Proof of $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}$

Question

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 1

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 an isomorphism. To see $g$ is well-defined, you may use the equality that $\gcd(m,n)=am+bn$ for some integers $a,b\in\mathbb{Z}$.

Answer 2

The part that is missing is pretty much the essence of the following (incomplete) alternative proof.

Determine the kernel of $$ \begin{array}{rlrl} g: & \mathbb{Z} & \rightarrow & (\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \\ & z & \to & z (1 \otimes 1) \end{array} $$ That is: When is it true that $z (1 \otimes 1)$ is null? Since it is true for $z \in m\mathbb{Z} \cup n\mathbb{Z}$, then you know that it is true for the ideal generated by it: $\langle \mathrm{gcd}(m,n)\rangle \subset \mathrm{ker}(g)$.

You know that the map is surjective because $1 \otimes 1$ is a generator. If you show that $\mathrm{ker}(g) \subset \langle \mathrm{gcd}(m,n) \rangle$, you will have the isomorphism you claim. This is the part you are missing. It is equivalent to showing that your $f$ is well-defined.

So, the conclusion is that it is not right.

Answer 3

Let $M$ be a $R$-module and let $I$ be an ideal of $R$. Then we have $\frac RI\otimes_RM\cong\frac M{IM}$.

First proof: consider the exact sequence $0\to I\to R\to\frac RI\to0$. Tensoring with $M$ and using right-exactness of the tensor product we get $I\otimes_RM\to R\otimes_RM\to\frac RI\otimes_RM\to0$. We know that $R\otimes_RM\cong M$ via $r\otimes m\mapsto rm$; under this isomorphism the image of $I\otimes_RM\to R\otimes_RM$ is precisely $IM$, and since this image is precisely the kernel of $R\otimes_RM\to\frac RI\otimes_RM$, the claim follows from the isomorphism theorem.

Second proof: the mapping $\frac RI\times M\to\frac M{IM}$ given by $(r+I,m)\mapsto rm+IM$ is well-defined and $R$-bilinear, so it induces a $R$-linear mapping $\frac RI\otimes_RM\to\frac M{IM}$, which is obviously surjective, and the inverse mapping is given by $m+IM\mapsto(1+I)\otimes m$; this mapping is well-defined because if $m\in IM$, say $m=\sum_k i_km_k$, with $i_k\in I, m_k\in M$, then by $R$-bilinearity of $\otimes$ we have $(1+I)\otimes m=\sum_k\bigl[i_k(1+I)\bigr]\otimes m_k=\sum_k0_{R/I}\otimes m_k=0$.

In general, if $N$ is a $R$-submodule of a module $M$, then

$$I\frac MN=\frac{IM+N}N\,,$$

so

$$m\mathbb Z\,\frac{\mathbb Z}{n\mathbb Z}=\frac{(m\mathbb Z)\mathbb Z+n\mathbb Z}{n\mathbb Z}=\frac{m\mathbb Z+n\mathbb Z}{n\mathbb Z}=\frac{\gcd(m,n)\mathbb Z}{n\mathbb Z}\,.$$

Finally, applying the first result we get

$$\frac{\mathbb Z}{m\mathbb Z}\otimes_{\mathbb Z}\frac{\mathbb Z}{n\mathbb Z}\cong\frac{\frac{\mathbb Z}{n\mathbb Z}}{m\mathbb Z\,\frac{\mathbb Z}{n\mathbb Z}}=\frac{\frac{\mathbb Z}{n\mathbb Z}}{\,\frac{\gcd(m,n)\mathbb Z}{n\mathbb Z}\,}\cong\frac{\mathbb Z}{\gcd(m,n)\mathbb Z}\,.$$

ADDENDUM

The same reasoning shows that for any ideals $I,J$ in $R$ we have $\frac RI\otimes_R\frac RJ\cong\frac R{I+J}$.

Answer 4

Claim: $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}$

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Proof:

We show the claim by showing that $(\mathbb{Z}/ \gcd(m,n)\mathbb{Z}, b)$, where $$b: \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z} / n \mathbb{Z} \to \mathbb{Z}/ \gcd(m,n)\mathbb{Z}$$ is the map $(a,b) \mapsto ab \bmod \mathrm{gcd} (n,m)$ satisfies the universal property of $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) $, that is, for any $\mathbb Z$-module $P$ and any bilinear map $b^\prime : \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z} / n \mathbb{Z} \to P$ there exists a unique linear map $l$ such that $l \circ b = b^\prime$.

Define $l: (a \bmod \mathrm{gcd} (n,m)) \mapsto b^\prime (1,a) $. Linearity of $l$ directly follows from bilinearity of $b^\prime$. For $(a,b) \in \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z} / n \mathbb{Z}$ we have $l(b(a,b)) = l(ab) = b^\prime (1,ab) = b^\prime (a,b) $ so that $l$ indeed makes the diagram commute.

To finish the proof we check that $l$ is unique. To this end let $l^\prime$ be such that $l^\prime \circ b = b^\prime$. Then $l(a) = l(b(1,a)) = b^\prime (1,a) = l^\prime (b(1,a)) = l^\prime (a)$, hence the claim follows.

Answer 5

The only thing that you are missing is to show that your map is well-defined. Indeed, we implicitly make a choice of representatives when we write $a \otimes b$ or $ab(1 \otimes 1)$ in $\mathbb{Z}/m\mathbb{Z} \otimes \mathbb{Z}/n\mathbb{Z}$. So we first need to show that if $a \equiv c \bmod m$ and $b \equiv d \bmod n$, then the image of $c \otimes d$ is the same under your map. That is, we need to show $cd \equiv ab \bmod gcd(m, n)$.