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The answer is no. In fact, even if $R$ is a (possibly noncommutative) right self-injective ring with no nontrivial idempotents, then $R$ is a local ring, and the unique maximal (left or right) ideal …

Suppose $R,S,T$ are rings. Let $_RM_S$ and $_SN_T$ be bimodules. The tensor product $M\otimes_S N$ is naturally an $R$-$T$-bimodule. The "middle" $S$-structure is gone. In your case, $E$ is an $R$ …

Here is my, admittedly, ad hoc way of proving they are distinct. It comes from trying to make it concrete that the graded pieces have different sizes. First, you better assume that $n\geq 2$ as thes …

This is just an extended comment: If $Ra$ is essential in a direct summand of $R,$ then $a$ has this property. To see this, fix some idempotent $e\in R$ such that $Ra$ is essential in $Re.$ Then if …

Let $R$ be the subring of $\prod_{i=1}^{\infty}\mathbb{Q}$ of sequences which are eventually constant. This ring has the "obvious" maximal ideals $M_i$ of sequences which are zero in the $i$th coordi …

It is a well-known and open problem to determine whether there exists a rectangular cuboid where the distance from any corner to any other corner is an integer. Such a beast, if it exists, is called …

Let $R=\mathbb{Q}[x_{i,j}\,:\, 1\leq i,j\leq n]$. Let $M$ be the $n\times n$ matrix $(x_{i.j})$. Let $\chi(M)$ be the characteristic polynomial of $M$. Finally, let $I$ be the ideal of $R$ generate …

Several nice characterizations for these rings are worked out as Exercise 4.15 in Lam's book "Exercises in Classical Ring Theory." These include (for commutative rings): (A) $R$ is reduced and $K$-d …

For this type of problem, I find that free algebras modulo the relations you want often suffice. That is the case here. Take $R=\mathbb{Z}\langle r,s : sr=rs,r^2s=rs\rangle$. We check that $$ r^2(s …

My general rule of thumb is to ask myself the following questions: (1) If I pretend that my paper was written by someone else, and I hadn't seen it before, and I want to verify its accuracy, would …

Sketch: First, if $e$ is an idempotent in $A$, show that $B$ can be replaced with $B+Re$, and the hypotheses still holds. Use this to reduce to the case that $A$ and $B$ contain the same idempotents …

I know this question is old, and has an accepted answer, but this excellent paper by Manny Reyes gives some further thoughts about possible Spec's for noncommutative rings.

Let us say that an ideal $I\leq R$ has property $(\ast)$, by way of definition, if whenever an intersection of ideals is contained in $I$ then one of the ideals in the intersection is contained in $I$ …

Fix $I_2=\langle x^2\rangle \vartriangleleft \mathbb Z[x]$. Let $f=x$. For each polynomial $g\in \mathbb{Z}[x]$ we have $h(1-gf)\in 1+I$ when taking $h=1+gf$. Your question: "Which polynomials $g\in …

In January, Vesselin Dimitrov posted to the arXiv a preprint showing that Mochizuki's work, if correct, would be effective. While this doesn't validate Mochizuki's work it does do a few things: It …