Abstract Nonsense

The Tensor Algebra and the Exterior Algebra (Pt. II)

Point of Post: This is a continuation of this post.

Let us now compute an example.

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Probably the simplest example is to compute $\mathcal{T}(\mathbb{Z}/n\mathbb{Z})$ when we think about $\mathbb{Z}/n\mathbb{Z}$ as an abelian group. Indeed, we claim that, as $\mathbb{Z}$-algebras (which is the same thing as rings),  $\mathcal{T}(\mathbb{Z}/n\mathbb{Z})\cong \mathbb{Z}[x]/(nx)$. Indeed, define $f:\mathbb{Z}[x]\to\mathcal{T}(\mathbb{Z}/n\mathbb{Z})$ by

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$\displaystyle \sum_i a_i x^i\mapsto \sum_i \underbrace{a_i\otimes 1\otimes\cdots\otimes 1}_{\text{in }\mathcal{T}^i(\mathbb{Z}/n\mathbb{Z})}$

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Now, the important thing to note is that since we can move elements of $\mathbb{Z}$ in between the tensors that

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\begin{aligned}f(a_i x^i b_jx^j) &=(a_i b_j)\otimes 1\otimes \cdots\otimes 1 \\ &=a_\otimes 1\otimes\cdots\otimes\underbrace{b_j}_{i+1}\otimes 1\otimes\cdots\otimes 1\\ &= (a_i\otimes 1\otimes\cdots\otimes 1)\otimes(b_j\otimes 1\otimes\cdots\otimes 1)\\ &= f(a_i x^i)f(b_j x^j)\end{aligned}

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and so it easily follows that $f$ is actually a ring map. That said, we know that $(\mathbb{Z}/n\mathbb{Z})^{\otimes k}\cong\mathbb{Z}/n\mathbb{Z}$ and thus, clearly, as groups $\mathcal{T}(\mathbb{Z}/n\mathbb{Z})\cong\mathbb{Z}\oplus (\mathbb{Z}/n\mathbb{Z})^{\oplus\mathbb{N}}$ via the isomorphism that, on simple tensors, is just

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$a_1\otimes \cdots\otimes a_i\mapsto \underbrace{a_1\cdots a_i}_{i^{\text{th}}\text{ coordinate}}$

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So, it clearly suffices to find $\ker f$ to find

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$\ker\left(\mathbb{Z}[x]\to\mathcal{T}(\mathbb{Z}/n\mathbb{Z})\to \mathbb{Z}\oplus(\mathbb{Z}/n\mathbb{Z})^{\oplus\mathbb{N}}\right)$

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But, following this map, it’s clear that $\displaystyle \sum a_i x^i$ is in the kernel if and only if $a_0=0\in\mathbb{Z}$ and $a_i=0\in\mathbb{Z}/n\mathbb{Z}$ for $i\geqslant 1$. So, this just tells us that $a_0=0$ and $n\mid a_i$ for $i\geqslant 1$ which tells us merely that $\displaystyle \sum a_ix^i$ is in $(nx)$. Thus, by the first ring isomorphism theorem we may conclude that $\mathcal{T}(\mathbb{Z}/n\mathbb{Z})\cong\mathbb{Z}[x]/(nx)$ as desired.

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Note that this worked for $n=0$, i.e. $\mathbb{Z}$, as well, so that $\mathcal{T}(\mathbb{Z})\cong\mathbb{Z}[x]$.

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In fact, I don’t think it’s at all hard to see that $\mathcal{T}(R)\cong R[x]$ (where the left tensor algebra is taken with respect to thinking of $R$ as the left regular $R$-module).

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We now discuss the notion of graded algebras, homgenous ideals ideals,  graded quotients, and graded morphisms and how this applies to our study of the tensor algebra of a module.

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Let $R$ be a ring. We say that $R$ is graded if there is distinguished decomposition $\displaystyle R=\bigoplus_{n\geqslant 0} R_n$ (where each $R_n$ is a subgroup of $R$) with the property that $R_i\cdot R_j\subseteq R_{i+j}$ (where by definition $XY=\left\{xy:x\in X\text{ and }y\in Y\right\}$–i.e. the image of the multiplication map restricted to $X\times Y$). We call $R_n$ the homogenous part of degree $n$ and obviously call $r\in R_n$homogenous element of degree $n$.

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graded $R$-algebra $A$ is a graded ring such that the homogenous parts are $R$-submodules.

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The classic example of a graded $R$-algebra is the polynomial ring $R[x]$ with the elements of $n$ being the monomials of polynomial degree $n$ along with $0$. This is clearly a graded ring for if $\deg p(x)=n$ and $\deg q(x)=m$ are monomials then $\deg(p(x)q(x))=m+n$ or $p(x)q(x)=0$.

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homogenous ideal of a ring $R$ is an ideal $\mathfrak{a}$ such that $\displaystyle \mathfrak{a}=\bigoplus\mathfrak{a}_n$ with $\mathfrak{a}_n=\mathfrak{a}\cap R_n$.

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Now, if $\mathfrak{a}$ is a homogenous ideal of the ring $R$ then $R/\mathfrak{a}$ is naturally a graded ring with homogenous part of degree $n$ just being $R_n/\mathfrak{a}$. I leave this to you to check. If $A$ is a graded $R$-algebra then this ideal $\mathfrak{a}$ will also be a submodule and so the quotient will also be a graded $R$-module with the already specified grading.

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Now, suppose that $A$ and $B$ are graded $R$-algebras. An $R$-algebra map $A\to B$ is called graded if it preserves homogeneous degrees–i.e. if $f(A_n)\subseteq B_n$.

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For example, if $A$ is a graded $R$-algebra and $\mathfrak{a}$ a graded ideal then the quotient map $\pi:A\to A/\mathfrak{a}$ is a graded $R$-algebra map.

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Now, the reason we are discussing this in the tensor algebra post should be obvious–the tensor algebra is a graded $R$-algebra. Indeed, the homogeneous part of degree $i$ in $\mathcal{T}(M)$ is just $\mathcal{T}^i(M)$–these are all submodules and they satisfy $\mathcal{T}^i(M)\mathcal{T}^j(M)\subseteq\mathcal{T}^{i+j}(M)$.

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.

[5] Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.