle a}$ and ${\displaystyle b}$.¹

Similarly all the polynomials of the form

$${\displaystyle \tau _{p}^{n}(x)=x^{p^{n}}}$$

are additive, where ${\displaystyle n}$ is a non-negative integer.¹

The definition makes sense even if ${\displaystyle k}$ is a field of characteristic zero, but in this case the only additive polynomials are those of the form ${\displaystyle ax}$ for some ${\displaystyle a}$ in ${\displaystyle k}$.

It is quite easy to prove that any linear combination of polynomials ${\displaystyle \tau _{p}^{n}(x)}$ with coefficients in ${\displaystyle k}$ is also an additive polynomial.¹ An interesting question is whether there are other additive polynomials except these linear combinations. The answer is that these are the only ones.³

One can check that if ${\displaystyle P(x)}$ and ${\displaystyle M(x)}$ are additive polynomials, then so are ${\displaystyle P(x)+M(x)}$ and ${\displaystyle P(M(x))}$. These imply that the additive polynomials form a ring under polynomial addition and composition. This ring is denoted⁴

$${\displaystyle k\{\tau _{p}\}.}$$

This ring is not commutative unless ${\displaystyle k}$ is the field ${\displaystyle \mathbb {F} _{p}=\mathbb {Z} /p\mathbb {Z} }$ (see modular arithmetic).¹ Indeed, consider the additive polynomials ${\displaystyle ax}$ and ${\displaystyle x^{p}}$ for a coefficient ${\displaystyle a}$ in ${\displaystyle k}$. For them to commute under composition, we must have

$${\displaystyle (ax)^{p}=ax^{p},\,}$$

and hence ${\displaystyle a^{p}-a=0}$. This is false for ${\displaystyle a}$ not a root of this equation, that is, for ${\displaystyle a}$ outside ${\displaystyle \mathbb {F} _{p}.}$¹

Let ${\displaystyle P(x)}$ be a polynomial with coefficients in ${\displaystyle k}$, and ${\displaystyle \{w_{1},\dots ,w_{m}\}\subset k}$ be the set of its roots. Assuming that the roots of ${\displaystyle P(x)}$ are distinct (that is, ${\displaystyle P(x)}$ is separable), then ${\displaystyle P(x)}$ is additive if and only if the set ${\displaystyle \{w_{1},\dots ,w_{m}\}}$ forms a group with the field addition.⁵