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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.

Every Brauer class has a central division algebra representative $D$, and every central simple algebra is a matrix algebra over such a $D$. Put $d=\mathrm{deg}D$; its reduced norm is a homogeneous degree-$d$ polynomial on the $d^2$-dimensional vector space $D$.

If $d>1$, then $d^2>d$, so the $C_1$ property supplies a nonzero $x\in D$ with $\mathrm{Nrd}(x)=0$. But every nonzero element of a division algebra is invertible, and its reduced norm cannot be zero. Thus $d=1$, and a central division algebra of degree one is $K$ itself. Every central simple algebra over $K$ is therefore a matrix algebra over $K$, givingThis proves the trivial Brauer group of a C1 field. The essential ingredient is the anisotropy of a division algebra's reduced norm, not the norm polynomial of an arbitrary matrix algebra, which certainly can vanish on nonzero singular matrices.

Put $d=\mathrm{deg}D$, choose a $K$-basis of $D$, and let $N(X_1,\dots ,X_{d^2})$ be its reduced norm polynomial. For $a\in K^{\times }$, considerThis has degree $d$ and $d^2+1>d^2$ variables. The $C_2$ property gives a nontrivial zero $(x,t)$. Necessarily $t\ne 0$: otherwise $N(x)=0$ forces $x=0$ in the division algebra, making the whole coordinate vector zero. Homogeneity now gives $N(x/t)=a$. The value zero is achieved by the zero element. HenceThis is the surjectivity of reduced norms over C2 fields, and the proof also gives the multiplicative surjection $D^{\times }\to K^{\times }$. The argument works for degree one as well.

The reduced norm of an element of a central simple algebra of degree $d$ is its determinant after any splitting isomorphism to $M_d$. Inner automorphisms of a matrix algebra preserve that determinant, while Galois invariance descends its polynomial coefficients to the center. It is a multiplicative homogeneous degree-$d$ polynomial, vanishing exactly on noninvertible elements. On a central division algebra it is anisotropic.

For a central division algebra $D$ of degree $d$ over a C2 field, the form $\mathrm{Nrd}(X)-a{T}^d$ has degree $d$ in $d^2+1$ variables. Its nontrivial zero must have $T\ne 0$, because the reduced norm of a division algebra is anisotropic. Dividing $X$ by $T$ gives norm $a$. Zero is the norm of zero.

A central division algebra of degree $d>1$ has an anisotropic reduced norm of degree $d$ in $d^2>d$ variables. That contradicts the C1 field property. Thus only degree one is possible, and every central simple algebra is split.