Editing Commutative ring (section)

== Ring homomorphisms ==
{{Main|Ring homomorphism}}
A ''ring homomorphism'' or, more colloquially, simply a ''map'', is a map {{nowrap|''f'' : ''R'' → ''S''}} such that
{{block indent|1= ''f''(''a'' + ''b'') = ''f''(''a'') + ''f''(''b''), ''f''(''ab'') = ''f''(''a'')''f''(''b'') and ''f''(1) = 1. }}
These conditions ensure {{nowrap|1=''f''(0) = 0}}. Similarly as for other algebraic structures, a ring homomorphism is thus a map that is compatible with the structure of the algebraic objects in question. In such a situation ''S'' is also called an ''R''-algebra, by understanding that ''s'' in ''S'' may be multiplied by some ''r'' of ''R'', by setting
{{block indent|1= ''r'' · ''s'' := ''f''(''r'') · ''s''. }}

The ''kernel'' and ''image'' of ''f'' are defined by {{nowrap|1=ker(''f'') = {{mset|1=''r'' ∈ ''R'', ''f''(''r'') = 0}}}} and {{nowrap|1=im(''f'') = ''f''(''R'') = {{mset|''f''(''r''), ''r'' ∈ ''R''}}}}. The kernel is an [[ring ideal|ideal]] of ''R'', and the image is a [[subring]] of ''S''.

A ring homomorphism is called an isomorphism if it is bijective. An example of a ring isomorphism, known as the [[Chinese remainder theorem]], is
<math display="block">\mathbf Z/n = \bigoplus_{i=0}^k \mathbf Z/p_i ,</math>
where {{nowrap|1=''n'' = ''p''<sub>1</sub>''p''<sub>2</sub>...''p''<sub>''k''</sub>}} is a product of pairwise distinct [[prime number]]s.

Commutative rings, together with ring homomorphisms, form a [[category (mathematics)|category]]. The ring '''Z''' is the [[initial object]] in this category, which means that for any commutative ring ''R'', there is a unique ring homomorphism '''Z''' &rarr; ''R''. By means of this map, an integer ''n'' can be regarded as an element of ''R''. For example, the [[binomial formula]]
<math display="block">(a+b)^n = \sum_{k=0}^n \binom n k a^k b^{n-k}</math>
which is valid for any two elements ''a'' and ''b'' in any commutative ring ''R'' is understood in this sense by interpreting the binomial coefficients as elements of ''R'' using this map.

[[File:Tensor product of algebras.png|thumb|The [[universal property]] of {{nowrap|''S'' ⊗<sub>''R''</sub> ''T''}} states that for any two maps {{nowrap|''S'' &rarr; ''W''}} and {{nowrap|''T'' &rarr; ''W''}} which make the outer quadrangle commute, there is a unique map {{nowrap|''S'' ⊗<sub>''R''</sub> ''T'' &rarr; ''W''}} that makes the entire diagram commute.]]
Given two ''R''-algebras ''S'' and ''T'', their [[tensor product of algebras|tensor product]]
{{block indent|1= ''S'' ⊗<sub>''R''</sub> ''T'' }}
is again a commutative ''R''-algebra. In some cases, the tensor product can serve to find a ''T''-algebra which relates to ''Z'' as ''S'' relates to ''R''. For example,
{{block indent|1= ''R''[''X''] ⊗<sub>''R''</sub> ''T'' = ''T''[''X'']. }}

=== Finite generation ===
An ''R''-algebra ''S'' is called [[finitely generated algebra|finitely generated (as an algebra)]] if there are finitely many elements ''s''<sub>1</sub>, ..., ''s''<sub>''n''</sub> such that any element of ''s'' is expressible as a polynomial in the ''s''<sub>''i''</sub>. Equivalently, ''S'' is isomorphic to
{{block indent|1= ''R''[''T''<sub>1</sub>, ..., ''T''<sub>''n''</sub>] / ''I''. }}

A much stronger condition is that ''S'' is [[finitely generated module|finitely generated as an ''R''-module]], which means that any ''s'' can be expressed as a ''R''-linear combination of some finite set ''s''<sub>1</sub>, ..., ''s''<sub>''n''</sub>.