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Write $d=\mathrm{deg}A$ for the degree of a central simple algebra, so ${\dim }_{K}A=d^2$. Choose a finite Galois splitting field of a central simple algebra $E/K$ for $A$ and an isomorphism $\varphi :A{\otimes }_{K}E\to M_d(E)$. The reduced norm isThe existence of a finite separable splitting field is a standard structural property of a central simple algebra; passing to its Galois closure supplies $E$.

Every $E$-algebra automorphism of $M_d(E)$ is inner. Here is the matrix-unit argument for the inner automorphisms of a matrix algebra result. For an automorphism $h$, choose $w\ne 0$ in the image of $h(E_{11})$, and set $w_i=h(E_{i1})w$. Then $h(E_{ij})w_k={\delta }_{jk}{w}_i$. The $w_i$ are nonzero and independent, and their number is $d$, so they form a basis. Relative to that basis $h$ acts on each matrix unit in the usual way. Thus $h(T)=PT{P}^{-1}$ for some $P\in {\mathrm{GL}}_d(E)$. Determinants are unchanged by conjugation, proving independence of $\varphi$.

For $\sigma \in \mathrm{Gal}(E/K)$, compare $\varphi$ with the isomorphism obtained by applying $\sigma$ to matrix entries and ${\sigma }^{-1}$ to the scalar factor of $A\otimes E$. Their difference is again inner. For $a\in A$, this shows $\sigma (\det \varphi (a))=\det \varphi (a)$. The determinant therefore belongs to $K$. In fact, for a $K$-basis $a_1,\dots ,a_{d^2}$, the same comparison shows that all coefficients ofare fixed by the Galois group, so the reduced norm is a homogeneous polynomial of degree $d$ over $K$. This coefficient argument also applies over finite fields, where equality merely as functions would not identify polynomials.

Finally, two finite splitting fields embed into a common finite splitting extension. Determinant commutes with scalar extension, and over that common extension the two matrix identifications differ by an inner automorphism. Hence the resulting polynomials agree over $K$. The reduced norm is independent of both the splitting field and the matrix identification. It is multiplicative, and $\mathrm{Nrd}(a)=0$ exactly when $a$ is not invertible: a matrix with nonzero determinant is invertible after scalar extension, and invertibility descends by the invertibility of the $K$-linear multiplication map. In a division algebra the only element of reduced norm zero is zero.