# Well definedness of $p:\mathbb{Z}/\gcd(m,n)\mathbb Z\rightarrow P$, universal property of the tensor product.

I am trying to get used to the universal property of the tensor product. I have tried to prove this common fact using the universal property

$$\mathbb{Z}/n\mathbb Z\otimes \mathbb{Z}/m\mathbb Z \cong \mathbb{Z}/\gcd(m,n)\mathbb Z$$

After checking that there is well-defined bilinear map (which i will call $$b$$) from $$\mathbb{Z}/n\mathbb Z\otimes \mathbb{Z}/m\mathbb Z$$ to $$\mathbb{Z}/\gcd(m,n)\mathbb Z$$ we need to show that we can factor any other bilinear map of $$\mathbb Z$$-modules $$p:\mathbb{Z}/n\mathbb Z\otimes \mathbb{Z}/m\mathbb Z\rightarrow P$$ over $$b$$.

The obvious thing to do is do define $$i:\mathbb{Z}/\gcd(m,n)\mathbb Z\rightarrow P$$ by $$x\mod mn\mapsto p(x,1)$$. I am having difficulty checking that this is well defined. We need to show that $$p(x,1)=p(x',1)$$ where $$x'=x+k\cdot \gcd(m,n)$$. But the only things we know are that

• $$p(x,1)=p(1,x)$$
• $$p(x+n,1)=p(x,1)$$
• $$p(1,y)=p(1,y+m)$$

I have not been able to use these to prove well definedness.

This exercise has been asked many times on this website but I am asking specifically about why in a certain method a particular map is well defined. I don't want an answer using a different method.

We have for some $$a, b \in \mathbb{Z}$$ $$p(x + k \cdot \gcd(m,n),1) = p(x + k(am + bn), 1)$$ $$= p(x + kam, 1) =p(1, x + kam) = p(1,x) = p(x,1)$$ because $$\gcd(m,n)$$ is an element which generates an ideal $$(m,n)$$ in PID $$\mathbb{Z}$$.
• Ah excellent, $\gcd(m,n)=am+bn$ was the key i was missing. I knew $\gcd(m,n)$ was in the ideal $(m,n)$ but never thought about it! This is a great way to remember it. May I ask, is it also true that $gcd(m,n)$ is the smallest element in $(m,n)$? Feb 3, 2021 at 0:39