Is there an element-free proof that $\mathbb{Z}/(m)\otimes_\mathbb{Z}\mathbb{Z}/(n)\cong\mathbb{Z}/(m,n)$? I'm aware of the proof which goes about showing the tensor product is generated by $1\otimes 1$, which has order $\gcd(m,n)$. 
Out of curiosity, is there an element-free proof?
 A: Yes, there is an abstract nonsense proof. You can show this by only using the universal properties of the tensor product and quotient algebras. Let more generally $A,B$ be some commutative algebras over a commutative ring $R$ with ideals $I \trianglelefteq A$ and $J \trianglelefteq B$. Then we can show $A/I \otimes_R B/J \cong (A \otimes_R B)/(I'+J')$ as an isomorphism of $R$-algebras, where $I'$ and $J'$ denote the canonical embeddings of $I$ and $J$ into $A \otimes_R B$.
Proof: Let $T$ be an arbitrary $R$-algebra. Then we have bijections $\DeclareMathOperator{\hom}{Hom_R} \hom(A/I \otimes_R B/J,T) \cong \hom(A/I,T) \times \hom(B/I,T) \cong \{f \in \hom(A,T) : f(I) = 0\} \times \{g \in \hom(B,T) : g(J) = 0\} \cong \{h \in \hom(A \otimes_R B,T) : h(I') = h(J') = 0\} \cong \hom((A \otimes_R B)/(I'+J'),T)$ which are natural in $T$. Hence the claim follows from the Yoneda lemma.
Of course you can also avoid the Yoneda Lemma by arguing directly with the universal properties, though the proof will become slightly more extensive then.
