# Why are elements of $GF(p^n)$ represented as polynomials mod some irreducible polynomial, while elements of $GF(p)$ are integers mod $p$?

My background is in computer engineering, not pure math. I've been studying the application of Finite Fields. Wikipedia on Finite field arithmetic says:

GF(p), where p is a prime number, is simply the ring of integers modulo p.

...

Elements of GF($$p^n$$) may be represented as polynomials of degree strictly less than n over GF(p).

I would really appreciate a simple, accessible explanation as to why this is the case? What fundamental property of Finite Fields makes it that the elements of $$GF(p)$$ and $$GF(p^n)$$ behave so differently and yet are both considered Finite Fields?

• Polynomials of degree less than 1 are the constants. – Empy2 Dec 10 '20 at 12:04
• They are both considered finite fields because they are both finite, and they are both fields. It would be perverse to consider some finite fields to be finite fields and consider other finite fields not to be finite fields. – Gerry Myerson Dec 10 '20 at 12:15
• What Gerry said! You may also benefit from checking out this older thread. I don't think I want to repeat any parts of my answer to that question here. I realize the focus here is different, I should have an answer prepared with CS people in mind in particular? Somewhere? Couldn't find it right away. – Jyrki Lahtonen Dec 10 '20 at 15:57
• Thanks for these comments. @JyrkiLahtonen yes an answer for CS people would be great. I appreciate that math is in some ways a language, not one I speak very well, whereas you guys do, and so having some kind of constructive conversation is not easy. – eddydee123 Dec 10 '20 at 16:22

The field $$GF(p)$$, $$p$$ prime, is a so-called prime field and given by the integers modulo $$p$$.
The field $$GF(p^n)$$ is a field extension of $$GF(p)$$; i.e., it contains $$GF(p)$$ as subfield.
The practical way to construct $$GF(p^n)$$ is to choose a polynomial $$f(x)$$ in $$GF(p)[x]$$ of degree $$n$$, which is irreducible over $$GF(p)$$. Then the field $$GF(p^n)$$ is given by the quotient ring $$GF(p)[x]$$ modulo the ideal $$\langle f(x)\rangle$$; practically, the computations are modulo the polynomial $$f(x)$$.
There are other ways to consider $$GF(p^n)$$, e.g. as the zero-set of a polynomial - but this would be more abstract and not suitable for practical computations.