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I am having trouble parsing the following problem:

Let $F$ be a field, $A$ an $F$-algebra with 1 (not necessarily commutative) and let $M$ be an $F$-vector space which is also an $A$-bimodule. Let $A\ltimes M$ be the vector space $A\oplus M$ equipped with the multiplication $(a,m)(a',m')=(aa',am'+ma')$. Show that $A\ltimes M\cong T_A(M)/I$ where $T_A(M)$ is the tensor algebra and $I$ is the two sided ideal generated by $m\otimes m'$ with $m,m'\in M$.

Any hints or ideas how to grasp this problem would be helpful.

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First a general remark, that really helps when trying to build isomorphisms between quite abstract objects : some objects are made for constructing morphisms from them, and some are made for contructing morphisms to them. Often this is can be seen in terms of a left/right adjunction.

Tensor algebras are made for constructing morphisms from them. What is the universal property of $T_A(M)$ ? It is that for any $F$-algebra $B$ with an algebra morphism $\varphi: A\to B$, and any $A$-bimodule morphism $f:M\to B$ (where $B$ is seen as a $A$-bimodule through $\varphi$ and its internal multiplication), $f$ and $\varphi$ extend uniquely to a $F$-algebra morphism $T_A(M)\to B$ (which makes sense because $A$ and $M$ are subspaces of $T_A(M)$).

And furthermore, quotients are also made for constructing morphisms from them, as their classical universal property shows.

So what you want is an isomorphism $T_A(M)/I\to A\ltimes M$, so a surjective morphism $T_A(M)\to A\ltimes M$ such that the kernel is $I$.

If we take our $B$ above to be $A\ltimes M$, then our $\varphi: A\to A\ltimes M$ and $f: M\to A\ltimes M$ are pretty clear : $\varphi(a) = (a,0)$ and $f(m) = (0,m)$. Then $\varphi$ is very clearly a $F$-algebra morphism, and $f$ a $F$-linear map. It remains to check it's a $A$-bimodule morphism. Now $\varphi(a)f(m) = (a,0)(0,m) = (0,am) = f(am)$ and likewise you can check that $f(ma)=f(m)\varphi(a)$. So from $\varphi$ and $f$ we find by the universal property of $T_A(M)$ a well-defined $F$-algebra morphism $\psi: T_A(M)\to A\ltimes M$.

Our $\psi$ is clearly surjective since by definition all elements $(a,0)$ and $(0,m)$ are in the image. We just have to check that $I\subset \ker(\psi)$ and that the induced map $\tilde{\psi}: T_A(M)/I\to A\ltimes M$ is injective.

It's easy to see that $I\subset \ker(\psi)$ : $\psi(m\otimes m') = \psi(m)\cdot \psi(m') = (0,m)(0,m') = (0,0)$ by definition of the product in $A\ltimes M$. Now in $T_A(M)$, a supplementary space of $I$ as $F$-vector spaces is $A\oplus M$, so $\tilde{\psi}$ is injective iff the restriction of $\psi$ to $A\oplus M\subset T_A(M)$ is injective (if we want to be very formal, we have to invoke the fact that the injectivity of an algebra morphism can be checked on the vector space level, which is obvious). But now the restriction of $\psi$ to $A\oplus M$ is by construction pretty much the identity : if $a\in A$ and $m\in M$, then $\psi(a+m) = (a,m)\in A\ltimes M$. So it's injective and the proof is over.

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  • $\begingroup$ This is great. Thank you! $\endgroup$ – Jimmy2Goons Apr 5 '16 at 20:37
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Consider the map $\phi : T_A(M) \longrightarrow A\ltimes M$ defined by $\phi(\sum_{i=0}^{l}m_i)=m_0+m_1$, where $m_i\in M^{\otimes i}$. Clearly this map is onto with $ker(\phi)= I$. Therefore we have an $F-$ linear isomorphism between $T_A(M)/I$ and $A\ltimes M$. That this isomorphism preserves the multiplication can be easily checked.

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  • $\begingroup$ @Captain : Sorry I did not notice your solution when I posted my answer. $\endgroup$ – Anirban Bose Apr 5 '16 at 10:56
  • $\begingroup$ No problem, it gives a shorter answer for people who don't want to read my long answer. $\endgroup$ – Captain Lama Apr 5 '16 at 20:35

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