After receiving important comments, which show that my original question has already been asked and answered, I now change my question to the following non-commutative setting:
Let $A \subseteq B \subseteq C$ be three non-commutative left and right Noetherian domains over $\mathbb{C}$.
Assume that $A$ and $C$ are simple.
Recall that a simple (left) Noetherian domain $R$ has a (left) division ring of fractions, denote it by $D(R)$, so we have $D(A) \subseteq D(C)$.
$D(C)$ is a free $D(A)$-module ($D(A)$ is a division ring, hence we have analogous results to the results concerning vector spaces over fields); assume that $D(C)$ is of finite rank over $D(A)$.
Should $B$ also be a simple ring?
An example: Let $f: A_1(\mathbb{C},x,y)\to A_1(\mathbb{C},x,y)$ be a $\mathbb{C}$-algebra endomorphism of the first Weyl algebra $A_1(\mathbb{C},x,y)$ over $\mathbb{C}$, namely, the $\mathbb{C}$-algebra generated by $x$ and $y$ such that $yx-xy=1$. Denote $p=f(x)$ and $q=f(y)$. The image of $A_1(\mathbb{C},x,y)$ under $f$ is $A_1(\mathbb{C},p,q)$. Now take $A:=A_1(\mathbb{C},p,q)$ and $C:=A_1(\mathbb{C},x,y)$, and take $B$ to be the (non-commutative) $\mathbb{C}$-algebra generated by $A$ and $x$. It is well known that the first Weyl algebra (in our case, $A$ and $C$) is simple left and right Noetherian domain, see Theorem 2.19, Corollary 3.9.1 and Corollary 2.11.1. Then I am asking if $B$ is also simple. ($B$ is Noetherian since it is affine over $\mathbb{C}$. $B$ is a domain since $C$ is a domain).
Remark: Another version of my original question can be the following, which I know how to answer: Let $A \subseteq B \subseteq C$ be three non-commutative rings, with $A$ and $C$ division rings, and $C$ is left algebraic over $A$. Is $B$ also a division ring? The answer is YES: Take $b \in B \subseteq C$. $C$ is left algebraic over $A$, so in particular $b$ is left algebraic over $A$: $a_mb^m+\cdots+a_1b+a_01=0$, for some $m \in \{1,2,3\ldots\}$ and $a_m,\ldots,a_1,a_0 \in A$. Now continue with similar arguments as in this answer, and apply this.
See also this question.
