[proofplan]
We first forget the $F$-algebra structure and regard $A$ as a ring. Since $A$ is finite-dimensional over $F$, it is Artinian, and since $A$ is simple as an $F$-algebra, it is simple as a ring; hence the Artin-Wedderburn theorem for simple Artinian rings gives a ring isomorphism $A \cong M_r(E)$ for a division ring $E$. We then recover the $F$-algebra structure by transporting the central copy of $F$ through this isomorphism and identifying the centre of $M_r(E)$ with $Z(E)$. Finally, an $F$-algebra isomorphism between two matrix presentations identifies a primitive corner in one presentation with a primitive corner in the other. Since primitive corners of $M_r(D)$ and $M_s(D')$ are respectively $F$-isomorphic to $D$ and $D'$, this gives the required $F$-algebra uniqueness of the division factor.
[/proofplan]
[step:Show that the underlying ring of $A$ is simple Artinian]
Let
\begin{align*}
\iota_A:F \to A
\end{align*}
denote the structure homomorphism of the $F$-algebra $A$. Since $A$ is central simple over $F$, the map $\iota_A$ identifies $F$ with $Z(A)$, the centre of $A$, and $A$ has no nonzero proper two-sided ideals.
Because $A$ is finite-dimensional as an $F$-[vector space](/page/Vector%20Space), every descending chain of left ideals of $A$ is a descending chain of $F$-subspaces of the finite-dimensional vector space $A$. Such a chain stabilizes. Therefore the underlying ring of $A$ is left Artinian. The same argument applies to right ideals, so $A$ is Artinian as a ring. Since $A$ has no nonzero proper two-sided ideals, the underlying ring of $A$ is a simple Artinian ring.
[/step]
[step:Apply Artin-Wedderburn to obtain a matrix algebra over a division ring]
By the Artin-Wedderburn theorem for simple Artinian rings, applied to the simple Artinian ring underlying $A$, there exist an integer $r \in \mathbb N$, a division ring $E$, and a ring isomorphism
\begin{align*}
\Phi:A \to M_r(E).
\end{align*}
Here $M_r(E)$ denotes the ring of $r \times r$ matrices with entries in $E$.
[guided]
The reason for passing to simple Artinian rings is that the Artin-Wedderburn theorem is stated at the level of rings, so we must verify its ring-theoretic hypotheses without using any extra algebra structure. First, the underlying ring of $A$ is simple: if $I \trianglelefteq A$ is a two-sided ideal of the underlying ring, then $I$ is stable under multiplication by every element of $A$, and in particular under multiplication by every scalar $\iota_A(a)$ with $a \in F$. Hence $I$ is an $F$-subspace and an $F$-algebra two-sided ideal of $A$. Since $A$ is simple as an $F$-algebra, $I=0$ or $I=A$.
Second, the underlying ring of $A$ is left Artinian. Let
\begin{align*}
L_1 \supset L_2 \supset L_3 \supset \cdots
\end{align*}
be a descending chain of left ideals of $A$. Each $L_j$ is an $F$-subspace of the finite-dimensional $F$-vector space $A$, because left ideals are closed under multiplication by the central scalar elements $\iota_A(a)$. A descending chain of subspaces in a finite-dimensional vector space stabilizes, so the chain $(L_j)_{j\in\mathbb N}$ stabilizes. Therefore $A$ is left Artinian. The same argument applied to right ideals shows that $A$ is right Artinian, although the simple left Artinian form of Artin-Wedderburn already suffices.
Thus the Artin-Wedderburn theorem for simple Artinian rings applies to the underlying ring of $A$. It gives an integer $r \in \mathbb N$, a division ring $E$, and a ring isomorphism
\begin{align*}
\Phi:A \to M_r(E).
\end{align*}
At this point $\Phi$ is only a ring isomorphism. It is not yet an $F$-algebra isomorphism, because the theorem has not recorded how the central copy of $F$ inside $A$ is represented inside $M_r(E)$. The next step is precisely to recover that central $F$-structure from the image of $Z(A)$ under $\Phi$.
[/guided]
[/step]
[step:Transport the central copy of $F$ to the division ring factor]
For each $a \in F$, the element $\iota_A(a)$ lies in $Z(A)$. Since $\Phi$ is a ring isomorphism, $\Phi(\iota_A(a))$ lies in $Z(M_r(E))$.
We use the standard computation of the centre of a matrix algebra: for any division ring $E$ and any $r \in \mathbb N$,
\begin{align*}
Z(M_r(E))=\{zI_r:z\in Z(E)\}.
\end{align*}
Therefore, for each $a \in F$, there is a unique element $\lambda(a)\in Z(E)$ such that
\begin{align*}
\Phi(\iota_A(a))=\lambda(a)I_r.
\end{align*}
This defines a map
\begin{align*}
\lambda:F \to Z(E).
\end{align*}
Because $\Phi$ and $\iota_A$ are unital ring homomorphisms and scalar matrices multiply entrywise, the map $\lambda$ is a unital field homomorphism.
Moreover, $\lambda$ is an isomorphism. Indeed, $\Phi$ restricts to an isomorphism
\begin{align*}
Z(A) \to Z(M_r(E)).
\end{align*}
Since $\iota_A:F\to Z(A)$ is an isomorphism and $Z(M_r(E))\cong Z(E)$ via $zI_r\mapsto z$, the map $\lambda$ is a field isomorphism.
Use $\lambda$ to make $E$ into an $F$-algebra: define the scalar action of $a\in F$ on $x\in E$ by
\begin{align*}
a\cdot x=\lambda(a)x.
\end{align*}
With this $F$-algebra structure, $E$ is a central division $F$-algebra, because its centre is exactly $\lambda(F)$, identified with $F$ through $\lambda$. Denote this central division $F$-algebra by $D$.
[/step]
[step:Promote the ring isomorphism to an $F$-algebra isomorphism]
We now regard $M_r(D)$ as an $F$-algebra through scalar matrices:
\begin{align*}
a \mapsto \lambda(a)I_r.
\end{align*}
For every $a\in F$ and every $x\in A$, the ring homomorphism property of $\Phi$ gives
\begin{align*}
\Phi(\iota_A(a)x)=\Phi(\iota_A(a))\Phi(x).
\end{align*}
By the definition of $\lambda$, this becomes
\begin{align*}
\Phi(\iota_A(a)x)=(\lambda(a)I_r)\Phi(x).
\end{align*}
This is exactly the condition that $\Phi:A\to M_r(D)$ is $F$-linear. Since $\Phi$ is already a unital ring isomorphism, it is an $F$-algebra isomorphism:
\begin{align*}
A \cong M_r(D).
\end{align*}
[/step]
[step:Identify the division algebra factor through primitive corners]
Suppose also that there is an $F$-algebra isomorphism
\begin{align*}
\Psi:A \to M_s(D')
\end{align*}
where $s\in\mathbb N$ and $D'$ is a central division $F$-algebra. Define the $F$-algebra isomorphism
\begin{align*}
\Theta:M_r(D) \to M_s(D')
\end{align*}
by
\begin{align*}
\Theta=\Psi\circ\Phi^{-1}.
\end{align*}
Here $\Phi:A\to M_r(D)$ is the $F$-algebra isomorphism constructed above.
Let $e\in M_r(D)$ denote the matrix unit with $1_D$ in the $(1,1)$ entry and $0_D$ in all other entries. The corner ring $eM_r(D)e$ is $F$-isomorphic to $D$ by the map
\begin{align*}
D &\to eM_r(D)e
\end{align*}
that sends $d\in D$ to the matrix with $d$ in the $(1,1)$ entry and $0_D$ elsewhere. This map is multiplicative, additive, unital onto the corner identity $e$, and $F$-linear because scalar multiplication on $M_r(D)$ is through central scalar matrices.
Put
\begin{align*}
p=\Theta(e)\in M_s(D').
\end{align*}
Since $e$ is a primitive idempotent and $\Theta$ is a ring isomorphism, $p$ is a primitive idempotent of $M_s(D')$. The restriction of $\Theta$ gives an $F$-algebra isomorphism
\begin{align*}
eM_r(D)e \to pM_s(D')p.
\end{align*}
Indeed, for every $a\in F$ and every $x\in eM_r(D)e$, the $F$-linearity of $\Theta$ gives $\Theta(a x)=a\Theta(x)$.
We use the following standard idempotent fact for full matrix algebras over division rings: if $E$ is a division ring, $n\in\mathbb N$, and $q\in M_n(E)$ is a primitive idempotent, then there is a unit $u\in GL_n(E)$ such that $uqu^{-1}$ is the matrix unit with $1_E$ in the $(1,1)$ entry. Its hypotheses apply here with $E=D'$, $n=s$, and $q=p$, because $D'$ is a division ring and $p$ is a primitive idempotent of $M_s(D')$. Choose such a unit $u\in GL_s(D')$. Conjugation by $u$ gives an $F$-algebra isomorphism
\begin{align*}
pM_s(D')p \to e'M_s(D')e'
\end{align*}
where $e'$ is the $(1,1)$ matrix unit in $M_s(D')$. This conjugation is $F$-linear because every scalar matrix $aI_s$ with $a\in F$ lies in the centre of $M_s(D')$.
Finally, the corner $e'M_s(D')e'$ is $F$-isomorphic to $D'$ by the map sending an element $d'\in D'$ to the matrix with $d'$ in the $(1,1)$ entry and $0_{D'}$ elsewhere. Composing the $F$-algebra isomorphisms
\begin{align*}
D \cong eM_r(D)e \cong pM_s(D')p \cong e'M_s(D')e' \cong D'
\end{align*}
gives an $F$-algebra isomorphism $D\cong D'$.
This proves both the existence of a decomposition
\begin{align*}
A \cong M_r(D)
\end{align*}
with $D$ a central division $F$-algebra and the uniqueness of $D$ up to $F$-algebra isomorphism.
[/step]
