Abstract Nonsense

Crushing one theorem at a time

The Tensor Algebra and Exterior Algebra (Pt. V)


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

Exterior Powers of Maps

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Not surprisingly, the construction \Lambda is functorial. In other words, we should not only be able to be able to take exterior powers of modules but also take exterior powers of maps. So, if f:M\to N is an R-map we should be able to construct some R-map \Lambda^k(f) mapping \Lambda^k(M)\to\Lambda^k(N) which respects compositions and identities.

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The idea of this map is simple, but powerful. Namely, we know that from our map f:M\to N we get a map \mathcal{T}^k(f):\mathcal{T}^k(M)\to\mathcal{T}^k(N) such that \mathcal{T}(v_1\otimes\cdots\otimes v_k)=f(v_1)\otimes\cdots\otimes f(v_k). Now, considering the composition \wedge\circ\mathcal{T}^k(f)\circ\otimes where \wedge is the universal k-linear map M^k\to\mathcal{T}^k(M) and \wedge:\mathcal{T}^k(N)\to\Lambda^k(N) is the quotient map, we get an alternating k-linear map M^k\to\Lambda^k(N) which satisfies (v_1,\cdots,v_k)\mapsto f(v_1)\wedge\cdots\wedge f(v_k). Thus, by the universal characterization of the exterior power we are afforded a map

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\Lambda^k(f):\Lambda^k(M)\to\Lambda^k(N):v_1\wedge\cdots\wedge v_k\mapsto f(v_1)\wedge\cdots\wedge f(v_k)

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as we would expect such a map to satisfy.  I think it’s fairly obvious that \Lambda^k(g\circ f)=\Lambda^k(g)\circ\Lambda^k(f) and that \Lambda^k(\text{id}_M)=1_{\Lambda^k(M)}. Thus, we may conclude that

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Theorem: The association of f:M\to N to \Lambda^k(f):\Lambda^k(M)\to\Lambda^k(N) is a functor R\text{-}\mathbf{Mod}\to R\text{-}\mathbf{Mod}.

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In fact, it’s pretty easy to see that:

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Theorem: Let f:M\to N be a surjective R-map. Then, \Lambda^k(f):\Lambda^k(M)\to\Lambda^k(N) is surjective.

Proof: Let S'\subseteq N be a generating set for N. We know then that there is some S\subseteq M such that f(S)=S'. But, it’s trivial that \left\{v_{i_1}\wedge\cdots\wedge v_{i_k}:v_{i_1},\cdots,v_{i_k}\in S'\right\}=T' is a generating set for \Lambda^k(N), and moreover that if T=\left\{v_{i_1}\wedge\cdots\wedge v_{i_k}:v_{i_1}\cdots,v_{i_k}\in S\right\} then \Lambda^k(f)(T)=T'. Thus, \Lambda^k(f) hits a generating set of \Lambda^k(N) and thus \Lambda^k(f) is surjective. \blacksquare

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What is also true is that:

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Theorem: Let f:M\to N be an injective R-map with \text{im }f a direct summand of N. Then, \Lambda^k(f):\Lambda^k(M)\to\Lambda^k(N) is injective. Moreover, \text{im }\Lambda^k(f) is a direct summand of N.

Proof: Since \text{im }f is a direct summand we have a split short exact sequence

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0\to M\xrightarrow{f}\underbrace{\text{im }f\oplus S}_{N}\xrightarrow{\pi}S\to0

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where S is the complement of \text{im }f in N. We know that \Lambda^k(f) is injective from the simple fact that functors take retractions to retractinos. Or, said differently, since our sequence splits we get a retraction map s:N\to M such that s\circ f=\text{id}_M. Then, \Lambda^k(s)\circ\Lambda^k(f)=\text{id}_{\Lambda^k(M)} and thus \Lambda^k(f) is injective. Now, to prove that we actually have that  \text{im }\Lambda^k(f) is a direct summand of \Lambda^k(N) we merely note that we get the exact sequence

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0\to\Lambda^k(M)\xrightarrow{\Lambda^k(f)}\Lambda^k(N)\xrightarrow{\pi}\Lambda^k(N)/\text{im }\Lambda^k(M)\to 0

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and since we have a retraction for \Lambda^k(f) the conclusion follows from the splitting lemma. \blacksquare

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Let us define a ring R to be semisimple if every submodule of a left R-module is a direct summand. For example division rings, finite dimensional matrix rings over division rings, and finite products of finite dimensional matrix rings over division rings are semisimple. It is the content of the famed Wedderburn-Artin theorem that these precisely the semisimple rings (i.e. that the rings described are semisimple, and they are the only semisimple rings). Anyways, we are getting too far off course. The above allows us to conclude the following:

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Theorem: Let R be a semisimple ring and f:M\to N a map of left R-modules. Then, \Lambda^k(f) is surjective (resp. injective) if f is surjective (resp. injective). 

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Note that this does NOT say that if F is a field then \Lambda^k:\mathbf{Vect}_F\to\mathbf{Vect}_F is exact! The reason it should be “obvious” that our functor isn’t even right exact is a simple fact about binomial coefficients. Indeed, functors that are right exact commute with quotients, and so if \Lambda^k was right exact then \Lambda^k(V/W)\cong \Lambda^k(V)/\Lambda^k(W) where V,W are F-vector spaces. But, counting dimensions both sides (saying that \dim V=n and \dim W=m) this says that

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\displaystyle {n-m \choose k}={n\choose k}-{m\choose k}

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a decidedly untrue fact.

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Now, back to our main discussion about facts concerning the exterior power of maps. The above told us that we need, in general, the image of our map f:M\to N to be a direct summand to say that the induced exterior power map is injective. That said, if we assume that both M,N are free then this assumption is not necessary. Indeed:

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Theorem: Let f:M\to N be an R-map with M,N free. Then, \Lambda^k(f):\Lambda^k(M)\to\Lambda^k(N) is injective.

Proof: Let \Psi_M,\Psi_N be the embeddings \Lambda^k(M)\hookrightarrow \mathcal{T}^k(M) and \Lambda^k(N)\hookrightarrow\mathcal{T}^k(N) as we constructed earlier. Then, we have the obvious relation \Psi_N\circ\Lambda^k(f)=\mathcal{T}^k(f)\circ\Psi_M. That said, since free modules are flat we know that \mathcal{T}^k(f) is injective since \mathcal{T}^k(f) is the composition of the maps

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(f\otimes 1\otimes\cdots\otimes 1)\circ (1\otimes f\otimes\cdots\otimes 1)\circ\cdots\circ(1\otimes 1\otimes\cdots\otimes f)

all of which are injective. \blacksquare

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

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


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