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It is well known that all finite fields of size $p^n$ are isomorphic to the same finite field $GF(p^n)$. In that sense, you could say there is only one field.

However, if we are required to label the elements, there can be more than one such labelling. Trivially this might give you $p^n!$ possibilities, but I would not consider all of those fundamentally different. I don't know what the correct term for this property is, or even to precisely define it. I think what I wish to know is best phrased as a game:

Alice and Bob are in separate rooms, each presented with a randomly shuffled but otherwise identical row of $p^n$ paper cards. On the back of each card (hidden to Alice and Bob) is written an element of $GF(p^n)$. There is also an arbiter that can see the back of the cards, which will answer any requests for addition/multiplication, using the same field arithmetic for both.

They are then asked to write a unique number $1$ through $p^n$ on the front of each card. What is the probability that Alice and Bob write the same number on their corresponding cards for all field elements, assuming no communication and optimal strategy?

Essentially, 1 divided by the probability is the number I am interested in, as it represents the number of choices one has while labelling a field.

An upper bound is the number of primitive elements of $GF(p^n)$, as one can ask questions to identify the primitive elements, and choose one arbitrarily as element $1$, followed by each generated element in turn, followed by the additive identity element.

However, this is not tight, as it turns out there is no free choice at all for the prime fields. One can simply identify the multiplicative identity element, label it $1$, and then use repeated addition to enumerate the entire field mod $p$.

For $GF(4)$ the answer must be 2, as you can identify the additive and multiplicative identity but otherwise can not distinguish between the two remaining elements.

As I'm not really well-versed in algebra, I don't know how to continue. It's not homework, just personal interest.

orlp
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    As you observed with $GF(4)$, the automorphisms of $GF(p^n)$ cannot be distinguished by Alice and Bob. That leaves $n$ ways of labelling the elements (as there are exactly $n$ automorphisms). – Jyrki Lahtonen May 03 '23 at 08:49
  • Related. – Jyrki Lahtonen May 03 '23 at 10:18
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    I hazard a guess that field automorphisms are the only possible source of disagreement between Alice and Bob. At least, if the oracle/arbiter can be used to evaluate polynomials with integer coefficients at a chosen point (that reduces to repeated addition/multiplication). – Jyrki Lahtonen May 03 '23 at 10:25
  • @JyrkiLahtonen Yes, the oracle/arbiter may be consulted as often as desired. Each query is essentially Alice/Bob pointing at two paper cards asking for the addition or multiplication, after which the arbiter points at the card corresponding to the result of the operation. – orlp May 03 '23 at 10:37
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    Ok. From the size of the pile of cards both Alice and Bob know $p^n$, so $p$ and $n$. Then they can pick an irreducible polynomial $p(x)$ of degree $n$ from $GF(p)[x]$ (say, lexicographically the first). Then they can both find a zero of $p(x)$ by random poking (this may take a while). Then the isomorphism comes from the choice of the zero (one of $n$), but Alice and Bob will have no way of ascertaining that they both used the same zero of $p(x)$ to label the elements etc... – Jyrki Lahtonen May 03 '23 at 10:46
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    Basically, if Alice finds a root $\alpha$ of $p(x)$, and Bob finds the root $\beta$ (possibly distinct from $\alpha$). Alice then labels with $(a_0,a_1,\ldots,a_{n-1})$ the element $\sum_{i=0}^{n-1}a_i\alpha^i$ with $a_i\in{0,1,\ldots,p-1}$. Similarly Bob gives that label to the element $\sum_{i=0}^{n-1}a_i\beta^i$. The two labellings give an automorphism of $GF(p^n)$, and there are $n$ of them. The probability that this scheme leads to identical labellings is then $1/n$. Is that what you are asking? – Jyrki Lahtonen May 03 '23 at 10:55
  • @JyrkiLahtonen Yes, that is a tighter bound that I think would work. But how would one show that you can not do better using some other strategy? – orlp May 03 '23 at 11:04

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