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I was curious about P-adics and finite fields, and I came up with the idea to create a finite field based off of balanced ternary, let's call it GF(3)[-1]. The elements of this field are {-1, 0, 1}. This field has the following multiplication and addition properties. 1 + -1 = 0 1 + 1 = -1 1 + 0 = 1 -1 + 0 = -1

1 * -1 = -1 1 * 0 = 0 -1 * 0 = 0 -1 * -1 = 1*1 = 1

You could also define fields like GF(5)[-1] with elements {-2, 1, 0, 1, 2}, and GF(7)[-1] with {-3, -2, -1, 0, 1, 2, 3}, and so on. My question is if these fields are isomorphic to their non balanced counterparts(for example, is GF[3] isomorphic to GF(3)[-1]?), and if not, are these fields well studied? I am very new to abstract algebra, so please don't judge me too hard if this is a stupid question.

Bill Dubuque
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Colonizor48
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    Yes, in fact they are the same field if we regard $-1 \in GF(3)[-1]$ and $2 \in GF(3)$ as representing the same element. In fact, for each prime power $q > 1$ there is a field of order $q$ and it is unique up to isomorphism. – Travis Willse Oct 21 '23 at 20:00
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    @Travis Wilse -Duh, I am stupid. Should I just close this question? – Colonizor48 Oct 21 '23 at 20:08
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    @Colonizor48, I don't think it should be closed. This question usefully illustrates the fact that "different-looking" things can nevertheless "be the same" (=isomorphic). – paul garrett Oct 21 '23 at 20:21

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This is exactly the same. Recall that GF(p) is defined* as the quotient ring Z/pZ. Its elements are not integers, even when they are denoted like that. They are equivalence classes of integers. We identify two integers when they are congruent modulo p, which means their difference is divisible by p. So for example, when p=5, the numbers ...-5,0,5,10,... all belong to the same equivalence class.

In general, it doesn't matter which element in an equivalence class we choose to represent it. Surely, for GF(p) the most common way to represent the equivalence classes is via the numbers 0,...,p-1, but you can really pick any set of p integers which are pairwise non-congruent modulo p.

Now, let me explain the connection to your question: you have just chosen the set of representatives -(p-1)/2,...,+(p-1)/2. This works as well, and it does not change the field at all, it is still GF(p), just with different "names" for the elements.

*When you find yourself in a situation where finite fields are defined as sets {0,...,p-1} with certain operations: run!

Martin Brandenburg
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    I legitimately feel embarrassed for not seeing that. Drawing the multiplication table then GF(3) in a different order threw me off. – Colonizor48 Oct 21 '23 at 20:11
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Oct 21 '23 at 21:19
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    @BillDubuque Do you really expect everyone on this site for search for duplicates before answering? Given that the duplicates may be hard to find? (Here, "balanced finite field" has no search results except for the given one.) – Martin Brandenburg Oct 22 '23 at 06:55
  • Yes, that is site policy - see the linked meta post. Of course balanced residue systems have been discussed here many times in the past and posts on such are easy to locate by search. It is best to wait a bit before answering questions that are likely dupes so that others have a chance to search. – Bill Dubuque Oct 22 '23 at 07:02
  • "Of course balanced residue systems have been discussed here many times in the past" - Why "of course"? Do you include people in this statement who haven't read every post? "posts on such are easy to locate by search" - With which search terms? Again, I have only used "balanced field". "... that are likely dupes" - why should something be "likely"? – Martin Brandenburg Oct 22 '23 at 07:14
  • Did you seriously think that balanced residue systems had never been discussed here? Have you never encountered them? Btw, you need to use at-user syntax or I won't know you have replied. – Bill Dubuque Oct 23 '23 at 03:00