# Questions tagged [noncommutative-algebra]

For questions about rings which are not necessarily commutative and modules over such rings.

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### Subalgebra of $M_n(\mathbb{C})$ generated by two elements (along with unity)

Let $M_n(\mathbb{C})$ denote the algebra of $n\times n$ matrices over the field of complex numbers $\mathbb{C}$. Let $h_1,h_2\in M_n(\mathbb{C})$ be two Hermitian matrices. Suppose that $h_1,h_2$ are "...
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### what if matrix multiplications were commutative [duplicate]

Would that make things easier in any science if matrix multiplication were commutative? I mean are researchers working to find more exceptions to the general rule of matrix multiplication being not ...
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### Is the quotient ring $R/I$ always Dedekind-finite, where $I$ is the two-sided ideal generated by all elements of the form $xy-1$ where $yx=1$?

I wonder whether the quotient ring $R/I$ is always Dedekind-finite for any ring $R$ if $I$ is the two-sided ideal of $R$ generated by all elements of the form $xy-1$ where $yx=1$. One might think ...
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### In a non-commutative ring (possibly without identity) with no nontrivial automorphisms, do nilpotent elements form an ideal?

In an old exam appeared this statement: True/False: "Let $R$ be a ring with the property that the unique ring automorphism is the identity. Then the set of all nilpotent elements form an ideal". I'...
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### Is there any example of a simple Abelian ring which is not domain?

A ring $R$ is called: simple if it has no two-sided ideal; a domain if it has no zero divisor; abelian if each idempotent of $R$ is central. Is there any example of a simple abelian ring which is ...
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### Rotational Symmetry Groups of Tetrahedron and Hexagonal Plate

Firstly, I am confused on how to show clearly how the Tetrahedron ($T$) and Hexagonal Plate ($H$) are not abelian (commutative). And as a follow on, how would I find $i)$ a pair of elements of $H$, ...
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### $(x,yz,t) = (x,y,t)z + (x,z,t)y$ holds in every Jordan algebra.

The identity $$(x,yz,t) = (x,y,t)z + (x,z,t)y$$ holds in every Jordan algebra. Remember that a Jordan algebra satisfies $xy=yx$ and $(x^2,y,x) = 0$ for all $x,y$. Here $(a,b,c) = (ab)c - a(bc)$ is ...