Additive polynomial

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

Definition

Let k be a field of finite characteristic p, with p a prime number. A polynomial $P(x)$ with coefficients in k is called additive, if

$P(a+b)=P(a)+P(b)$

for all a and b in k. One says that $P(x)$ is absolutely additive, or Frobenius, if it is additive over the algebraic closure of k.

The absolutely additive polynomials are the interesting ones. It will be described below how they relate to the ordinary additive polynomials.

Examples

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

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

Since p is prime, one can prove that for all n = 1, ..., p−1 the binomial coefficient

${p \choose n}$

is divisible by p, which implies that

$(a+b)^{p}=a^{p}+b^{p}$

over the algebraic closure of k.

It can be shown in similar manner that all the polynomials of the form

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

are absolutely additive, where n is a non-negative integer.

The ring of absolutely additive polynomials

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

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

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

It can be shown that this ring is not commutative unless k equals the field $\mathbb {F} _{p}=\mathbf {Z} /p\mathbf {Z}$ (see modular arithmetic). Indeed, consider the absolutely additive polynomials $ax$ and $x^{p}$ for a coefficient $a$ in $k$. For them to commute under composition, we must have

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

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

Additive vs. absolutely additive

From their definition, it follows quickly that any absolutely additive polynomial is also an additive polynomial. But they are not equivalent. The polynomial

$x^{p+1}-x^{2}=x(x^{p}-x)\,$

over $\mathbb {F} _{p}$ is trivially additive, as it takes only the value 0 over this field according to Fermat's little theorem, but it is not absolutely additive, since it is not a linear combination of the polynomials $\tau _{p}^{n}(x)$.

Another way of emphasizing the difference between these two types is the following: for an additive polynomial $P(x)$ which is not absolutely additive, the equality

$P(a+b)=P(a)+P(b)$

holds over k, but will fail over a bigger field.

One can show however, that if the field k is infinite, then any additive polynomial is absolutely additive.

The fundamental theorem of additive polynomials

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

References

• David Goss, Basic Structures of Function Field Arithmetic, 1996, Springer, Berlin.