Making Sense of Finite Fields
This is going to be a series of questions (that I had) with answers (that I found out/looked up) regarding Galois fields. Our last post discussed that every Galois Field has \(p^n\) elements for some \(n \in \mathbb{N}\).
Q. Consider the Galois Field \(GF(p)\). This is isomorphic to (ie, can be thought of as) \(\mathbb{Z}_p\). What about \(GF(p^n)?\) Is that isomorphic to \(\mathbb{Z}_{p^n}?\)
A. No! This is because \(\mathbb{Z}_{p^n}\) has zero divisors, but \(GF(p^n)\) cannot, as it is a field.
Q. Okay, so what are the elements of \(GF(p^n)?\) How would you describe them?
A. They can be thought of as polynomials, each of whose coefficients come from \(\mathbb{Z}_p\).
Q. What are these polynomials? Where do they come from?
A. Actually, there is no ‘canonical’ representation of the elements of \(GF(p^n)\) as polynomials. This is because \(GF(p^n)\) is isomorphic to \(GF(p)[x]/{\langle f(x) \rangle}\) where \(f(x)\) is an irreducible polynomial of degree \(n\). Depending on what \(f\) is, you get different representations of each element!
Q. Interesting, but what is \(f\) here? Why does it have to be irreducible? Why does it have to be of degree \(n?\)
A. Well, we can’t work with \(GF(p)[x]\) directly, because it is not a field! For example, the polynomial \(x\) has no multiplicative inverse. (Evaluate at \(0\) to show impossibility.)
Note that the elements of \(GF(p)[x]/{\langle f(x) \rangle}\) are in fact, sets containing polynomials. Polynomials \(a(x)\) and \(b(x)\) are in the same set iff the difference \(a(x) - b(x)\) is a multiple of \(f(x)\). As \(f\) has degree \(n\), we can characterize each set by a representative element \(c_{n - 1}x^{n - 1} + c_{n - 2}x^{n - 2} + \cdots + c_0\). There are \(p\) choices for each coefficient \(c_i\) (as these come from \(\mathbb{Z}_p\)) and \(n\) of them totally, so we have \(p^n\) elements, as required. Each choice of \(c_i\) gives rise to a distinct set.
We also need no zero divisors in this construction. This follows exactly because \(f\) is irreducible!
The choice of \(f\) actually gives us a basis to represent \(GF(p^n)\) - look at the coefficients \(c_i\) above. As we know, changing the basis only changes the way we view elements, not the actual elements themselves.
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