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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!!

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Careful: $\mathbb{Z}/m\mathbb{Z}$ is not free as $\mathbb{Z}$-module! However, you can say that the module is generated by $1 \in \mathbb{Z}/m\mathbb{Z}$ (though technically, $\{1\}$ is not a basis, because the module is not free). –  Shaun Ault Oct 13 '11 at 13:04
Hint: If $d=mx+ny$, then $d(1\otimes 1) = x(m1 \otimes 1) + y(1\otimes n1) = 0$. –  Thomas Andrews Oct 13 '11 at 13:13
@ShaunAult: thanks! I'm not sure I understand why I cannot treat $1$ as a basis though... Could you tell me more please? –  Rudy the Reindeer Oct 13 '11 at 14:51
$\{1\}$ is not a basis because it is not linearly independent over $\mathbb{Z}$ : $m \cdot 1 = 0$. –  Ted Oct 13 '11 at 15:57
@Ted is correct: To be a basis, a set has to be spanning and independent. $\{1\} \subset \mathbb{Z}/m\mathbb{Z}$ is spanning, but not independent. However, you can still use $1$ in the way you want to. You don't need to use linear independence; all that you need is that it generates the module. –  Shaun Ault Oct 14 '11 at 2:57

5 Answers 5

up vote 11 down vote accepted

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 $n \in m\mathbb{Z} \cup n\mathbb{Z}$, 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.

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Well let $x \in \mathrm{ker}(g)$. Then $g(x) = x(1 \otimes 1) = 0 \otimes 0$. Hence if $x$ was not already zero it is either a multiple of $n$ or a multiple of $m$ and hence in $\langle \mathrm{gcd}(m,n) \rangle$. –  Rudy the Reindeer Jul 26 '12 at 20:04
@Matt: Careful: If $m=6$ and $n=4$, then $x = 2\in \ker(g)$ but $x$ isn't a multiple of $4$ or $6$. What's happening is $2(1\otimes 1) = (6-4)(1\otimes 1) = 6(1\otimes 1) -4(1\otimes 1) = 6\otimes 1 - 1\otimes 4 = 0-0=0$ –  Jason DeVito Jul 26 '12 at 20:29
@Matt: I don't know a particularly elegant solution, but here goes. Consider the bilinear map $\mathbb{Z}_m\times \mathbb{Z}_n\rightarrow \mathbb{Z}_{\gcd(m,n)}$ sending $(x,y)$ to $xy \bmod{\gcd(m,n)}$. By universal property of tensor products, this factors through $\mathbb{Z}_m\otimes\mathbb{Z}_n$. Using this, one can show the element $k(1\otimes 1)\neq 0$ for every $0<k < \gcd(m,n)$. Finally, going back to $\subset$, if $x$ isn't a multiple of $\gcd(m,n)$, write $x = s\gcd(m,n) + k$ with $0<k<\gcd(m,n)$. Show that $g(x) = g(k)$ and conclude that $g(k)\neq 0$, so $g(x)\neq 0$. –  Jason DeVito Jul 26 '12 at 21:16
@Matt: I'm just using the universal property to better understand the elements of the tensor product. The point is you have this element $k(1\otimes 1)$ with $0<k<\gcd(m,n)$ and you don't know if this is actually $0$ due to some weird combination of relations. To rule this out, I find a linear map out of the tensor product which maps $k(1\otimes 1)$ to something nonzero. Since a linear map always maps $0$ to $0$, this proves $k(1\otimes 1)$ is nonzero. Once you know this is nonzero, go back to $g(x)$. (continued) –  Jason DeVito Jul 27 '12 at 11:46
@Matt: (continued). If $x$ is not in the ideal generated by $\gcd(m,n)$, then $x = s\gcd(m,n) + k$ with $0<k<\gcd(m,n)$. First show $g(x) = g(k)$, but then $g(k) = k(1\otimes 1)$ is nonzero by the previous comment. This shows $g(x)$ is not in the kernel. Contrapostively, what we've proved is if $x$ is in the kernel of $g$, then $x$ is in the ideal generated by $\gcd(m,n)$. –  Jason DeVito Jul 27 '12 at 11:47

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 $cd \equiv ab \bmod gcd(m, n)$.

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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 $R\otimes_RM\cong R$ 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

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


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

Finally, applying the first result we get

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


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

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


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 $(a,b) \mapsto ab \mod \mathrm{gcd} (n,m)$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 \mod \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.

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I like this because it lets me avoid fiddling with numbers. –  Rudy the Reindeer Aug 5 '12 at 10:07
Why do you know that $l$ is well-defined? Why don't you call your functions $f$ and $f'$ instead of $b$ and $b'$? –  André Caldas Aug 12 '12 at 2:50
@AndréCaldas Yes, should of course verify that it's well-defined! $b$ stands for bilinear. –  Rudy the Reindeer Aug 12 '12 at 5:10

Suppose that in the above commutative diagram with exact rows and columns, $A$ is an arbitrary $R$-module. Furthermore, let $F$ be a flat module and $E$ the injective hull (= weak-injective envelope) of $F$. Then $F$ is the flat cover of $A$ if and only if $W$ is the weak-injective envelope of $A$.

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