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Let $R$ be a ring with identity. A matrix $A=[a_{ij}] \in M_n(R)$ is called a J-matrix if for any $i$, $a_{ii} \not \in J(R)$ but for any $i \not = j$, $a_{ij} \in J(R)$. Now suppose that every J-matrix is non-singular. Is it true that the ground ring $R$ is local ? (Note that the converse is clearly true)

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    $\begingroup$ What is $J$? The radical? $\endgroup$ – abx Nov 20 '13 at 7:54
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I might be interpreting your question wrong, but if not then I believe the answer is yes. This boils down to the fact that a ring is local if and only if the Jacobson radical $J(R)$ is maximal if and only if every element $x \notin J(R)$ is a unit. This characterization of local comes from the fact that the Jacobson radical is the intersection of all maximal left ideals. Then the first if and only if statement is clear and the second follows from the fact that any nonunit is contained in a maximal ideal.

Now, for the proof of your statement. Suppose for some $n$, every $J$-matrix in $M_n(R)$ is nonsingular. Then for any $a \notin J(R)$, consider the matrix $A = \text{diag}(a,1,\ldots,1) \in M_n(R)$. By construction, $A$ is a $J$-matrix so $A$ is nonsingular. However, $A$ has an inverse if and only if $a$ is invertible. Thus, any $a \notin J(R)$ is invertible so $J(R)$ is maximal and $R$ is a local ring.

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