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If a not necessarily commutative $R$ has a unit $1$, consider the set $\mathscr I$ of all bilateral ideals of $R$ which do not contain $1$. It is more or less obvious that the union of a totally ordered subset of $I$ is again an element of $\mathscr I$, so $\mathscr I$ has maximal elements —this is just an application of Zorn's lemma, of course.


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Whereas Mariano Suarez-Alvarez's answer covers the abstract generalities, there has yet to be provided a "concrete example" of a unital, non-commutative ring $R$ having no (non-trivial) maximal ideals; by "non-trivial ideal" here I mean an bilateral ideal which is neither the zero ideal nor the entire ring $R$. It is clear from the thread of comments to ...


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All direct products of projective right $R$ modules are projective iff $R$ is right perfect and left coherent. I believe it appears in Lam's Lectures on rings and modules or Anderson & Fuller's book (or both). If you know the theorem that all products of flat right modules are flat iff a ring is left coherent, and that the flat and projective modules ...


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Yes, your reasoning is valid that if $M_n(I)\subsetneq M_n(K) \subseteq M_n(R)$, then $I\subsetneq K\subseteq R$, and so if $I$ is maximal, $M_n(I)$ is maximal. As for the title question, a maximal ideal in a noncommutative ring is always prime, unless you mean to apply the commutative definition of prime ideals to noncommutative rings. The general ...


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For any field $F$, $R=\prod_{i=1}^\infty F$. It has $J(R)=\{0\}$ (since it is von Neumann regular) and is clearly not Artinian. Then if you want a version with a nonzero radical, you can take the 2 by 2 upper triangular matrix ring over this ring. Another example would be to take the subring of this matrix ring of elements with constant diagonal.


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Recall that any $f \in C^\infty(\mathbb{T}^n)$ can be uniquely written as a convergent Fourier series $$ f = \sum_{\mathbb{k} \in \mathbb{Z}^n} f_{\mathbb{k}} U_{\mathbb{k}}, \quad f_{\mathbb{k}} := \int_{\mathbb{T}^n} e^{-2\pi i \langle \mathbb{k},t \rangle}f(t)\,dt, $$ where for each $\mathbb{k} \in \mathbb{Z}^n$, $$ \forall t \in \mathbb{T}^n, \quad ...


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Here are some examples coming from linear algebra. Let $R$ be a ring ( commutative or not). Let $n$ and $1\le k \le n-1$ natural numbers. Consider the ring $N_{n,k}(R)$ of $n\times n$ matrices with elements in $R$ that have zero entries below the diagonal $\{ (i,j)\ | \ j-i =k\}$. Then $N_{n,k}(R)$ has nilpotency index $\le (n-k+1)$, and if $R$ has a ...



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