Minimum extension over Q to get the quintics solvable

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

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.

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• Thank you very much for your answer, I have edited the question to be a bit clearer, please have a look at it, and let me know if you think it is clearer and if you can help me by answering it Commented yesterday