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.


1 Answer 1


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 $\mathbb{Z}$.

  • $\begingroup$ 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 $(m,n)$? $\endgroup$ Commented Feb 3, 2021 at 0:39
  • 1
    $\begingroup$ The smallest positive (but it does not make sense for an arbitrary ring). $\endgroup$ Commented Feb 3, 2021 at 0:41

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