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Problem of the Week #254 - Aug 01, 2017

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Euge

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Jun 20, 2014
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,896
No one answered this week's problem. You can read my solution below.


For all $k \ge 1$, the $k$th tensor algebra of $\Bbb Z/n\Bbb Z$ is $(\Bbb Z/n\Bbb Z)^{\otimes_{\Bbb Z} k}$, which is isomorphic to $\Bbb Z/n\Bbb Z$. So the tensor algebra $\mathcal{T}(\Bbb Z/n\Bbb Z)$ of $\Bbb Z/n\Bbb Z$ is isomorphic to $M=\Bbb Z \oplus \Bbb Z/n\Bbb Z \oplus \Bbb Z/n\Bbb Z \oplus \cdots$. The mapping $\Bbb Z[x] \to M$ mapping $p(x) = \sum_{i = 0}^m a_i x^i$ to $(a_0,a_1,\cdots,a_m,0,0,0,\ldots)$ is a surjective morphism with kernel $(nx)$, so $M \approx \Bbb Z[x]/(nx)$.
 
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