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The current Finite field Wikipedia article introduces a finite/Galois field as:

[…] a field that contains a finite number of elements. As with any field, a finite field is a set on which the operations of multiplication, addition, subtraction and division are defined and satisfy certain basic rules. The most common examples of finite fields are given by the integers mod p when p is a prime number.

If the integers mod p "when p is a prime number" are considered a finite field, would the integers "mod s" (where s is a power of two, e.g. 2^8 or 2^32) also be an example of this type of field?

Do the properties of a simple computer byte like uint8_t and/or its signed counterpart int8_t technically count as a Galois field — even if, perhaps, a fairly uninteresting one as far as mathematic or cryptographic analysis goes?

[For IEEE 754 float/double types the answer is, iiuc: no. But I'm assuming this is due to either their basic structure or at least their denormalization semantics which are a lot more complicated than int-like types.]

natevw
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    The integers mod a composite number are not a field, because there are then elements that do not have multiplicative inverses. In particular, $2$ itself has no multiplicative inverse mod any power $2^k$. As for the set of bytes, the answer depends on what you consider the addition and multiplication operations to be. – anon Mar 25 '21 at 22:11
  • @runway44 Ah, thanks for the succinct answer. If understand this correctly, in "mod 7" the multiplicative inverse of 2 would be 4 [since (2 * 4) % 7 == 1]? Whereas in "mod 8" all of the even numbers are missing their multiplicative inverses, so we say — field-wise — that not all "the operations of multiplication" follow sufficient rules? – natevw Mar 25 '21 at 22:22
  • No, we say he usual multiplication operation for the integers mod $n$ does not satisfy the field axioms. We don't say "the operations of multiplication" when we're only talking about one multiplication operation. – anon Mar 25 '21 at 22:25
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    Did you read the Finite field WIkipedia article section "Field with four elements"? For any prime $p$ there is only one field with $p^n$ elements, and for $n>1$ it is not mod $p^n$. – Somos Mar 25 '21 at 22:44
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    No, but the fact that there is a finite field whose elements fit neatly into a $8$-bit integer makes it an attractive candidate for implementations that map elements of $\mathrm{GF}(2^{8})$ to values of uint8. Addition can be implemented as a simple XOR, while multiplication takes a bit more work. This is used, for instance, in AES encryption. More examples here: https://en.wikipedia.org/wiki/Finite_field_arithmetic – Erick Wong Mar 25 '21 at 23:26

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