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## Hot answers tagged modules

3

You're right to be skeptical. As you've already noted at your other question, Artinian rings are ruled out from consideration since the Krull-Schmidt theorem can be used to prove they have the aforementioned property. Happily, I can demonstrate a ring without the property that is even von Neumann regular. I was inspired by a theorem in Goodearl's book von ...

3

$R=k[x]/(x^n)$ is an artinian local ring. Let $M$ be an $R$-module having a finite free resolution, or equivalently $\operatorname{pd}_RM<\infty$. Now we can apply the Auslander-Buchsbaum formula and get $\operatorname{pd}_RM=0$, that is, $M$ is projective, hence free.

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Since then $D^i$ would be a finite dimensional $\Bbb R$ division algebra, the Frobenius theorem says $D^i$ has to be $\Bbb R$, $\Bbb C$ or $\Bbb H$. But $\Bbb R$ and $\Bbb H$ are not $\Bbb C$ algebras because their centers are both $\Bbb R$, and so neither center can contain a copy of $\Bbb C$. So $D^i=\Bbb C$ for every $i$. That means each matrix ring has ...

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For non-finite-dimensional algebras certainly not. Take $A=k[X]$ and $V=k[X],$ the regular module. Then $\mathrm{rad} V=0,$ basically since $k[X]$ is a PID, but $V$ is not semisimple (because, for example, all simple $A$-modules are easily seen to be finitely dimensional. However, $A$ is not, and if $A$ was semisimple, it would be a sum of finitely many ...

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Yes, the trivial homomorphism. But apart from that, there is no natural homomorphism. Proof. Suppose that $\eta_{A,B} : A \otimes B \to A \times B$ is a natural homomorphism, meaning that $\eta_{A,B}$ defined for all $R$-modules $A,B$ and $\eta = (\eta_{A,B})_{A,B}$ is a natural transformation. Let $(r_1,r_2) := \eta_{R,R}(1 \otimes 1) \in R \times R$. If ...

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The Krull-Schmidt theorem literally says: If you decompose a module with finite composition length into a direct sum of indecomposables in two ways, then the two decompositions are the same length, and there's some permutation that pairs up factors from each decomposition into isomorphic pairs. So the idea is that you take the two decompositions ...

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As you say, take any non-free module, for example $\mathbb{Z}/2$ over $\mathbb{Z}$. But even for free modules the claims (i) and (ii) are wrong. (i) $\{2,3\}$ is a generating set of the $\mathbb{Z}$-module $\mathbb{Z}$, but it does not contain a basis. (ii) $\{2\}$ is a linearly independent subset of the $\mathbb{Z}$-module $\mathbb{Z}$, but it cannot be ...

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It is not the case that: $$\int |g(x)| \, dx = \left|\int g(x) \, dx\right|$$ Instead, to get rid of the absolute values, we use the fact that: $$|x - 1| = \begin{cases} x - 1 &\text{if } x \geq 1 \\ 1 - x &\text{if } x < 1 \end{cases}$$ We thus split the integral into two separate ones:  \int_{-5}^1 [(\tfrac{-1}{5}x + 7) - (2 + (1 - x))] \, ...

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Let $I = \langle(2,1+\sqrt{-5}),(1-\sqrt{-5},2)\rangle$. As you've probably seen, $2 \times 2 - (1+\sqrt{-5})(1-\sqrt{-5}) = -2 \neq 0$ so $I$ is a free $R$-module of rank $2$, and we even have an $R$-basis of $I$. More importantly this calculation also shows that $(0,2)$ and $(2,0)$ are in $I$ (just do the combination that cancels each component in the ...

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Any valuation domain $R$ satisfies the condition "P". If $I$ is a proper ideal of a valuation domain $R$, then $\sqrt I$ is a prime ideal. Let $a,b\in R$ such that $ab\in\sqrt I$. Then there is $n\ge 1$ such that $(ab)^n\in I$. If $a\mid b$ write $b=ax$ and note that $b^{2n}\in I$, so $b\in\sqrt I$. The conclusion: it is not possible to bound the ...

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By the Artin-Wedderburn theorem, a semisimple ring is a direct product of simple rings, i.e. $R=\prod_{i=1}^n R_i$. The maximal ideals are of the form $M_j=\prod_{i=1}^n I_i$ where $I_i=R_i$ for $i\neq j$ and $I_j=\{0\}$ for a particular $j$, $1\leq j\leq n$. Thus the only semisimple rings with a unique maximal ideal are the simple ones.

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In the unlikely case by semisimple you mean "has zero radical" (this is sometimes called "Jacobson semisimple" ) then yes, it's always true that $\mathrm{rad}(V/\mathrm{rad}(V))=\{0\}$. But if you mean "is a direct sum of simple modules" then (as PavelC has already pointed out) there are many counterexamples. The polynomial ring is nice, but I'll try to ...

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Hints: For (i): consider any ring $R$ with nonzero zero divisors. (In other words, the example you gave was fine. It sounds like you are forgetting to specify which ring the module is over, and that will be critical in deciding if it is free or not.) For (ii): consider any ring $R$ without zero divisors and with a nonprojective ideal $I$. $I$ is your ...

1

Associativity may or may not be an axiom, depending on your context. Requiring associativity results in a strictly smaller class of objects. While associative algebras are common, the study of nonassociative algebras is also full of important topics (Lie theory is a good example.) A basic example to keep in mind is $\Bbb R^3$ with the cross product. That ...

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Here's a more elementary explanation (from Exercise 19.7 in Eisenbud): Proposition: Let $(R,P)$ be an Artinian local ring, not a field. Then no submodule of $PR^n$ is free. Proof: $P$ is nilpotent in $R$, say $P^k = 0$, $P^{k-1} \ne 0$ for some $k > 1$, so any submodule of $PR^n$ is annihilated by $P^{k-1}$, hence has nontrivial annihilator, and thus ...

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You should look at a much more general statement: For any ideal $I\subset R$, and any homomorphism $f: N\to N'$ of $R$-modules, $f(IN) \subset IN'$. This statement is immediate from the definition of a module homomorphism. If $f$ is an isomorphism, then we can apply the same claim to $f^{-1}$, and see that $f$ restricts to an isomorphism $IN\to IN'$. ...

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