Editing Symmetric algebra (section)

===From polynomial ring===
The symmetric algebra {{math|''S''(''V'')}} can also be built from [[polynomial ring]]s. 

If {{mvar|V}} is a {{mvar|K}}-vector space or a [[free module|free {{mvar|K}}-module]], with a basis {{mvar|B}}, let {{math|''K''[''B'']}} be the polynomial ring that has the elements of {{mvar|B}} as indeterminates. The [[homogeneous polynomial]]s of degree one form a vector space or a free module that can be identified with {{mvar|V}}. It is straightforward to verify that this makes {{math|''K''[''B'']}} a solution to the universal problem stated in the introduction. This implies that {{math|''K''[''B'']}} and {{math|''S''(''V'')}} are canonically isomorphic, and can therefore be identified. This results also immediately from general considerations of [[category theory]], since free modules and polynomial rings are [[free object]]s of their respective categories.

If {{mvar|V}} is a module that is not free, it can be written <math>V=L/M,</math> where {{mvar|L}} is a free module, and {{mvar|M}} is a [[submodule]] of {{mvar|L}}. In this case, one has 
:<math>S(V)=S(L/M)=S(L)/\langle M\rangle,</math>
where <math>\langle M\rangle</math> is the ideal generated by {{mvar|M}}. (Here, equals signs mean equality [[up to]] a canonical isomorphism.) Again this can be proved by showing that one has a solution of the universal property, and this can be done either by a straightforward but boring computation, or by using category theory, and more specifically, the fact that a quotient is the solution of the universal problem for morphisms that map to zero a given subset. (Depending on the case, the [[kernel (algebra)|kernel]] is a [[normal subgroup]], a submodule or an ideal, and the usual definition of quotients can be viewed as a proof of the existence of a solution of the universal problem.)