Additive polynomial

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

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

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

as polynomials in ${\displaystyle a}${\displaystyle a} and ${\displaystyle b}${\displaystyle b}. It is equivalent to assume that this equality holds for all ${\displaystyle a}${\displaystyle a} and ${\displaystyle b}${\displaystyle b} in some infinite field containing ${\displaystyle k}${\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 ${\displaystyle P(a+b)=P(a)+P(b)}${\displaystyle P(a+b)=P(a)+P(b)} for all ${\displaystyle a}${\displaystyle a} and ${\displaystyle b}${\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 ${\displaystyle q}${\displaystyle q} any multiple ${\displaystyle P}${\displaystyle P} of ${\displaystyle x^{q}-x}${\displaystyle x^{q}-x} will satisfy ${\displaystyle P(a+b)=P(a)+P(b)}${\displaystyle P(a+b)=P(a)+P(b)} for all ${\displaystyle a}${\displaystyle a} and ${\displaystyle b}${\displaystyle b} in the field, but will usually not be (absolutely) additive.

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

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

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

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

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

Similarly all the polynomials of the form

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

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

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

It is quite easy to prove that any linear combination of polynomials ${\displaystyle \tau _{p}^{n}(x)}${\displaystyle \tau _{p}^{n}(x)} with coefficients in ${\displaystyle k}${\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 ${\displaystyle P(x)}${\displaystyle P(x)} and ${\displaystyle M(x)}${\displaystyle M(x)} are additive polynomials, then so are ${\displaystyle P(x)+M(x)}${\displaystyle P(x)+M(x)} and ${\displaystyle P(M(x))}${\displaystyle P(M(x))}. These imply that the additive polynomials form a ring under polynomial addition and composition. This ring is denoted^[4]

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

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

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

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

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

• Drinfeld module
• Additive map

• Weisstein, Eric W. "Additive Polynomial". MathWorld .

• Algebraic number theory
• Modular arithmetic
• Polynomials

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