Can one construct a ${\rm GF}(p^m)$ without using Polynomials?

Question

I am just learning about the finite field theory for channel codes. My understanding is that when constructing a ${\rm GF}(p^m)$, where $p$ is a prime number and $m$ is a positive integer, one has to use polynomials as some of the elements. In contrast, if one is constructing a ${\rm GF}(p)$, where $p$ is a prime number, the set is simply $\{0,1,...,p-1\}$.

However, for a low order finite field, for example, ${\rm GF}(2^2)$, I don't understand why can't one just use $\{0,1,...,3\}$ with mod-4 addition and multiplication, but has to use $\{0,1,x,x+1\}$ with mod-$g(x)$, where $g(x)$ is the primitive polynomial of order 2.

Thank you.

Did you work out whether your proposed operations produce a field? — Tobias Kildetoft, Apr 26 '20 at 15:23
It looks to me that they satisfy the three field axiom, hence produce a field. But there might be something wrong. — Xianhe, Apr 26 '20 at 15:30
How can you boil down the axioms of a field to just 3? I think you might be using the wrong definition. — Tobias Kildetoft, Apr 26 '20 at 15:34
oh, I just realised that there is no inverse for element 2 with mod-4 multiplication. Is that the reason? — Xianhe, Apr 26 '20 at 15:34
We can construct these objects in other ways, too. But it is often not very enlightening (sometimes it can be). I elaborated on this here. — Jyrki Lahtonen, Apr 27 '20 at 20:26

score 1 · Accepted Answer · answered Apr 26 '20 at 15:35

The ring $\mathbb Z/n$ is a field if and only if $n$ is prime, this is why you cannot construct $\mathrm{GF}(4)$ using the usual modulo $4$ addition on $\{0, 1, 2, 3\}$.

To see that it's not a field just note that if $n$ factors then there is a nonzero number that you cannot divide by. For example, $4 = 2\cdot2$ so there cannot be a multiplicative inverse to $2$ because modulo $4$ we already have $4 = 0$. If $2^{-1}$ existed then multiplying both sides by $2^{-1}$ would give $2 = 0$, but this is false modulo $4$.

So generally you will always use polynomials to construct finite fields whose orders are not prime. Of course, for small finite fields you don't have to explicitly use polynomials, you could just write down the addition and multiplication tables explicitly without calling anything a polynomial. But this isn't a general construction that you can base proofs on, so it's not a very useful thing to do.

Can one construct a ${\rm GF}(p^m)$ without using Polynomials?

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