# Tag Info

## Hot answers tagged noncommutative-algebra

3

The most modern term for this is $R$ satisfies the $SI$ condition This is seen in papers by Greg Marks, which I have found to be the most thoughtful and comprehensive recent papers discussing these things. For example , see A taxonomy of 2-primal rings. Earlier works used the following terms: $R$ is zero-insertive (zi) $R$ satisfies the ...

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You almost answer your own question, since $M_2(\mathbb{Z})$ is a PI-ring. See https://en.wikipedia.org/wiki/Amitsur%E2%80%93Levitzki_theorem

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If your nil ideals are two-sided proceed like this: Let $I,J$ be two-sided nil ideals of a Ring $R$. We want to show that $I+J$ is also nil. To do this consider $R/J$ and $(I+J)/J$. You can see rather easily that $(I+J)/J$ is a two-sided nil ideal of $R/J$: Let $[x] \in (I+J)/J$, then there exists a representative $a$ of $[x]$ that lies in $I$. Since then ...

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Note that since $A$ is not only semi-simple but also simple, there is a unique simple $A$-module. Call it $I$. Then any finitely generated $A$-module is isomorphic to $I^n$ for some $n$ (in particular they are all projective). Then $\operatorname{End}_A(P) \simeq \operatorname{End}_A(I^n) = M_n(\operatorname{End}_A(I))$ (the last equality is valid for any ...

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No, it's not finitely generated, and so in particular it cannot be Noetherian. The point is that the submodule generated by any finite set of elements consists of elements where the denominator contains only finitely many possible irreducible factors in $F[x]$ (namely, the irreducible factors in the denominators of the generators), and there are infinitely ...

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There is a tensor-hom adjunction for the tensor product of algebras, but it exists at the level of the Morita 2-category, rather than the 1-category of algebras. Namely, the Morita 2-category has objects $k$-algebras, and the category of morphisms $A \to B$ is the category $\text{Mod}(A^{op} \otimes B)$ of $(A, B)$-bimodules, where composition is given by ...

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A right ideal of a ring with identity which is a direct summand is always idempotent: if it is $eR$, then $e=e^2\in (eR)^2$, so $eR\subseteq (eR)^2$, and $(eR)^2\subseteq eR$ trivially. (Are you really asking about PI rings without identities?)

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1) $S$ is also a left ideal, in fact, if $r\in R$ and $I$ is a right ideal of $R$ so is $rI$ (Evident!). Let $x\in I$, we first observe that $(xr)^k=0$ for some integer $k$. Therefore, we have $(rx)^{k+1}=0$, showing that $rI$ is a nil right ideal of $R$. Now, if $y\in S$ we can write $y=y_1+...y_n$, where each $y_j$ belongs to some nil right ideal $I_j$ ...

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First off, $rI$ is certainly some right ideal, and we hope it is contained in $I$ for al $r$ so that it's a left ideal. So you're looking for a reason that $rI\neq R$. If to the contrary $rI=R$, then $ri=1$ for some $i$. Now $ir$ is an idempotent element, so $R=irR\oplus(1-ir)R$ (just as you have for any idempotent element.) Considering there is only one ...

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Is the arbitrary sum of nilpotent right ideals nilpotent? The counterexample writes itself: Take $R=F[x_1,x_2,\ldots]/(\{x_n^n\mid n\in \Bbb N\})$ for a field $F$. While the individual ideals $(x_i)$ are nilpotent, their sum is merely a nil ideal. Is the arbitrary sum of nil right ideals nil? This is not even known for finite sums. The statement ...

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