K-Algebras and Bimodules

K-algebras

A k-algebra A is a (possibly noncommutative) ring with identity that is also a k-vector space, such that for α ∈ k and a,b ∈ A,

(1.1) α(ab) = (αa)b = a(αb).

Note that scalar commutes with ring elements.

Examples:

• Field extensions such as $F/E$.
• Polynomial ring $k[X,Y,Z]$.
• Matrix $M_{nm}(k)$ ring (under addition and multiplication) is $k$ algebra. Here we can see that $k$ can commute with elements of $M_{nm}(k)$ but the ring multiplication is non-commutative.
• The set $Hom_k(V,V)$ of $k$-linear maps of $k$ vector spaces forms a $k$-algebra under addition and composition of linear maps.
  

Since center of ${\cal H}$ consists of real numbers, ${\cal H}$ is a ${\cal R}$ algebra.

Finite dimensional $k$ algebra it is a finite dimensional vector space over $k$.

Note ${\cal C}$ is 2 dimensional over ${\cal R}$ etc.,

Bimodules

If $R,S$ are two rings, an $R-S$ bimodule is an abelian group $(M,+)$ such that

• $M$ is a left $R$ module, and a right $S$ module.
• for all $r \in R$, $s \in S$ and $m \in M$ \begin{equation}
  \label{eq:rs}
  (rm)s=r(ms)
  \end{equation}
  

An $R-R$ bimodule is known as $R-$bimodule.

For positive intergers $m,n$, the set of $n \times m$ matrices $M_{nm}(\mathcal{R})$. Here the $R$-module is $n \times n$ matrices $M_{nn}(\mathcal{R})$. And the $S$-module is $m \times m$ matrices $M_{mm}(\mathcal{R})$.

Addition and multiplication are carried out using the usual rules of matrix addition and matrix multiplication; the heights and widths of the matrices have been chosen so that multiplication is defined.

The crucial bimodule property, that $(rx)s = r(xs)$, is the statement that multiplication of matrices is associative.

A ring $R$ is a $R-R$ module.

For $M$ an $S-R$ bimodule and $N$ a $R-T$ bimodule then $M \otimes N$ is a $S-T$bimodule.

Bimodule homomorphism:

For $M,N$ $R-S$ bimodule, bimodule homomorphism $f:M \rightarrow N$ is an right $R$ module homomorphism as well as an right $S$ modules homomorphism.

An $R-S$ bimodule is same as left module over ring $R \otimes_Z S^{op}$ where $S^{op}$ is opposite ring of $S$. Note in opposite ring multiplication is performed in opposite direction of the original ring.

This caused me some confusion initially. $S^{op}$ is a ring with multiplication reversed. Denote multiplication in the righ $S$ by "." and opposite multiplication by "*". So, how does all this work to define a bimodule?
Lets look at $R \otimes_Z S^{op}$ operating on $m \in M$.
$$\begin{equation} r s*m = r m.s = (rm)s = r(ms) \end{equation}$$
Thus the definition is satisfied. Using $R \otimes_Z S^{op}$ is more nicer.