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I am given the following exercise: Find the units of the ring $R=M_n(\mathbb{Q})$.

That's what I have thought: $$$$ $E \in M_n(\mathbb{Q})$ is an unit of the ring $R=M_n(\mathbb{Q})$,if $\exists $ $E' \in M_n(\mathbb{Q})$ such that $EE'=E'E=I_n$,that means that $E$ must be inversible.So,$det(E) \neq 0$. $$$$ How can I continue or is that the answer?

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    $\begingroup$ That's the answer; because of the explicit expression of the inverse by means of minors of $E$, the inverse of a matrix with rational coefficients has rational coefficients. $\endgroup$
    – egreg
    Mar 30 '14 at 13:29
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You have shown a necessary condition. You have not yet shown it is sufficient.

That is to say, it's certainly true that every unit in $M_n(\mathbb Q)$ has nonzero determinant. But is every element with nonzero determinant a unit?

Now, presumably you already know that a real-valued matrix with nonzero determinant has an inverse. Moreover, this inverse is unique. So, given a rational-valued matrix, if it has an inverse, it must be the same as the real-valued inverse. So your only job is to check if the real-valued inverse of a rational matrix is also rational. Any of the standard methods for computing inverses should help you do this.

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