Exercise 10.6

Proof. If $M={(a_{i,j})}_{1\le i\le n,1\le j\le n}\in M_n(F)$,

if $f(x_{1,1},\dots ,x_{n,n})=\underset{\sigma \in S_n}{\sum <!--\; FUNCTION\; APPLICATION\; -->}\mathrm{sgn}(\sigma )x_{\sigma (1)1}\cdots x_{\sigma (n)n}-1$, then $M\in {\mathrm{SL}}_n(F)$ if and only if $f(a_{1,1},\dots ,a_{n,n})=0$, where $f$ is a non zero polynomial, since it contains the non zero term $x_{1,1}\cdots x_{n,n}$. Therefore ${\mathrm{SL}}_n(F)$ is an hypersurface of $M_n(F)$.

Since a matrix $M\in M_n(F)$ is inversible if and only if its columns $(C1,\dots ,C_n)$ is a basis of $F^n$, the number of matrices in ${\mathrm{GL}}_n(F)$ is

Indeed we choose $C_1$ between $(q^n-1)$ non zero scalars, then we choose $C_2$ between the $q^n-q$ vectors $v\notin ⟨C_1⟩$. If $C_1,\dots ,C_k$ are chosen, we take $C_{k+1}$ between the $q^n-q^k$ vectors $v\notin ⟨C_1,\dots ,C_k⟩$. At last, we choose $C_n\notin ⟨C_1,\dots C_{n-1}⟩$. This gives

Moreover, ${\mathrm{SL}}_n(F)$ is the kernel of the group homomorphism

Therefore $F^{\text{∗}}\simeq {\mathrm{GL}}_n(F)∕{\mathrm{SL}}_n(F)$. This gives

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