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Math Help - Zm tensor Zn

  1. #1
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    Zm tensor Zn

    This is a standard problem I've seen, but I am new to tensors. If I could see the proof I think it would help me greatly to compute tensors in general.

    The problem statement here is from Hungerford GTM section IV.5.2

    Show Z_m \otimes Z_n \cong Z_c where c is gcd(m,n).

    (everything is regarded as an add. abelian group-- the tensor is over Z)

    A lead question is to show A \otimes Z_m \cong A/mA, which I think I managed with the map f sending a \otimes (n+mZ) \mapsto (an+mA) showing the kernel was 0 and was surjective. For what it's worth, as a bijection this implies elements of A \otimes Z_m are required to have the form simply a \otimes 1-- whether or not this will help with the problem stated above has not been something I can reconcile on my own.

    Any insights would be appreciated greatly.
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  2. #2
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    Quote Originally Posted by gosualite View Post
    This is a standard problem I've seen, but I am new to tensors. If I could see the proof I think it would help me greatly to compute tensors in general.

    The problem statement here is from Hungerford GTM section IV.5.2

    Show Z_m \otimes Z_n \cong Z_c where c is gcd(m,n).

    (everything is regarded as an add. abelian group-- the tensor is over Z)

    A lead question is to show A \otimes Z_m \cong A/mA, which I think I managed with the map f sending a \otimes (n+mZ) \mapsto (an+mA) showing the kernel was 0 and was surjective. For what it's worth, as a bijection this implies elements of A \otimes Z_m are required to have the form simply a \otimes 1-- whether or not this will help with the problem stated above has not been something I can reconcile on my own.

    Any insights would be appreciated greatly.
    To show A \otimes Z_m \cong A/mA, the first isomorphism theorem can be used.

    First define a surgective map f such that

    f:A \otimes Z_m \rightarrow A given by (a,x) \mapsto a , where a \in A , x \in Z_m.

    f is clearly surgective. Now, you need to find a kernel of f.

    Since ma \otimes x = a \otimes mx = a \otimes 0 = 0 , mA is a kernel of f.

    To show Z_m \otimes Z_n \cong Z_c, you can use the above result and replace A with Z_n.

    Since Z_m \otimes Z_n \cong Z_n/mZ_n, you remain to show that Z_n/mZ_n \cong Z_c. In a similar manner with the above, you can find a surgective map and its kernel and apply the first isomorphism theorem.
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  3. #3
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    Quote Originally Posted by aliceinwonderland View Post
    To show A \otimes Z_m \cong A/mA, the first isomorphism theorem can be used.

    First define a surgective map f such that

    f:A \otimes Z_m \rightarrow A given by (a,x) \mapsto a , where a \in A , x \in Z_m.

    f is clearly surgective. Now, you need to find a kernel of f.

    Since ma \otimes x = a \otimes mx = a \otimes 0 = 0 , mA is a kernel of f.
    that's not correct! here's how to do it: the map f_0: A \times \mathbb{Z}_m \longrightarrow A/mA defined by f_0(a,k)=ka + mA is \mathbb{Z} bilinear and thus it induces f: A \otimes_{\mathbb{Z}} \mathbb{Z}_m \longrightarrow A/mA, which maps a generator

    a \otimes k to ka + mA. now define g:A/mA \longrightarrow A \otimes_{\mathbb{Z}} \mathbb{Z}_m by g(a + mA)=a \otimes 1. it's clear that g is a \mathbb{Z} homomorphism. it's well-defined because if a + mA=0, then a=ma', and therefore

    a \otimes 1 = (ma') \otimes 1 = a' \otimes m=0. now we have fg(a+mA)=f(a \otimes 1)=a+mA. hence fg= \text{id}_{A/mA} and gf(a \otimes k)=g(ka + mA)=(ka) \otimes 1=a \otimes k, i.e. gf= \text{id}_{A \otimes \mathbb{Z}}. \ \Box

    ... you remain to show that Z_n/mZ_n \cong Z_c.
    we have: m(\mathbb{Z}/n\mathbb{Z})=(m\mathbb{Z}+n\mathbb{Z})/n \mathbb{Z}=c \mathbb{Z}/n \mathbb{Z}. thus \mathbb{Z}_n / m \mathbb{Z}_n = (\mathbb{Z}/n \mathbb{Z})/(c \mathbb{Z}/n \mathbb{Z}) \cong \mathbb{Z}/c \mathbb{Z}=\mathbb{Z}_c.


    Notation: you should use \mathbb{Z}/n or \mathbb{Z}/n \mathbb{Z} instead of \mathbb{Z}_n.

    Remark: a trivial generalization of this problem is that for any unitary commutative ring R, ideals I,J and R module A we have: \frac{R}{I} \otimes_R A \cong \frac{A}{IA}, and \frac{R}{I} \otimes_R \frac{R}{J} \cong \frac{R}{I+J}.


    by the way i'm not completely back yet! i'll try to help whenever i can (unfortunately not as often as before! don't blame me, blame my advisor!)
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  4. #4
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    Thank you both. It was this one fact:

    Quote Originally Posted by NonCommAlg View Post
    we have: m(\mathbb{Z}/n\mathbb{Z})=(m\mathbb{Z}+n\mathbb{Z})/n \mathbb{Z}=c \mathbb{Z}/n \mathbb{Z}.
    that gcd is the least linear combination, that had eluded my attempts to prove this. I appreciate both replies greatly.
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