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Came here to see if anyone noticed this. You can actually make a field of Z using a bijection to the rationals Q, which do exist. But then you aren't using the standard operations, but using the ones inherited from the particular mapping to Q.

True, perhaps the author meant rationals.

It's usually integers modulo p (p prime), which is a field.

Edit: but yes, the OP writes "integers (Z) are a field", which is wrong. Integers form a ring.

I think integers mod p (p prime) is a field