Relationship between trivial Brauer group and commutative division algebras

Let $k$ be an algebraically closed field and $F$ a finite field extension of the field of rational functions $k(t)$. I've heard two different statements of Tsen's theorem under these conditions:

Statement 1. The Brauer group of $F$ is trivial. i.e. Every central simple algebra over $F$ is isomorphic to a matrix algebra over $F$.

Statement 2. Every division algebra $D$ which is finite dimensional as a vector space over $F$ is commutative.

I have the following questions:

1. Does the center of $D$ (in statement 2) need to have $F$ as it's center, or is it enough that the center of $D$ contains $F$?
2. Does statement 2 imply $D=F$?
3. I understand the proof of statement 1. In section 7.3 of Cohn's Algebra Volume 3, they show $F$ and $k(t)$ are $C_1$ fields and that every $C_1$ field has a trivial Brauer group. How does statement 1 imply statement 2?
4. Does statement 2 still hold if instead $F$ is any field with trivial Brauer group?

For question $1$, in this case it is enough that the center of $D$ contains $F$, because since $D$ is finite dimension vector space over $F$ and it is a division ring, its center, say $L$ is a finite dimensional field extension over $F$. Now we can show that finite extensions of $C_1$ fields are $C_1$, so $L$ is $C_1$ and thus has trivial Brauer group. Proof of above fact: Pick a basis $v_1, v_2,... v_n$ of $L$ as an $F$-vector space. Then expressing elements of $L$ as $\sum a_iv_i$ with $a_i$ in $F$, we see that the norm map from $L$ to $F$ is a homogenous polynomial (embed $L$ into $M_n(F)$ with $n=[L:F]$ via the regular representation and use the determinant definition of norm) $P(a_1, a_2, a_3,...., a_n)$ of degree $n$ with coefficients in $F$, we call this the norm form. Now, since for $l$ in $L$, $N_{L/F}(l)=0$ iff $l=0$, $P$ has the property that, $P(a_1, a_2, a_3,...., a_n)=0$ for $a_i$ in $F$, then all $a_i$ are equal to $0$.

Now we use a small trick. Assume to the contrary $L$ is not $C_1$. Then there is a homogenous polynomial $Q$ of degree $d$ in $e$ variables with coefficients in $L$ with no nontrivial zeroes in $L^e$, where $e>d>0$. Now, writing each element of $L$ as $\sum v_ia_i$, identifying $L^e$ with $F^{en}$ as $F$-vector spaces, we can show that the composition $P(Q)$ is a homogenous polynomial function of degree $dn$ in $en$ variables with no nontrivial zeroes in $F^{en}$, and $en>dn>0$. This contradicts that $F$ is not $C_1$. Thus, we obtain that $L$ must be $C_1$. Proved.

So we have $L$ is $C_1$, thus $B(L)$ is trivial. So the central simple $D$ over $L$ is trivial in the Brauer group. Thus it is isomorphic to $M_r(L)$ for some $r$ positive integer. As $D$ is an integral domain, we see that $r=1$ and $D=L$. This answers question $3$, why statement $1$ implies statement $2$. Also, the answer of $2$ is no. We can just take $D$ a nontrivial field extension of $F$ of finite dimension. For instance take $F=k(t), D=k(t^{\frac{1}{2}})$ of degree $2$ over $F$.

For question $4$, for fields $F$ with trivial Brauer group, a similar proof shows a bit different version of statement $2$: Let $D$ be a division ring with center $Z(D)=F$, with $D$ being finite dimensional vector space over $F$. Then $D=F$. In particular $D$ is commutative.