Module over algebra is free iff it has a basis

abstract-algebraalgebrasmodulestensor-productsvector-spaces

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Let $\k$ be a field and $A$ be associative algebra with unity (over $\k$).

Def. An $A$-module $V$ is a vector space over k with a bilinear map $(A,V)\to V,~(a,v)\mapsto av$, such that $1v=v$ and $a(bv)=(ab)v$ for all $a,b\in A$ and $v\in V$.

Def. An $A$-module isomorphism between $A$-modules V and U is a linear map $f\cn V\to U$ such that $f(av)=af(v)$ for all $a\in A$ and $v\in V$.

Def. A Free module is $A$-module isomorphic to $A\oti V$, where $V$ is some vector space over $\k$ and $a(a'\oti v)=aa'\oti v$.

Def. A Basis of $A$-module $M$ is $\{x_i\}_{i\in I}\ss M$ such that mapping $g\cn\bigoplus_{i\in I}A\ni(a_i)\mapsto\sum a_ix_i\in M$ is a $A$-module isomorphism.

My question is: whyis an $A$-module $M$ is free iff it has a basis? (End of section 2.1 from chapter 2 of Kassel's Quantum Groups.)

  1. If it has a basis then $M\cong \bop_I A$ as $A$-modules. Also $\bop_I A\cong\bop_I(A\oti\k)\cong A\oti\left(\bop_I\k\right)$ as vector spaces. So it is sufficient to show that mapping $f\cn (a_i)_I\mapsto (a_i\oti 1)_I\mapsto \sum_I a_i\oti 1$ is $A$-linear: $f(a(a_i)=(aa_i))=\sum aa_i\oti 1$. But why it is equal to $af((a_i))=a\sum a_i\oti 1=\sum a(a_i\oti 1)$?
  2. If it is free, let us define $(x_{(i,j)}=a_i\oti v_j)_{(i,j)\in I}$ as a basis, where $(a_i\oti v_j)_I$ is a basis in $A\oti V$. Now it is evident from properties of vector space basis that $g$ is bijection.

So the question is: why any $A$-module $A\oti_\k \left(\bop_I\k\right)$ must have the property $a(b\oti 1)=ab\oti 1$?

Best Answer

The point is that you define an $A$-module structure on $A\otimes (\bigoplus_I k)$ by $a(a'\otimes v)=(aa')\otimes v$. Using this $A$-module structure, then your map $f$ is $A$-linear. So, $M\cong A\otimes V$ where $V=\bigoplus_I k$ and the $A$-module structure on $A\otimes V$ is exactly as specified in your definition of a free module, so this proves $A$ is free.

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