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Cool Edit Pro !!HOT!! Download Keygen

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Cool Edit Pro Download Keygen

Cool Edit Pro is a sound editing software developed by Syntrillium Software. It is a standalone piece of software.It was released in 2007.
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Deriving a formula for a (bijective) morphism of algebras

Let $A\subset B$ be subalgebras of a finite dimensional algebra $B$ over a field $K$.
For a $K$-algebra $C$, a $B$-$A$-bimodule $M$, and $b_1, b_2 \in B$, how can one derive an expression for $(b_1b_2)_M – b_1(b_2_M)$? This question was formulated as problem 4.2.14 from the notes from a course on representation theory. I have written down all the definitions and how to solve the problem, but I am not able to get to a solution which takes care of the special details.
Thanks in advance for your help!

A:

It is given that $A$ is a right ideal of $B$. I will write
$$
b\,\cdot\,m=b_1m
$$
for “$m\in M$” and “$b\in B$” written on the right. You are given $b,b_1\in B$, $m\in M$. What you want is to get rid of $b$ and $m$ by choosing $b_2$ and $m_1$ such that $b_2\cdot m_1=m$. Writing $m$ for $m_1$, we get
$$
b_1m-m_1=b_2(m_1-m).
$$
That says that you have to take
$$
b_2=b_1-m_1,
$$
and $m_1$ such that
$$
b_2\cdot m_1=m-m_1=m-m.
$$
The first condition is equivalent to $b_2=b_1+m_2$ for some $m_2\in M$, the second implies $m_1
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