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Their descriptions seem pretty similar. Is a ring the same as a field except it uses abstracted versions of addition and multiplication instead of only classical addition and multiplication as in a field?

I've also already seen the proposed "other answer" and it didn't really answer my question, it was much more convoluted than necessary. Honestly all I'm looking for is a one-line answer.

Bangarang
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  • Both rings and fields use "abstracted" addition and multiplication. In fact, fields are a special case of rings, namely the ones in which you require all nonzero elements to have multiplicative inverses. – Wojowu Mar 09 '20 at 08:35
  • So a ring is like a field except with no guarantee of the existence of inverse elements under + and x, kind of like the difference between a group and semigroup, which makes rings more generalized. – Bangarang Mar 09 '20 at 08:39
  • If $R$ is a ring then it is a field iff $R-{0}$ equipped with multiplication is a commutative group. – drhab Mar 09 '20 at 08:42
  • Okay, so the main difference between a field and ring is that a field's elements commute under multiplication but that you ignore the 0 element of the set under multiplication. – Bangarang Mar 09 '20 at 08:58

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