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Algebraic $K_0$ group for an algebra $A$ may be defined in terms of stable isomorphism classes of idempotents in $M_n(A)$ or equivalently in terms of isomorphism clasess of finitely generated projective modules. If $f:A \to B$ is a morphism of algebras and $e=(e_{ij})_{i,j}$ is an idempotent in $M_n(A)$ then we define $f_*([e]):=[(f(e_{ij}))_{i,j}]$. In the "module picture" given $P$-projective finitely generated $A$ module we form $f_*([P]):=[P \otimes_A B]$ where the right $B$-module structure on $P \otimes_A B$ is via $(x \otimes b)b'=x \otimes bb'$. You can wonder where the map $f$ is involved: well, it is in the relation $xa \otimes b=x \otimes f(a)b$.
Now one can show that any finitely generated projective $A$ module is of the form $eM_n(A)$ for some idempotent $e=(e_{ij})_{i,j}$. So my question is

Why these two description are equivalent, i.e. why is $(f(e_{ij}))_{i,j}M_n(B) \cong P \otimes_A B$ where $P=eM_n(A)$ is finitely generated projective $A$ module?

Please note that I posted a similar question on stck exchange however I didn;t received a satisfactory answer.

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    $\begingroup$ Have you looked in Bass, Algebraic K-theory, or one of the other standard foundational books on (lower) K-theory? $\endgroup$
    – David Handelman
    Commented Sep 23, 2016 at 23:10

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It seems that by $M_n(A)$ you denote two different objects: actual $M_n(A)$ -- which is the same as $\text{Hom}(A^n, A^n)$ -- and $A^n$ itself.

One can show that any finitely generated projective $A$-module $P$ can be written in the form $P = eA^n$ for some idempotent $e = (e_{i,j}) \in \text{Hom}(A^n,A^n)$. And, of course, $P \otimes_A B = (eA^n) \otimes_A B \simeq f(e)B^n$. Indeed, we have an isomorphism $\alpha : A^n \otimes_A B \longrightarrow B^n$ given by $(a_j)\otimes 1 \mapsto (f(a_j))$. Take now a vector $(a_j) \in eA^n$, so $(e_{i,j})(a_j) = (a_j)$. Its image in $B^n$ is preserved by $(f(e_{i,j}))$: $(f(e_{i,j})) (f(a_j)) = f ((e_{i,j}) (a_j)) = f((a_j))$ since $f$ is a morphism of rings. On the other hand, every object in $f(e)B^n$ can be written as $f(e)f((a_j)) \cdot b$ for some $(a_j) \in A^n$ and $b \in B$; if $(f(e_{i,j})) (f(a_j) \cdot b) = f ((e_{i,j}) (a_j)) \cdot b$ does not equal $f((a_j)) \cdot b$, then $(a_j) \notin eA^n$. So the image of $(eA^n) \otimes_A B$ under the isomorphism $\alpha$ is exactly $f(e)B^n$.

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    $\begingroup$ Thank you for your answer. Of course I should have written $A^n$ and $B^n$ instead of $M_n(A)$ and $M_n(B)$. Let me also ask: you showed that if you pick an element in $eA^n$ the image will lie in $f(e)B^n$. However I don't see why the image should be exactly $f(e)B^n$, it seems to me that you have just repeated the same argument as for the fact that $(a_j)_j \in eA^n$ gives that $(f(a_j))_j \in f(e)B^n$ (arguing by contraposition) $\endgroup$
    – truebaran
    Commented Sep 23, 2016 at 23:04
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    $\begingroup$ There was an inaccuracy there, thank you! However, there is a simpler argument. Let $v_j \in A^n$ be vectors of the form $(0, \ldots , 0, 1, 0, \ldots , 0)$ with "1" in the position $j$ and zeroes elsewhere, and $w_j$ be analogous vectors in $B^n$. We have $\alpha(v_j) = w_j$ and $\alpha(e v_j) = f(e) w_j$. $f(e)B^n$ is generated by the vectors $f(e)w_j$, which lie in the image of $\alpha$, so we are done. $\endgroup$
    – A K
    Commented Sep 24, 2016 at 1:08

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