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I have seen that for every ring $R$ a fraction field $\text{Frac}(R)$ exists. What would be the fraction field of the set of the matrix $n\times n$, i.e what would be $$\text{Frac}(\mathcal M_n(\mathbb R))\ \ ?$$ I think this can be reduced to looking for the sums of invertible matrices which are not necessarily invertible. Or it it possibly a subset of the diagonalizable matrices ?

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    $\begingroup$ The ring $\mathcal{M}_n (\mathbb{R})$ has zero divisors. If you try the usual fraction field construction on a ring with zero divisors, the whole structure will collapse, this is why it is not defined for this. You might want to look at the total quotient ring: en.wikipedia.org/wiki/Total_ring_of_fractions $\endgroup$ – sebigu Oct 26 '15 at 9:43
  • $\begingroup$ An example of a zero divisor in $\mathcal{M}_2(\mathbb{R})$ would be $\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$. Analagous for any $n\geq1$. $\endgroup$ – Eman Nov 21 '19 at 11:25
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Since a field has no nonzero zero divisors, it is impossible to put a ring inside a field unless it is an integral domain. So you have definitely misunderstood that "all ring R has a fraction field."

A matrix ring could not have a fraction field since it contains nonzero zero divisors, and also it is not commutative.

For some rings, you can pull off the same type of construction to get a ring of fractions, but there is really no hope of inverting zero divisors without everything collapsing. The very best you can do is to give everything that is not a zero divisor an inverse.

In fact, a square matrix ring over a field already has the property that everything is either a unit of a zero divisor. So, you would have to say that it is already an optimal ring of quotients for itself.

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