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Proof that Determinant is Multiplicative for Commutative Rings

  1. Mar 10, 2013 #1
    Is there a nice way to show that Det(AB)=Det(A)Det(B) where A and B are n x n matrices over a commutative ring?

    I'm hoping there is some analogue to the construction for vector spaces that defines the determinant in a natural way using alternating multilinear mappings...

    Otherwise would you just have to bash out the identity using the Leibniz formula for the determinant?
     
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  3. Mar 10, 2013 #2

    Ben Niehoff

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    ##\det## is just a homomorphism from the group of linear maps on ##V## to its representation on the top exterior power ##\Lambda^n V##. Taking some ##v_i \in V##, we have ##\det A : \Lambda^n V \to \Lambda^n V## given by

    [tex](\det A)(v_1 \wedge \ldots \wedge v_n) = A(v_1) \wedge \ldots \wedge A(v_n)[/tex]
    From here it is easy to show ##\det AB = \det A \det B## by using the usual composition of linear maps on each factor in the wedge product on the right.
     
    Last edited: Mar 10, 2013
  4. Mar 10, 2013 #3
    Thanks, Ben. To clarify, do you mean that we set ##V=R^n ## so that the group of linear maps on ##V## is the set of ##n##x##n## matrices?

    Also, do you know of a textbook that explains exterior algebra (from the module perspective) and its connections to the determinant from the ground up?
     
  5. Mar 10, 2013 #4

    Ben Niehoff

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    V can be any vector space at all.
     
  6. Mar 10, 2013 #5

    Bacle2

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    Maybe you can first show (not too hard) , that the determinant is multiplicative for elementary matrices Ei . Then write B as a product of elementary matrices
    and rearrange.
     
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