Is the Tensor Algebra of $\Bbb Z/n\Bbb Z$ Isomorphic to $\Bbb Z[x]/(nx)$?

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  • Thread starter Thread starter Euge
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    2017
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SUMMARY

The tensor algebra of $\Bbb Z/n\Bbb Z$ is definitively isomorphic to $\Bbb Z[x]/(nx)$. This conclusion is drawn from the structural properties of both algebraic constructs, where the tensor algebra encapsulates the elements of $\Bbb Z/n\Bbb Z$ and their interactions, while $\Bbb Z[x]/(nx)$ represents polynomials modulo the ideal generated by \(nx\). The isomorphism indicates a deep relationship between these two algebraic frameworks, confirming their equivalence in terms of algebraic structure.

PREREQUISITES
  • Understanding of tensor algebras
  • Familiarity with modular arithmetic in $\Bbb Z/n\Bbb Z$
  • Knowledge of polynomial rings, specifically $\Bbb Z[x]$
  • Concept of isomorphism in algebraic structures
NEXT STEPS
  • Study the properties of tensor algebras in detail
  • Explore the structure of polynomial rings and their quotients
  • Investigate isomorphisms in algebraic contexts
  • Review examples of modular arithmetic applications in algebra
USEFUL FOR

Mathematicians, algebra students, and educators interested in advanced algebraic structures and their interrelations will benefit from this discussion.

Euge
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Here is this week's POTW:

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Show that the tensor algebra of $\Bbb Z/n\Bbb Z$ is isomorphic to $\Bbb Z[x]/(nx)$.

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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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