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6

$R^{\omega}\cong R^{\omega}\times R^{\omega}$ for any ring $R$, commutative or not. This does not contradict the fact that a commutative ring always has the invariant basis number property because while we can construct an "artificial" isomorphism of rings, it is not an isomorphism of $A$-modules, as the action of $A$ is different. So $A\times A$ really is a ...


4

If you want $\operatorname{Hom}_R(P,R)\simeq P$ for $P$ a finitely generated projective $R$-module, then forget it. $\bullet$ If $R$ is an integral domain, and $I\subset R$ is a non-zero ideal, then $\operatorname{Hom}_R(I,R)\simeq I^{-1}$, where $I^{-1}=\{x\in Q(R):xI\subseteq R\}$. (Here $Q(R)$ stands for the field of fractions of $R$.) $\bullet$ If ...


3

Let $R\subset\mathbb Z_2^{\mathbb N}$ be the set of sequences with elements in $\mathbb Z_2$ which are constant from a rank on. Then $R$ is a boolean ring. If $R\simeq\prod_{i\in I}R_i$ with $R_i$ indecomposable, then $R_i$ is isomorphic to a subring of $R$, hence $R_i$ is boolean. Moreover, since $R_i$ is indecomposable we must have $R_i\simeq\mathbb Z_2$. ...


2

If I understand your question, the answer is yes - given any set $X$ with more than one element, let $*$ be left projection: $x*y=x$. If we ask for a noncommutative operation satisfying some other restrictions, of course, then the answer can be quite surprising. My favorite example: there are no noncommutative division rings of finite cardinality! That ...


2

$xR\cap yR=\{0\}$, so your ideal is the direct sum of $xR$ and $yR$, both of which are isomorphic to $R$ as right ideals, so its endomorphism algebra is isomorphic to $M_2(R)$.


1

This follows from the long exact sequence for Tor. If $B$ and $C$ are, say, flat left $R$-modules, and $M$ is a right $R$-module, then there is an exact sequence $$\def\Tor{\operatorname{Tor}}\Tor^R_2(M,C)\to\Tor^R_1(M,A)\to\Tor^R_1(M,B)$$ and $\Tor^R_2(M,C)$ and $\Tor^R_1(M,B)$ are zero by hypothesis. It follows that $\Tor^R_1(M,A)=0$ for all right ...


1

Let k be a field, let A be the enveloping algebra of the three-dimensional Heisenberg Lie algebra g, and let I be the ideal of A generated by any nonzero element of the center of g. This satisfies all your conditions.


1

In the book Rings with generalized identities by Beidar, Martindale and Mikhalev, Section 1.4 (from page 24 onwards) you can find the definition from commutative diagrams, the characterization with proof (which is the one given by Chindea Filip in his answer) and several properties. The main purpose of the coproduct in this book is to serve as the "home" ...


1

Pretty much the same proof as in the commutative case shows that flat implies torsion free for modules over a noncommutative ring $R$: Let $a\in R$, not a right zero divisor. Then the map $\mu_a:R\to R$ given by $r\mapsto ra$ is an injective left $R$-module homomorphism. If $M$ is a flat right $R$-module, then $\operatorname{id}_M\otimes\mu_a:M\otimes_RR\to ...


1

According to a comment by Tom Goodwillie in this MO question, there exist rings $R,S$ such that $M_2(R) \cong M_4(S) = M_2(M_2(S))$ yet $R \ncong M_2(S)$.



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