In this, page 48, Exercies in chapter 1, there is a following exercise.
Exercise 1. Let $D$ be a division algebra which has finite dimension over the field $k.$ For each $a\in D$ show there is a monic polynomial in $k[x]$ which has a as a root. Conclude that if $k$ is algebraically closed, then $k=D.$
Suppose that $\dim_kD=n.$ For each $a\in D,$ the sequence $1,a,a^2,\cdots,a^n$ are linearly dependent. Then there exist scalars $a_0,a_1,a_2,\cdots,a_n,$ not all zero and belong to $k$, such that $$a_01+a_1a+a_2a^2+\cdots+a_na^n=0.$$ Thus we can choose a monic polynomial $f(x)$ such that $f(a)=0$ where $\deg f\leq n.$ If $k$ is algebraically closed, then we can write $$f(x)=(x-b_0)(x-b_1)(x-b_2)\cdots(x-b_t)$$ where $t\leq n,$ and the $b_i$'s are in $k.$ Since $D$ is a divison algebra, we must have $a-b_i=0.$ This implies $a\in k,$ and so $k=D.$
This is my solution. Can you help me see if it's reasonable? I am not sure about $k=D$ except $k\subseteq D.$ I think it must be $\dim_kD=1.$
Thanks to Kcd's comment, I have given 2 solution. Let's take a look together.
Solution 1. Suppose that $\dim_kD=n.$ For each $a\in D,$ the seqence $1,a,a^2,\cdots,a^n$ are linearly dependent. Then there exist scalars $a_i$'s, not all zero and belong to $k,$ such that $a_01+a_1a+a_2a^2+\cdots+a_na^n.$ Let $f(x)=a_01+a_1x+a_2x^2+\cdots+a_nx^n$ where $\deg f=t\leq n.$ We can choose $a_t=1.$ If $a_t\neq1,$ then $a_t$ has a inverse since $k$ is a field. Thus we can suppose that $f(x)$ is a monic polynomial.
We can embed $k$ into $D$ via a ring homomorphism $i:k\to D,$ is given by $r\mapsto r1.$ Indeed, it is easy to see that $i$ is well-defined. By the definition, we have $i(r_1+r_2)=(r_1+r_2)1=r_11+r_21=i(r_1)+i(r_2)$ and $i(r_1r_2)=(r_1r_2)1=r_1(r_21 )=r_1(r_2(1\cdot 1))=r_1(1(r_21))=(r_11)(r_21)=i(r_1)i(r_2)$ and $i(1)=1\cdot1=1$ for all $r_1,r_2\in k.$ Moreover, $i$ is injective since $k$ is a field. Thus we can embed $k$ as a subring of $D,$ and so we can assume that $k$ is a subring of $D.$ Let $a\in D.$ If $k$ is algebraically closed, then we can write $$f(x)=(x-b_0)(x-b_1)(x-b_2)\cdots(x-b_t)$$ where the $b_i$'s are in $k.$ Since $f(a)=0,$ we have $$(a-b_0)(a-b_1)(a-b_2)\cdots(a-b_t)=0.$$ This implies $a-b_i=0$ for some $i$ since $D$ is a division algebra. Thus $a\in k,$ and so $k=D.$
Solution 2. Suppose that $\dim_kD=n.$ For each $a\in D,$ consider the ring homomorphism $\Phi:k[x]\to k[a]\subseteq D,$ is given by $f(x)\mapsto f(a).$ If $\Phi$ is injective, then we can embed $k[x]$ into $k[a].$ Since $D$ has finite dimensional over $k,k[x]$ also has the same properties, a contradition. Thus $\Phi$ is not injective, implies $\ker\phi\neq0$. Furthermore, $k[x]$ is a principal ideal domain and $\ker\Phi$ is a nonzero ideal in $k[x],$ so there exists a minimal polynomial $p(x)\in k[x]$ such that $\ker\Phi=\langle p(x)\rangle,$ and $p(a)=0.$ It is easy to see that $p(x)$ is monic.
Similarly, we can assume that $k\subseteq D.$ Let $a\in D.$ By the First Isomorphism Theorem, we have $k[a]={\rm Im}\Phi\cong k[x]/\ker\Phi=k[x]/\langle p(x)\rangle.$ If $p(x)$ is reducible, then we can write $p(x)=m(x)n(x)$ where $\deg m$ and $\deg n$ are both less than $\deg p.$ Since $D$ is a division ring and $p(a)=0,$ we must have either $m(a)=0$ or $n(a)=0.$ This contradicts the property of $p(x).$ Thus $p(x)$ is irreducible, and so $\langle p(x)\rangle$ is a maximal ideal. This implies $k[x]/\langle p(x)\rangle$ is a field, and so $k[a]$ is a field. All of $k[a]$'s elements are algebraic over $k$ and $k$ is algebraically closed, so $k[a]=k.$ Therefore $a\in k,$ and so $D=k.$