A problem of central simple algebras: why $(E,s,\gamma)\cong M_n(F)$ only if $\gamma$ is the norm of an element of $E$? I am stuck in the following problem, which is the exercise 6 of section 4.6 of N. Jacobson's Basic Algebra II:

Problem. Prove that $(E,s,\gamma)\cong M_n(F)$ if and only if $\gamma$ is the norm of an element of $E$.



*

*Here $F$ is a field, and $E/F$ is a cyclic extension of degree $n$ with $\mathrm{Gal}(E/F)=\langle s\rangle$. $(E,s,\gamma)$ is an $n^2$-dimensional central simple algebra over $F$ generated by all matrices $\overline b=\mathrm{diag}\{b, s(b),\cdots,s^{n-1}(b)\}$ ( $\forall b\in E$ ) and 
$$
z=\left(\begin{matrix}
0&1& & &\\
0&0&1& &\\
\vdots& &\ddots&\ddots &\\
0& \cdots&\cdots& 0& 1\\
\gamma&\cdots&\cdots & 0& 0
\end{matrix}\right)
$$
for a fixed $\gamma\in F^\times$. 

*Actually the "if" part is quite obvious as a concrete isomorphism can be formulated (Indeed, the correspondence $\lambda_b\mapsto\overline b$, $\lambda_as\mapsto z$ gives an isomorphism from $\mathrm{End}_FE$ to $(E,s,\gamma)$, where $\gamma=N_{E/F}(a)$ and $\lambda_b$ is the $\require{enclose}\enclose{horizontalstrike}F$-isomorphism  $\enclose{horizontalstrike}{x=\sum_{i=0}^{n-1}c_i\alpha^{i}\mapsto \sum_{i=0}^{n-1}c_is^{i}(b)\alpha^i}$, in which $\enclose{horizontalstrike}\alpha$ is a primitive element over $\enclose{horizontalstrike}F$).  is the $F$-linear homomorphism $x\mapsto bx$.

*Where I am stuck is the "only if" part of this problem. Although the book gives a hint to use Skolem-Noether theorem, I can hardly find where to apply it.  So I would like to ask how to obtain the implication

$(E,s,\gamma)\cong M_n(F)$ $\Longrightarrow$ $\gamma=N_{E/F}(a)$ (i.e., $\gamma=as(a)\cdots s^{n-1}(a)$) for some $a\in E$.

Any help would be greatly appreciated.

Here are some attempts (I will continue to update until a solution is come up):


*

*I guess this is one probable way. If we denote the isomorphism by $f\colon (E,s,\gamma)\to M_n(F)$, and let $\enclose{horizontalstrike}g$ be the monomorphism from the algebra generated by all $\enclose{horizontalstrike}{\overline b}$ to $\enclose{horizontalstrike}{M_n(F)}$ that sends $\enclose{horizontalstrike}{\overline b}$ to the matrix of $\enclose{horizontalstrike}{\lambda_b}$ under a basis of $\enclose{horizontalstrike}{E/F}$, then the Skolem-Noether theorem tells us that there is a matrix $\enclose{horizontalstrike}{M\in \mathrm{GL}_n(E)}$ such that $\enclose{horizontalstrike}{f(\cdot)=M^{-1}g(\cdot)M}$. Then  we can define $\enclose{horizontalstrike}{g(z)}$ and it remains to show that $\enclose{horizontalstrike}{g(z)}$ is the matrix of $\enclose{horizontalstrike}{as}$ for some $\enclose{horizontalstrike}{a\in E}$. However, this way seems not working and the use of inner automorphism is still unclear...

*A previous exercise of this section asserts that for each $x\in(E,s,\gamma)$, it can be written in a unique way as
$$
x=\overline b_0+\overline b_1 z+\cdots+\overline b_{n-1}z^{n-1}.
$$
If we can prove that $\enclose{horizontalstrike}{M\in \mathrm{GL}_n(F)}$, then $\enclose{horizontalstrike}{g(z)}$ corresponds a homomorphism $\enclose{horizontalstrike}{\zeta\in\mathrm{End}_FE}$. Moreover, since if so, $\enclose{horizontalstrike}{g\colon(E,s,\gamma)\to M_n(F)}$ is an $\enclose{horizontalstrike}F$-linear isomorphism, and thus $\enclose{horizontalstrike}{s\in\mathrm{End}_FE}$ can be written in a unique manner as follows
$$
\enclose{horizontalstrike}{s=\lambda_{b_0}+\lambda_{b_1}\zeta+\cdots+\lambda_{b_{n-1}}\zeta^{n-1}}
$$
where $\enclose{horizontalstrike}{b_0,\cdots,b_{n-1}\in E}$. Hopefully by $\enclose{horizontalstrike}{s^n=1}$ we perhaps can get that all $\enclose{horizontalstrike}{b_i=0}$ but $\enclose{horizontalstrike}{b_1}$ (at least this holds for $\enclose{horizontalstrike}{n=2}$), and then $\enclose{horizontalstrike}{\zeta=\lambda_{b_1^{-1}}s}$, which indicates $\enclose{horizontalstrike}{\lambda_\gamma=\zeta^n=(\lambda_{b_1^{-1}}s)^n}$, namely $\enclose{horizontalstrike}{\gamma=N_{E/F}(b_1^{-1})}$.

*Let us denote by $\iota\colon(E,s,\gamma)\hookrightarrow M_n(E)$ the natural embedding. Then by Skolem-Noether theorem, $\iota(\cdot)=N^{-1}f(\cdot)N$ for some $N\in\mathrm{GL}_n(E)$. Since $f$ is isomorphic, each $x\in M_n(F)$ can be written as
$$
x=f(\overline b_0)+f(\overline b_1)f(z)+\cdots+f(\overline b_{n-1})f(z)^{n-1}
$$
in a unique manner with $b_i\in E$. Note that $s\in \mathrm{End}_FE$ corresponds to a matrix 
$$x_s=f(\overline b_0)+\cdots+f(\overline b_{n-1})f(z)^{n-1}\in M_n(F)$$ 
and $x_s^n=I\in M_n(F)$ (the identity matrix). If we can deduce that $f(\overline b_i)=0$ but $f(\overline b_1)$ (again, this holds at least for $n=2$), then it follows that $\enclose{horizontalstrike}{\det(f(\overline b_1z))^n=1}$, and thus $\enclose{horizontalstrike}{\det(\overline b_1z)^n=1\Longrightarrow N_{E/F}(b_1)^n\gamma^n=1}$. $x_s=f(b_1)f(z)$ and hence $\gamma I=N_{E/F}(b_1^{-1})I$.

*If it can be shown that $x_sf(\overline b)=f(s(\overline b))x_s$, then that $f(\overline b_i)=0$ but $f(\overline b_1)$ can be deduced, and the arguments before all make sense.

A failed attempt:
I tried to prove that the $m_b$ defined as follows corresponds to the matrix $f(\overline b)$ under some $F$-basis of $E$ but at last found that I made several fatal mistakes, having no idea how to fix them for the moment. I think this is very likely to be a working way.
Assume that $f\colon(E,s,\gamma)\to M_n(F)\subset M_n(E)$. For the inclusion mapping $\iota\colon(E,s,\gamma)\hookrightarrow M_n(E)$, the Skolem-Noether theorem implies that 
$$
f(\cdot)=N^{-1}\iota(\cdot)N
$$
for some $N\in\mathrm{GL}_n(E)$. For a primitive element $\alpha$ of $E/F$, $B:=\{1,\alpha,\cdots,\alpha^{n-1}\}$ is an $F$-basis of $E$. Define for each $b\in E$,
\begin{align}
m_b\colon E&\to E\\
x=\sum_{i=0}^{n-1}c_i\alpha^i&\mapsto\sum_{i=0}^{n-1}c_is^i(b)\alpha^i
\end{align}
and it follows that $m_b$ is $F$-linear, i.e., $m_b\in\mathrm{End}_FE$. We assert that there is an $F$-basis $B'$ of $E$, under which the matrix of $m_b\ (\forall b\in E)$ is $f(\overline b)$. Indeed, for each $b\in E$,
\begin{align}
m_b(1,\alpha,\cdots,\alpha^{n-1})N=&(b,s(b)\alpha,\cdots,s^{n-1}(b)\alpha^{n-1})N\\
=&(1,\alpha,\cdots,\alpha^{n-1})\overline bN\\
=&(1,\alpha,\cdots,\alpha^{n-1})NN^{-1}\iota(\overline b)N\\
=&(1,\alpha,\cdots,\alpha^{n-1})Nf(\overline b).
\end{align}
Thus it remains to show that $B':=(1,\alpha,\cdots,\alpha^{n-1})N$ is an $F$-basis of $E$. Let us consider the matrix
$$
V=\left(
\begin{matrix}
1&\alpha&\cdots&\alpha^{n-1}\\
1&s(\alpha)&\cdots&s(\alpha^{n-1})\\
\vdots&\vdots&&\vdots\\
1&s^{n-1}(\alpha)&\cdots&s^{n-1}(\alpha^{n-1})
\end{matrix}\right)\in M_n(E).
$$
Since $\mathrm{Gal}(E/F)=\langle s\rangle$ and $E=F(\alpha)$, $s^i(\alpha)\neq s^j(\alpha)$ for $i\neq j$ ($i, j=1,\cdots,n$). Therefore the Vandermonde matrix $V\in\mathrm{GL}_n(E)$. If $B'=(1,\alpha,\cdots,\alpha^{n-1})N$ is $F$-linear dependent, then the columns of $\enclose{horizontalstrike}{VN}$ are $\enclose{horizontalstrike}{F}$-(and of course $\enclose{horizontalstrike}{E}$-)linear dependent (since $\enclose{horizontalstrike}{s^i}$ fixes elements of $\enclose{horizontalstrike}{F}$). As $\enclose{horizontalstrike}{N}$ is invertible, it leads to a contradiction that $\enclose{horizontalstrike}{V\notin\mathrm{GL}_n(E)}$. Therefore $\enclose{horizontalstrike}{B'}$ is an $\enclose{horizontalstrike}{F}$-basis of $\enclose{horizontalstrike}{E}$.
Since $M_n(F)\cong \mathrm{End}_FE$, under the basis $B'$, $s$ corresponds to a matrix $x_s\in M_n(F)$. Then
$$
x_s=f(\overline b_0)+f(\overline b_1)f(z)+\cdots+f(\overline b_{n-1})f(z^{n-1})
$$
for some $b_0,\cdots,b_{n-1}\in E$. Note that $\enclose{horizontalstrike}{sm_b=m_{s(b)}s}$ and we have for each $\enclose{horizontalstrike}{b\in E}$, $\enclose{horizontalstrike}{x_sf(\overline b)=f(s(\overline b))x_s}$, while
\begin{align}
x_sf(\overline b)=&f(\overline b_0)f(\overline b)+f(\overline b_1)f(z)f(\overline b)+\cdots+f(\overline b_{n-1})f(z)^{n-1}f(\overline b),\\
f(s(\overline b))x_s=&f(s(\overline b))f(\overline b_0)+f(s(\overline b))f(\overline b_1)f(z)+\cdots+f(s(\overline b))f(\overline b_{n-1})f(z)^{n-1}.
\end{align}
The exercise 4 of section 4.6 of N. Jacobson's Basic Algebra II asserts that

Each $x\in (E,s,\gamma)$ can be written in a unique way as $$x=\overline b_0+\overline b_1z+\cdots+\overline b_{n-1}z^{n-1},$$ where $b_i\in E$.

Thus it follows that
\begin{align}
&f(\overline b_0)f(\overline b)=f(s(\overline b))f(\overline b_0),\\ &f(\overline b_1)f(s(\overline b))=f(s(\overline b))f(\overline b_1),\\
&\quad\vdots\\
&f(b_{n-1})f(s^{n-1}(\overline b))=f(s(\overline b))f(\overline b_{n-1}).
\end{align}
The arbitrariness of $b\in E$ forces that $b_0=b_2=\cdots=b_{n-1}=0$ (precisely speaking,  for each $i=0,2,\cdots,n-1$, pick a $b\in E\setminus F$ which is not fixed by $s^{i-1}$, and then $f(\overline b_i)f(s^i(\overline b))=f(s(\overline b))f(\overline b_i)=f(\overline b_i)f(s(\overline b))$; if $b_i\neq 0$, $f$ being isomorphic entails that $f(s^i(\overline b))=f(s(\overline b))$ and thence, $b=s^{i-1}(b)$, which is a contradiction). Hence $x_s=f(\overline b_1)f(z)$, and the invertibility of $x_s$ implies that $b_1\neq 0$. We thus have
\begin{align}
\gamma I=&f(z)^n=(f(\overline b_1)^{-1}x_s)^n=(f(\overline{b_1^{-1}})x_s)^n\\
=&f(\overline{b_1^{-1}})f(s(\overline{b_1^{-1}}))\cdots f(s^{n-1}(\overline{b_1^{-1}}))x_s^n\\
=&N_{E/F}(b_1^{-1})I,
\end{align}
i.e., $\gamma=N_{E/F}(a)$, where $a=b_1^{-1}\in E^\times$.
 A: The following argument is copied from the Lemma in Section 15.1 and Lemma in Section 14.2 of [2]. The result is also proven in Corollary 4.7.5 of [1] in a different way.
Let me denote $z_{\gamma}$ for your $z$, and set $\mathcal{A}_{\gamma} := (E,s,\gamma)$ for any $\gamma \in F^{\times}$. Suppose there is an $F$-algebra isomorphism $\varphi : \mathcal{A}_{\gamma} \to \mathcal{A}_{1_{F}}$. The map $b \mapsto \varphi(\overline{b})$ is an $F$-algebra homomorphism $E \to \mathcal{A}_{1_{F}}$, and by Skolem-Noether there exists an $F$-algebra automorphism $\alpha : \mathcal{A}_{1_{F}} \to \mathcal{A}_{1_{F}}$ such that $\alpha(\overline{b}) = \varphi(\overline{b})$ for all $b \in E$. Hence, after replacing $\varphi$ by $\alpha^{-1}\varphi$, we may assume that $\varphi(\overline{b}) = \overline{b}$ for all $b \in E$. Let $\xi : E \to \mathcal{A}_{1_{F}}$ denote the $F$-algebra map $b \mapsto \overline{b}$. Note that $z_{\gamma} \cdot \overline{b} \cdot z_{\gamma}^{-1} = \overline{s(b)}$ for all $\gamma \in F^{\times}$ and $b \in E$. Applying $\varphi$, we get $\varphi(z_{\gamma}) \cdot \overline{b} \cdot \varphi(z_{\gamma})^{-1} = \overline{s(b)}$, which implies $(z_{1_{F}}^{-1}\varphi(z_{\gamma})) \cdot \overline{b} \cdot (z_{1_{F}}^{-1}\varphi(z_{\gamma}))^{-1} = \overline{b}$ for all $b \in E$; in other words $z_{1_{F}}^{-1}\varphi(z_{\gamma}) \in C_{\mathcal{A}_{1_{F}}}(\operatorname{im} \xi)$. One may check $C_{\mathcal{A}_{1_{F}}}(\operatorname{im} \xi) = \operatorname{im} \xi$ (using the fact that each $a \in \mathcal{A}_{1_{F}}$ can be written in a unique way as $a = \overline{b_{0}} + \overline{b_{1}}z_{1_{F}} + \dotsb + \overline{b_{n-1}}z_{1_{F}}^{n-1}$ with $b_{i} \in E$). Thus $\varphi(z_{\gamma}) = z_{1_{F}}\overline{x}$ for some $x \in E^{\times}$. Taking $n$th powers gives $\overline{\gamma} = \overline{x} \dotsb \overline{s^{n-1}(x)}$ since $(\varphi(z_{\gamma}))^{n} = \varphi(z_{\gamma}^{n}) = \varphi(\overline{\gamma}) = \overline{\gamma}$.
References:
[1] P. Gille and T. Szamuely, "Central Simple Algebras and Galois Cohomology", Cambridge Studies in Advanced Mathematics, Vol. 101 (2006)
[2] R. S. Pierce, "Associative Algebras", Graduate Texts in Mathematics, Vol. 88, Springer-Verlag (1982)
