# Show that $\mathbb{Z}/(m_1) \otimes_{\mathbb{Z}/(n)} \mathbb{Z}/(m_2) \cong \mathbb{Z}/(m_1,m_2)$

Let $$m_1,m_2\in \mathbb{N}$$ s.t. $$m_1\mid n$$ and $$m_2\mid n$$. By that, both $$\mathbb{Z}/(m_1)$$ and $$\mathbb{Z}/(m_2)$$ are $$\mathbb{Z}/(n)$$-Algebras. Show that $$\mathbb{Z}/(m_1) \otimes_{\mathbb{Z}/(n)} \mathbb{Z}/(m_2) \cong \mathbb{Z}/(m_1,m_2)$$

## Please see my edit below

Hi, i was working on this problem and would like to have some feedback and maybe some general advice regarding my approach (since i didn't manage to solve it).

My attempt:

I tried to invoke the universal property of the tensor product by first defining a map

\begin{align*} \varphi \colon \mathbb{Z}/(m_1) \times \mathbb{Z}/(m_2) &\to \mathbb{Z}/(m_1,m_2) \\ (z_1+(m_1),z_2+(m_2)) &\mapsto z_1z_2 + (m_1,m_2) \end{align*}

This map is bilinear, thus it induces a unique map

\begin{align*} \widetilde{\varphi} \colon \mathbb{Z}/(m_1) \otimes \mathbb{Z}/(m_2) &\to \mathbb{Z}/(m_1,m_2) \end{align*}

which is indeed well defined (by the universal property of the tensor product).

I was then trying to somehow naively guess an inverse and my attempt was trying something like

\begin{align*} \psi \colon \mathbb{Z}/(m_1,m_2) &\to \mathbb{Z}/(m_1) \otimes \mathbb{Z}/(m_2) \\ z + (m_1,m_2) &\mapsto z+(m_1)\otimes z+(m_2)\end{align*}

where $$\overline{z_i} = z_i + (m_i)$$ are the respective cosets.

However, i didnt succeed in proving well definedness and since the solution we were given to this exercise was in fact quite different to what i've tried, i wanted to ask whether my attempt is actually reasonable of if i went completely the wrong way.

Question 1: Is this attempt fine so far? Or should i have thought about something completely different while appempting to solve this problem?

Question 2: If my attempt was not reasonable, would you mind elaborating how i should have approached this problem instead?

Question 3: What role does the property of being an $$\mathbb{Z}/(n)$$-Algebra play here actually? I've treated everything as if i'd be working with modules, is this ok?

Thanks for any help!

## Edit

I figured the following maps would actually do the trick:

\begin{align*} \varphi \colon \mathbb{Z}/(m_1) \otimes \mathbb{Z}/(m_2) &\to \mathbb{Z}/(m_1,m_2)\\ \left(z + (m_1)\right)\otimes \left(1 + (m_2) \right)&\mapsto z\cdot 1 + (m_1,m_2)\end{align*}

with the inverse map

\begin{align*} \psi \colon \mathbb{Z}/(m_1,m_2) &\to \mathbb{Z}/(m_1) \otimes \mathbb{Z}/(m_2) \\ z + (m_1,m_2) &\mapsto \left(z+(m_1)\right)\otimes \left(1+(m_2)\right)\end{align*}

It obviously holds that $$\varphi\circ\psi = \operatorname{id},\ \psi\circ\varphi = \operatorname{id}$$

Is my solution correct?

• Your formula for $\psi$ seems not to be "well defined": When choosing an element $x \in\mathbb{Z}/(m_1,m_2)$ how do you arrive at $x=z_1\otimes z_2$? You must choose (in a well defined way) two integers $z_1,z_2 \in \mathbb{Z}$ such that $z_1\otimes z_2$ is independent of choices. Commented Sep 10, 2021 at 8:17
• Thanks for your comment, i accidentally have cited the wrong map (copied the wrong map from my notes), i'm sincerely sorry. However, i didn't manage to confirm the well-definedness for this particular $\psi$, too. I'm mostly curious whether my general attempt is fine (assuming $\psi$ would be chosen correctly of course).
– Zest
Commented Sep 10, 2021 at 16:24
• I've edited my question and added a new attempt for which i hope to be correct.
– Zest
Commented Sep 10, 2021 at 16:50
• The $\varphi$ in your edit does not uniquely define a function. What happens if the element is not of the form $\overline z\otimes\overline 1$? If you take the $\widetilde\varphi$ from your original attempt instead, everything should work. Commented Sep 10, 2021 at 16:51
• Of course, thanks for the feedback. If i pick $\widetilde{\varphi}$ though, i'd need to look for another $\psi$, so far, $\psi$ is not an inverse to $\widetilde{\varphi}$. I'll try to find an inverse. But the approach is fine nonetheless?
– Zest
Commented Sep 10, 2021 at 16:55

Combining your two approaches, let's take \begin{align*}\Phi: \mathbb Z/(m_1)\otimes_{\mathbb Z/(n)}\mathbb Z/(m_2)&\to\mathbb Z/(m_1,m_2)\\(z_1+(m_1))\otimes_{\mathbb Z/(n)}(z_2+(m_2))&\mapsto z_1z_2+(m_1,m_2) \end{align*} and \begin{align*}\Psi:\mathbb Z/(m_1,m_2)&\to \mathbb Z/(m_1)\otimes_{\mathbb Z/(n)}\mathbb Z/(m_2)\\z+(m_1,m_2)&\mapsto(z+(m_1))\otimes_{\mathbb Z/(n)}(1+(m_2))\end{align*} (I am using capital letters for the functions to avoid possible confusion with the functions in the question. Also note that I will write $$\overline z$$ for the residue class of $$z$$ when the context is clear, and also drop the indeces from tensor products.)

It is straightforward to show that $$\Phi$$ is a well-defined homomorphism using the universal property of tensor products. To show that $$\Psi$$ is well-defined, take $$z,z'\in\mathbb Z$$ such that $$z-z'\in(m_1,m_2)$$ and show that $$\Psi(\overline z)=\Psi(\overline{z'}).$$ It should be clear that $$\Psi$$ is a homomorphism.

Now, all that is left to do is to show that these two functions are inverses of each other. One direction $$\Phi\circ\Psi=\mathrm{id}$$ is obvious, the other one needs a little more work. Take a general element of $$\mathbb Z/(m_1)\otimes\mathbb Z/(m_2)$$, it has the form $$z=\sum_{i=1}^n\overline x_i\otimes\overline y_i$$. Then calculate \begin{align*}\Psi(\Phi(z))&=\sum_{i=1}^n\Psi(\overline{x_iy_i})\\&=\sum_{i=1}^n(x_iy_i+(m_1))\otimes(1+(m_2))\\&=\sum_{i=1}^n(y_i+(n))\cdot((x_i+(m_1))\otimes(1+(m_2)))\\&=\sum_{i=1}^n(x_i+(m_1))\otimes(y_i\cdot1+(m_2))=z,\end{align*} as desired.

• Thanks a lot stefan, the reason we can factor out $y_i + (n)$ is due to $m_1$ dividing $n$, is that correct?
– Zest
Commented Sep 10, 2021 at 22:09
• Well, if $m_1$ did not divide $n$, then $\mathbb Z/(m_1)$ would not be a $\mathbb Z/(n)$-module. More explicitely, you can view $x_iy_i+(m_1)$ both as the product $(x_i+(m_1))(y_i+(m_1))$ in the ring/algebra $\mathbb Z/(m_1)$ and as $(y_i+(n))(x_i+(m_1))$, a scalar product in the $\mathbb Z/(n)$-module. This is used to move the $y_i$ around. Commented Sep 10, 2021 at 22:26
• Does that mean that we could also have written $(x_1+(m1))(y_i+(m_1))$ instead of $(x_1+(n))(y_i+(m_1))$
– Zest
Commented Sep 10, 2021 at 23:08
• Yes, that is correct too (but not useful in the proof). Commented Sep 11, 2021 at 9:23
• (+1) Fine answer, except that one should: * notice that quotienting by $(n)$ actually changes nothing: $$\mathbb{Z}/(m_1) \otimes_{\mathbb{Z}/(n)} \mathbb{Z}/(m_2) \cong \mathbb{Z}/(m_1) \otimes \mathbb{Z}/(m_2);$$ * give a thorough proof of well-definedness for $\Psi:$ apply Bézout to check that$$(\gcd(m_1,m_2)+(m_1))\otimes(1+(m_2))=0.$$ Commented Mar 24, 2023 at 17:19

Question 1: "Is this attempt fine so far? Or should i have thought about something completely different while appempting to solve this problem?"

Question 2: "If my attempt was not reasonable, would you mind elaborating how i should have approached this problem instead?"

Question 3: What role does the property of being an $$Z/(n)$$ -Algebra play here actually? I've treated everything as if i'd be working with modules, is this ok?

Answer Q3: Here is another approach using a formula for the tensor product: Since $$n=k_1m_1$$ it follows $$\mathbb{Z}/(n)\cong \mathbb{Z}/(n)/(\overline{m_1})$$ where $$\overline{m_1}\subseteq \mathbb{Z}/(n)$$ is the equivalence class of $$m_1$$. There is a general formula saying

$$FT.\text{ }A/I\otimes_A M \cong M/IM$$

for any ideal $$I \subseteq A$$ and any left $$A$$-module $$M$$. You may describe $$\mathbb{Z}/(m_1) \cong (\mathbb{Z}/(n))/(\overline{m_1})$$ since $$n=k_1m_1$$ and use $$FT$$.

From this you get isomorphisms

$$\mathbb{Z}(m_1)\otimes_{\mathbb{Z}/(n)} \mathbb{Z}/(m_2) \cong (\mathbb{Z}/(n))/(\overline{m_1})) \otimes_{\mathbb{Z}/(n)} \mathbb{Z}/(m_2) \cong$$

$$(\mathbb{Z}/(m_2))/\overline{m_1}\mathbb{Z}/(m_2) \cong \mathbb{Z}/(m_1,m_2).$$

Since there is an inclusion of ideals $$(n) \subseteq (m_1)$$, there is an exact sequence

$$0 \rightarrow (m_1)/(n):=(\overline{m_1}) \rightarrow \mathbb{Z}/(n) \rightarrow \mathbb{Z}/(m_1) \rightarrow 0.$$