# Why is addition in GF(p^n) defined as they are?

I am not familiar with field theory so please bear with me if this is obvious to you. I actually have two questions:

1. Why is addition and subtraction in GF(p^n) defined as:

Addition and subtraction are performed by adding or subtracting two of these polynomials together, and reducing the result modulo the characteristic.

Why modulo the characteristic p instead of p^n?

2. Why does multiplication in GF(p^n) require an irreducible polynomial? (Why can't other polynomial work?)

• The result of multiplication is reduced modulo an irreducible polynomial. You can multiply any two polynomials together. Your operations need to be well-defined. If you do multiplication modulo a reducible polynomial, you run into consistency problems. – Alvin Lepik Sep 28 '17 at 3:18

(1): I think "modulo the characteristic" here is a bit of a red herring. I will write $\mathbb{F}_p$ for the field with $p$ elements, where $p$ is a prime. You can think of each element in $\mathrm{GF}(p^n)$ as a polynomial with coefficients in $\mathbb{F}_p$, where the polynomial has degree less than $n$. All operations in $\mathrm{GF}(p^n)$ are just normal polynomial operations, but modulo some irreducible polynomial of degree $n$.
Forgetting about the modulo-the-irreducible part for the moment, computing in $\mathrm{GF}(2^n)$ should be just arithmetic using polynomials with coefficients in $\mathbb{F}_2$. For example:
\begin{aligned} (x^2 + 1)(x^3 + x + 1) + x &= x^5 + x^3 + x^2 + x^3 + x + 1 + x \\ &= x^5 + (1 + 1)x^3 + x^2 + (1 + 1)x + 1 \\ &= x^5 + x^2 + 1 \end{aligned}
The part where $(1 + 1) = 0$ is what the paragraph means by "reducing the result modulo the characteristic". The trick about $\mathrm{GF}(2^n)$ is that you then have to reduce this further using your chosen irreducible polynomial of degree $n$.
(2): If the polynomial you choose is not irreducible, the resulting system you end up with will not be a field. For example, let's suppose I take real polynomials modulo the reducible polynomial $x^2 -3x + 2 = (x-1)(x-2)$. Then the polynomial $x - 1$ is not zero, and $x - 2$ is not zero, but $(x-1)(x-2) \equiv 0 \pmod{x^2 - 3x + 2}$. So taking polynomials modulo a reducible polynomial does not leave us with a field.