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Let $R$ and $S$ be commutative rings. I need to show that there is a unique ring structure on the $\mathbf{Z}$-module $T := R \otimes_\mathbf{Z} S$ such that $$ (r_1 \otimes s_1)(r_2 \otimes s_2) = r_1r_2 \otimes s_1s_2. $$ So far, I've defined a map $$ (R \times S) \times (R \times S) \to R \times S : (r_1,s_1), (r_2,s_2) \mapsto (r_1r_2,s_1s_2). $$ We know that this is the product in the product ring $R \times S$, so it satisfies the right axioms. Now I want to show that this induces a bilinear map $m : T \times T \to T$, but that's where I'm stuck.

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Note, that if $R$ is a ring, then the right way to think about its multiplication is not to consider a mapping of sets: $$ R\times R\xrightarrow{\cdot_R} R, $$ but to consider a $\mathbb{Z}$-module homomorphism: $$ R\otimes_{\mathbb{Z}}R\xrightarrow{\mu_R} R. $$

The multiplication of $R\otimes_{\mathbb{Z}}S$ is the composition: $$ (R\otimes_{\mathbb{Z}}S)\otimes_{\mathbb{Z}}(R\otimes_{\mathbb{Z}}S)\to(R\otimes_{\mathbb{Z}}R)\otimes_{\mathbb{Z}}(S\otimes_{\mathbb{Z}}S)\to R\otimes_{\mathbb{Z}}S, $$ where the first morphism is the isomorphism, which sends $(r_1\otimes s_1)\otimes(r_2\otimes s_2)$ to $(r_1\otimes r_2)\otimes(s_1\otimes s_2)$ and the second morphism is $\mu_R\otimes_{\mathbb{Z}}\mu_S$, where $\mu_R$ and $\mu_S$ are multiplications of $R$ and $S$.

Appendix (ring theory via tensor products). There are two equivalent definitions of a ring. First is more standard for most textbooks on basic algebra, it uses set-theoretic notion of element. Second is more natural for algebraic considerations, it uses tensor products over $\mathbb{Z}$. For different purposes we may refer to any of these definitions, but the second approach has the advantage that it can be used in a much more general context.

Definition 1. A ring is a pair $(R,\cdot_R)$, where $R$ is an abelian group and $\cdot_R\colon R\times R\to R$ is a mapping, such that the following properties hold:

1). $(r_1+r_2)\cdot_Rr_3=r_1\cdot_Rr_3+r_2\cdot_Rr_3$ for every $r_1,r_2,r_3\in R$;

2). $r_1\cdot(r_2+r_3)=r_1\cdot_Rr_2+r_1\cdot_Rr_3$ for every $r_1,r_2,r_3\in R$.

A ring $R$ is called associative iff $(r_1\cdot_R r_2)\cdot_Rr_3=r_1\cdot_R(r_2\cdot_Rr_3)$ for every $r_1,r_2,r_3\in R$. A ring $R$ is called commutative iff $r_1\cdot_Rr_2=r_2\cdot_Rr_1$ for every $r_1,r_2\in R$. A ring $R$ is called unital iff there exists $1_R\in R$, such that $r\cdot_R1_R=1_R\cdot_Rr=r$ for every $r\in R$.

Definition 2. A ring is a pair $(R,\mu_R)$, where $R$ is an abelian group and $\mu_R\colon R\otimes_{\mathbb{Z}}R\to R$ is a $\mathbb{Z}$-module homomorphism.

A ring $R$ is called associative iff the following diagram is commutative: enter image description here

A ring $R$ is called unital iff there exists a $\mathbb{Z}$-module homomorphism $\nu_R\colon\mathbb{Z}\to R$, such that the following diagram is commutative:

enter image description here

Diagrammatic expression of commutativity is left to the reader.

Equivalence between such definitions means that there exists an isomorphism of categories between the category of rings by the first definition and the category of rings by the second definition. Indeed, if $R$ is a ring, then $\cdot_R$ is $\mathbb{Z}$-bilinear, therefore the canonical isomorphism: $$ \text{Hom}_{\mathbb{Z}}(R\otimes_{\mathbb{Z}}R,R)\cong\text{Bilin}_{\mathbb{Z}}(R\times R,R) $$ gives the desired isomorphism of categories. Obviously, associative (unital, commutative) rings correspond to associative (unital, commutative) rings by this isomorphism.

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  • $\begingroup$ I don't exactly see how a ring multiplication can be considered as a $\mathbb{Z}$-module homomorphism. Suppose we have $\mu_R: R \otimes_\mathbb{Z} R \to R$ given, do you define ring multiplication as $r \cdot s := \mu_R(r \otimes s)$? If so, how to ensure associativity, $(ab)c = a(bc)$ for all $a,b,c \in R$? since not necessarily $ab \otimes c = a \otimes bc$ right? It seems I didn't get the full idea of this correspondence. $\endgroup$ – Sigurd Apr 25 at 15:37
  • $\begingroup$ @Sigurd see Appendix. $\endgroup$ – Oskar Apr 25 at 17:49
  • $\begingroup$ Thanks for posting that! $\endgroup$ – Sigurd Apr 27 at 7:52

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