# Showing that $k(1\otimes 1) \neq 0$ when $k<\gcd(n,m)$ in $\mathbb{Z}/n\mathbb{Z}\otimes_{\mathbb{Z}} \mathbb{Z} / m \mathbb{Z}$.

I only recently learned about the tensor product of modules and came across the following exercise:

Show that $$\mathbb{Z}/n\mathbb{Z}\otimes_{\mathbb{Z}} \mathbb{Z} / m \mathbb{Z} \cong \mathbb{Z}/\gcd(n,m)$$ where $n,m\in\mathbb{N}$ and $\gcd(n,m)$ denotes the greatest common divisor of $m$ and $n$.

I proceeded by showing that

• any element of $\mathbb{Z}/n\mathbb{Z}\otimes_{\mathbb{Z}} \mathbb{Z} / m \mathbb{Z}$ has the form $k\otimes 1$, hence the tensor product is a cyclic group generated by $1\otimes 1$
• any element has at most order $\gcd(n,m)$

which leads to $$\left| \mathbb{Z}/n\mathbb{Z}\otimes_{\mathbb{Z}} \mathbb{Z} / m \mathbb{Z} \right|\leq \left| \mathbb{Z}/\gcd(n,m)\mathbb{Z} \right|.$$

I went on by constructing a surjective bilinear map $\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z} / m \mathbb{Z}\rightarrow \mathbb{Z}/\gcd(n,m)$, namely $$\left([a], [b]\right)\mapsto ab + \gcd(n,m)\mathbb{Z}$$ and using the universal property of the tensor product to get a surjective homomorphism of abelian groups $$\mathbb{Z}/n\mathbb{Z}\otimes_{\mathbb{Z}} \mathbb{Z} / m \mathbb{Z} \rightarrow \mathbb{Z}/\gcd(n,m)\mathbb{Z},$$ which shows that $\left| \mathbb{Z}/n\mathbb{Z}\otimes_{\mathbb{Z}} \mathbb{Z} / m \mathbb{Z} \right|\geq \left| \mathbb{Z}/\gcd(n,m)\mathbb{Z} \right|$. Hence both groups are cyclic and have the same order, so they are isomorphic. $\;\square$

Showing directly that $k(1\otimes 1) \neq 0$ when $k<\gcd(n,m)$ seems more elegant, but in fact I could not even manage to prove that $1\otimes 1 \neq 0$ (if $\gcd(n,m)\neq 1$) without using the universal property, let alone showing it for all the relevant multiples $k (1\otimes 1)$.

Is there an easy way to show directly that $k(1\otimes 1) \neq 0 \text{ when } k<\gcd(n,m)$, that is, without using the universal property?

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First of all,I don't know but I highly doubt there is something that isn't more or less, or closely, equivalent to the universal property which, by the way, I don't find that awful to use... –  DonAntonio Jan 10 '13 at 17:33
Thanks for your assessment. Actually, I don't mind the universal property at all, it seems to be a very useful tool to have. It's just that I still feel rather unfamiliar with tensor products so it seems kind of odd that even the simple question "is something equal to 0?" may not have a straightforward answer. –  Andy Brandi Jan 10 '13 at 17:55

You've got the right candidate map. Let $g=gcd(m,n)$. Call the biadditive map p you gave $B:\Bbb Z/(n)\times \Bbb Z/(m)\rightarrow \Bbb Z/(g)$ given by $(x,y)\rightarrow xy$. Let the tensor map be $\tau$.

To answer your question about $1\otimes 1$ being nonzero, notice that $B(1,1)=1$. So, it is impossible for $1\otimes 1=0$, because there must be an $f$ making the diagram commute, and that would say that $0=f(0)=f(\tau((1,1)))=B(1,1)=1$. This shows that $1\otimes 1\neq 0$, and furthermore that the map is onto.

Let's show that it's also injective. Find $p,q$ such that $pn+qm=g$. If $(a,b)$ is sent to 0 by the map, this means that $g|ab$, say $gx=ab$. Then $xpn+xqm=ab$. Now

$$a\otimes b=ab(1\otimes 1)=(xpn+xqm)(1\otimes 1)=xpn(1\otimes 1)+xqm(1\otimes 1)=0\otimes 0$$

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That wasn't exactly what I've been looking for as it still makes explicit use of the universal property, but you gave a nice alternate proof by showing directly that the induced homomorphism must be an isomorphism. I tried that approach before but failed at showing that the map is injective. This makes your answer very helpful, thanks! –  Andy Brandi Jan 10 '13 at 18:28
@AndyBrandi Ah, crap, yeah I didn't read it correctly that you didn't want to use the UP. I think Don's right though: the definition of the tensor is intimately tied up with the UP, so you can't avoid using it. –  rschwieb Jan 10 '13 at 18:30
You're probably right. I was hoping that an explicit construction of the tensor product would give an obvious answer, but things seem to get messy there. I'll have to get used to it. –  Andy Brandi Jan 10 '13 at 18:40