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=== Rings with generators and relations === The most general way to construct a ring is by specifying generators and relations. Let {{mvar|F}} be a [[free ring]] (that is, free algebra over the integers) with the set {{mvar|X}} of symbols, that is, {{mvar|F}} consists of polynomials with integral coefficients in noncommuting variables that are elements of {{mvar|X}}. A free ring satisfies the universal property: any function from the set {{mvar|X}} to a ring {{mvar|R}} factors through {{mvar|F}} so that {{math|''F'' β ''R''}} is the unique ring homomorphism. Just as in the group case, every ring can be represented as a quotient of a free ring.{{sfnp|Cohn|1995|loc=[https://books.google.com/books?id=u-4ADgUgpSMC&pg=PA242 pg. 242]|ps=}} Now, we can impose relations among symbols in {{mvar|X}} by taking a quotient. Explicitly, if {{mvar|E}} is a subset of {{mvar|F}}, then the quotient ring of {{mvar|F}} by the ideal generated by {{mvar|E}} is called the ring with generators {{mvar|X}} and relations {{mvar|E}}. If we used a ring, say, {{mvar|A}} as a base ring instead of {{tmath|\Z,}} then the resulting ring will be over {{mvar|A}}. For example, if <math>E = \{ xy - yx \mid x, y \in X \},</math> then the resulting ring will be the usual polynomial ring with coefficients in {{mvar|A}} in variables that are elements of {{mvar|X}} (It is also the same thing as the [[symmetric algebra]] over {{mvar|A}} with symbols {{mvar|X}}.) In the category-theoretic terms, the formation <math>S \mapsto \text{the free ring generated by the set } S</math> is the left adjoint functor of the [[forgetful functor]] from the [[category of rings]] to '''Set''' (and it is often called the free ring functor.)<!-- Note: we need to discuss a ring as a solution to a universal problem with some specific application. --> Let {{math|''A''}}, {{math|''B''}} be algebras over a commutative ring {{mvar|R}}. Then the tensor product of {{mvar|R}}-modules <math>A \otimes_R B</math> is an {{mvar|R}}-algebra with multiplication characterized by <math>(x \otimes u) (y \otimes v) = xy \otimes uv.</math> {{See also|Tensor product of algebras|Change of rings}}
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