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I'm trying to understand a few basic notions on the reduced norm of division algebras, and more specifically the relation between the norm of an algebra and the norm of algebras similar to it. Disclaimer- I am new to the subject, and there may be some mistakes in my reasoning here- I would very much appreciate if anyone who sees something awkward here can point it out for me.

Let $D$ be a central-simple division algebra over a (local) field $F$, and let $\operatorname{Nrd}_{D/F}$ denote the reduced norm of $D/F$- that is, given an algebra representation $\phi:D\to M_n(E)$ for some $n$, and $E/F$ a minimal field extension for which such an $F$-algebra homomorphism exists, let $\operatorname{Nrd}_{D/F}(x)=\det(\phi(x))$. Suppose $M$ is a division algebra which is similar to $D$ (i.e a matrix algebra over $D$), then $M$ is also central-simple over $F$, and hence also has a reduced norm over $F$- denoted $\operatorname{Nrd}_{M/F}$.

What can be said about the relation between $\operatorname{Nrd}_{D/F}$ and $\operatorname{Nrd}_{M/F}$? Namely- is it possible to find a multiplicative map $F:M\to D$ such that the following diagram commutes \begin{equation}\begin{matrix}M^\times\longrightarrow&D^\times\\ {}_{\operatorname{Nrd}_{M/F}}\searrow&{}\downarrow&{}_{\operatorname{Nrd}_{D/F}}\\ &F \end{matrix}? \end{equation}

At first I thought one might be able to obtain such a map by taking a determinant map from $M$ to $D$. But as was pointed to me, the determinant is not well-defined for matrix algebras over non-commutative rings..

I would very much appreciate if someone knows of any natural construction for such a map, or can direct me to understand why no such map exists.

Thank you very much


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I am not sure about following answer but I have tried on following line. But I am putting it as an answer because I am not able to comment on your question:

We have $\mathrm Nrd_{A\otimes B}(x\otimes y)={\mathrm Nrd_A(x)}^{\mathrm deg B}{\mathrm Nrd_B(x)}^{\mathrm deg A}$ [I seem to recall this, but check this statment].

Let $M=D\otimes_F M_n(F)$ and $T:=M_n(F)$. Define $f:M^{\times}\rightarrow D^{\times}$ by $x\otimes y\mapsto x^n\mathrm Nrd_T(y)$, where $n=\mathrm deg~T$. It works fine on elements of the form $x\otimes y$ but I don't know about what will happen for non-zero elements of the form $x_1\otimes y_1+x_2\otimes y_2$.

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Not sure, but I'm skeptical that such a map exists in general. On the other hand, the Dieudonne determinant is a pretty good substitute for what you want. It gives you a homomorphism $\det: M^\times \to D^\times_\mbox{ab}$ to the abelianized unit group of the division algebra. If $F$ is a local field, then you have the desired commutativity: $$\mbox{Nrd}_{M/F} = \mbox{Nrd}_{D/F}\circ\det;$$ this makes sense since $\mbox{Nrd}_{D/F}$ kills commutators.

Over a general field $F$, this compatibility may fail, but one still has the following result of Dieudonne (see 2.8.10 in these notes): $$\mbox{Nrd}_{M/F}(M^\times) = \mbox{Nrd}_{D/F}(D^\times).$$

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