Proof of (Z/mZ)⊗Z(Z/nZ)≅Z/gcd(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}

I’ve just started to learn about the tensor product and I want to show: (\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}.

Can you tell me if my proof is right:

\mathbb{Z}/m\mathbb{Z} and \mathbb{Z} / n \mathbb{Z} are both finite free \mathbb{Z}-modules with the basis consisting of one single element \{ 1 \}. So (\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) has the basis \{ 1 \otimes 1 \}.

Therefore, any element in (\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) is of the form (ab) 1 \otimes 1 and any element in \mathbb{Z}/ \gcd(m,n)\mathbb{Z} is of the form k 1 = k where k \in \{ 0, \dots , \gcd(n,m) \}.

I would like to construct an isomorphism that maps ab to some k. Let this map be ab (1 \otimes 1) \mapsto ab \bmod \gcd(n,m).

This is a homomorphism between modules: it maps 0 to 0 because it maps the empty sum to the empty sum. It also fulfills f(a + b) = f(a) + f(b) because there is only one element, a = 1.

It is surjective. So all I need to show is that it is injective. But that is clear too because if ab \equiv 0 \bmod \gcd(m,n) then both a \equiv 0 \bmod n and b \equiv 0 \bmod m so the kernel is trivial.

Many thanks for your help!!

Answer

The better way to define a homomorphism from \mathbb{Z}/m\mathbb{Z}\otimes \mathbb{Z}/n\mathbb{Z} to \mathbb{Z}/\gcd(m,n)\mathbb{Z} is via the universal property.

Note that the map \mathbb{Z}/m\mathbb{Z}\times \mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}/\gcd(m,n)\mathbb{Z} defined by (x+m\mathbb{Z},y+n\mathbb{Z})\mapsto xy+\gcd(m,n)\mathbb{Z} is well-defined and also bi-linear, thus by the universal property of tensor product, there is a linear map f:\mathbb{Z}/m\mathbb{Z}\otimes \mathbb{Z}/n\mathbb{Z}\to \mathbb{Z}/\gcd(m,n)\mathbb{Z} such that f(x+m\mathbb{Z}\otimes y+n\mathbb{Z})=xy+\gcd(m,n)\mathbb{Z}. Verify that the linear map g:\mathbb{Z}/\gcd(m,n)\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}\otimes\mathbb{Z}/n\mathbb{Z} defined by
g(z+\gcd(m,n)\mathbb{Z})=(z+m\mathbb{Z})\otimes (1+n\mathbb{Z})
is well-defined, and we also have g\circ f=1, f\circ g=1, thus f is isomprhism. To see g is well-defined, you may use the equality that \gcd(m,n)=am+bn for some integers a,b\in\mathbb{Z}.

Attribution
Source : Link , Question Author : Rudy the Reindeer , Answer Author : Xiang Yu

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