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As the question,

Is $\{0\}$ the only maximal ideal in $M_n(\mathbb{Q})$?

Intuitively it should be true. I think that if there is a bigger ideal, then it should be $M_n(\mathbb{Q})$ itself but I am not sure how to prove it. Thx

Edit: $M_n(\mathbb{Q})$ is the set of all $n\times n$ square matrices over $\mathbb{Q}$.

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  • $\begingroup$ What's $M_n(\Bbb{Q})$? The ring of $n \times n$ matrices over $\Bbb{Q}$? $\endgroup$ – user61527 Nov 22 '13 at 7:44
  • $\begingroup$ in what way is this intuitively true? $\endgroup$ – Mariano Suárez-Álvarez Nov 22 '13 at 7:46
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Let $R$ be a commutative ring with $1$. prove that $I$ is an ideal in $R$ implies $M_n(I)$ is an ideal in $M_n(R)$. actually converse is also true (Can you see what is the converse statement?). So we get a $1-1$ correspondence between the ideals of $R$ and the ideals of $M_n(R)$. $\mathbb{Q}$ is a field. So you have even better result then what you are claiming. Got it?

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If $M_n(\mathbb{Q})$ denotes the ring of matrices, the answer is "yes".

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