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  "body": "\n## Statement\n\nLet $R$ be a [[ring|0B1261B5ED]] with identity, let $n \\ge 1$ be an integer, and let $M_n(R)$ be the [[ring of $n \\times n$ matrices|0FHVPRAE60]] over $R$.\nThen $R$ and $M_n(R)$ are [[Morita equivalent|RD5WKA1VK7]].\nAn equivalence from left $R$-[[modules|0NZ8JA25WK]] to left $M_n(R)$-modules sends $N$ to the module $N^n$ of columns of length $n$ over $N$.\nOn $N^n$ matrices act by matrix multiplication.\n\n## Remarks\n\nA quasi-inverse functor sends a left $M_n(R)$-module $L$ to $e_{11} L$, where $e_{11}$ is the matrix with $1$ in position $(1, 1)$ and $0$ elsewhere.\nIt is a left $R$-module through the scalar matrices.\nThe matrix units $e_{ij}$ give mutually inverse isomorphisms between $e_{11} L$ and $e_{jj} L$, and $L = e_{11} L \\oplus \\cdots \\oplus e_{nn} L$, which shows that the two functors are inverse up to natural isomorphism.\nIn the language of [[progenerators|ZQ9ABBKF1N]], $P = R^n$ is a progenerator of right $R$-modules and its endomorphism ring is $M_n(R)$.\nExample: a module over $M_2(\\Z)$ is the same as a pair of copies of one abelian group $N$, glued by the matrix units, and the module is $N^2$.\nSo the ideal theory, the Artinian and Noetherian conditions and the homological dimensions of $M_n(R)$ match those of $R$.\nBeing commutative is not shared: $M_2(R)$ is not commutative for $R \\ne 0$.\n\n## Explanations\n\n### L1 human\n\nThe ring of $n \\times n$ matrices over $R$ looks much bigger than $R$, but its modules carry no new information.\nEach module over the matrix ring is just $n$ stacked copies of a module over $R$.\nSo anything about modules over $R$ can be read off from modules over the matrix ring, and back.\n\n### L2 human\n\nThis is the first example of Morita equivalence and the model for the general theory.\nFor a division ring $D$, it shows that $M_n(D)$, like $D$, has exactly one simple module up to isomorphism, namely $D^n$.\nMorita's theorem says that every ring Morita equivalent to $R$ is of the form $e M_n(R) e$ for some $n$ and some idempotent $e$ with $M_n(R) e M_n(R) = M_n(R)$.\nSo matrix rings and full corners of them give all rings with the same module category as $R$.\n",
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