# What is the fraction field of the ring of the matrix?

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 ?

• 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 – sebigu Oct 26 '15 at 9:43
• 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 $n\geq1$. – Eman Nov 21 '19 at 11:25