Abstract Nonsense

Crushing one theorem at a time

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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Graded Algebras and Graded Ideals

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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_nhomogenous 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.

[4] Lang, Serge. Algebra. Reading, MA: Addison-Wesley Pub., 1965. Print.

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

[6]  Conrad, Keith. “Exterior Powers.” Www.math.uconn.edu/~kconrad. Web. <http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf&gt;.

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May 10, 2012 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , ,

1 Comment »

  1. […] The Tensor Algebra and Exterior Algebra (Pt. III) Point of Post: This is a continuation of this post. […]

    Pingback by The Tensor Algebra and Exterior Algebra (Pt. III) « Abstract Nonsense | May 10, 2012 | Reply


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