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Field? Like a soccer field?
In abstract algebra, a field is a nonzero commutative division ring, or equivalently a ring whose nonzero elements form an abelian group under multiplication.
what is the ring?
the (Mathematics) ring on the link?
a ring is a group with another operation that is associative and has identity and the later operation distributes over the group operation
example. The real numbers abelian group with respect to addition: a+0=a=0+a for all a and 0 is real a+b is a real number for all a,b a+-a=0 for all real a a+b=b+a multiplication a(bc)=(ac)b 1 is a real number 1*a=a*1=a for all a distributes a(b+c) = ab+ac for all ab
the first comment I left should say "abelian group"
Put it simply, ring is something with two operations, addition and multiplication. Ring addition follows the same properties of real number addition, namely, - (a + b) + c = a + (b + c) for all a, b, c in R (+ is associative). - a + b = b + a for all a, b in R (+ is commutative). - There is an element 0 in R such that a + 0 = a for all a in R (0 is the additive identity). - For each a in R there exists −a in R such that a + (−a) = 0 (−a is the additive inverse of a). - a + b in R for all a, b in R. (Closure) However, the multiplication is different in a ring compared to the real numbers. The multiplication in ring satisfies the following properties. - (a ⋅ b) ⋅ c = a ⋅ (b ⋅ c) for all a, b, c in R (⋅ is associative). - There is an element 1 in R such that a ⋅ 1 = a and 1 ⋅ a = a for all a in R (1 is the multiplicative identity). - a ⋅ b in R for all a, b in R. (Closure) However, there is usually no multiplicative inverse for elements in ring, that is you cannot divide one element by another element in a ring. Furthermore, ring multiplication need not be commutative (a ⋅ b != b ⋅ a) and usually isn't.
A field is a ring satisfying addition properties. A field is something like real number (or should I say real number is a field). The addition axioms of a ring are unchanged for a field. However, the following multiplication axioms are added. - For each a in R there exists b in R such that a ⋅ b = b ⋅ a = 1 (b is the multiplicative inverse of a). - a ⋅ b = b ⋅ a for all a, b in R (⋅ is commutative). Since there exist a multiplicative inverse for each element in a field, you can divide one element by another in a field.