Minimum extension over Q to get the quintics solvable

Question

I noticed that when we extend $\mathbb{Q}$ to the Ring $M_5(\mathbb{Q})$, the ring of $5 \times 5$ matrices, we get that the quintics over $\mathbb{Q}$ become solvable. Namely, by the companion matrix of the polynomial. That is, if f(x) is a polynomial where $ x \in M_5(\mathbb{Q}) $ and the coefficients of $f(x) \in \mathbb{Q}$ and $f$ is a quintic then the companion matrix is a solution of that quintic ( easily verifiable by doing the actual multiplication. But since $M_5(\mathbb{Q})$ is just an extension of $M_5(\mathbb{Q})$(ring extension not field extension) . Then it is possible to extend $\mathbb{Q}$ to get the quintics solvable by radicals. Now the question is what is the minium degree of extension ( Ring not field) that Gives us the quintics to be solvable by radicals and can we create a general formula for that quintics in that extension

Edit: Since the question is not very clear, I am trying to clarify it a bit

Honestly, I am not 100% sure how to type my in a very coherent way, but I will try We start with the ring $\mathbb{Q}$ we add to it the generators $e_{11}, e_{12}, .. e_{15}, ... e_{55}$ such that they respect matrix multiplication, we get a new ring $M_5(\mathbb{Q})$. This new ring is just $\mathbb{Q}$ added to it 25 new generators. In this new ring, every quintic is solvable by radicals. That is, for every Quintic $f(x)$with coefficients in the base ring $\mathbb{Q}$ there is an element $\alpha \in \mathbb{Q}[e_{11}, e_{12}, .. e_{15}, ... e_{55}]$ such that $f(\alpha) = 0$ . Now the question is, what is the minimum size of generators that one has to add to $\mathbb{Q}$ such that the extended ring has a solution for every quintic

Answer 1

I don't know whether I understand you completely, but the reason why quintics have roots in $M_5(\mathbb{Q})$ can be explained as follows:

Let's assume that the polynomial $f\in\mathbb{Q}[x]$ is of degree 5 and also irreducible. Let $\alpha$ be a root of $f$ and set $L=\mathbb{Q}(\alpha)$. Then it is known that $[L:\mathbb{Q}]=5$ and consider the linear map $$M:x\mapsto \alpha x.$$

If we choose the ordered $\mathbb{Q}$-basis $B=(1,\alpha,\alpha^2,\alpha^3,\alpha^4)$ then the transformation matrix $M_B$ of $M$ under this basis is the companion matrix of $f$.

Now it can be shown that the map $\varphi:L\to M_5(\mathbb{Q})$ where $\gamma=a_1+a_2\alpha+a_3\alpha^2+a_4\alpha^3+a_5\alpha^4$ is mapped to $$\gamma\mapsto a_1M_B^0+a_2M_B^1+a_3M_B^2+a_4M_B^3+a_5M_B^4$$ is an embedding of $L$ into $M_5(\mathbb{Q})$.

If the polynomial $f$ is not irreducible and has an irreducible factor of degree $k$, then we can find a root in $M_k(\mathbb{Q})$ by the same procedure.