Products, Coproducts, and Tensor Products of Rings

We now define products and coproducts of rings.

Products of Rings

The product of rings can be defined without difficulty. Let a family \((A_i)_{i\in I}\) of rings be given. Then the product \(\prod_{i\in I}A_i\) of abelian groups is well-defined. On the other hand, the multiplication structure \(\mu_i: A_i\otimes A_i \rightarrow A_i\) on \(A_i\) is the same as a bilinear map \(A_i\times A_i \rightarrow A_i\), and through this, we can define a function between sets

\[\left(\prod_{i\in I} A_i\right)\times\left(\prod_{i\in I} A_i\right) \cong \prod_{i\in I} (A_i\times A_i) \overset{\prod \mu_i}{\longrightarrow} \prod_{i\in I}A_i\]

Proposition 1 The function defined above is a bilinear map from the abelian group \(\left(\prod A_i\right)\times\left(\prod A_i\right)\) to \(\prod A_i\), and thus induces an abelian group homomorphism \(\left(\prod A_i\right)\otimes\left(\prod A_i\right) \rightarrow \prod A_i\).

Proof

Writing the above function explicitly in terms of elements, elements of \(\prod A_i\) are tuples of the form \((\alpha_i)_{i\in I}\), and for two elements \((\alpha_i)_{i\in I}, (\beta_i)_{i\in I}\in \prod A_i\), the result of applying the above function to these two is

\[(\alpha_i)_{i\in I}(\beta_i)_{i\in I}=(\alpha_i\beta_i)_{i\in I}\]

giving the multiplication. That is, the given function multiplies two elements componentwise. Now bilinearity can also be verified componentwise.

Through this, \(\prod A_i\) also carries a ring structure. In this case, the additive identity of this ring is the element with all components being \(0\), and the multiplicative identity is the element with all components being \(1\). On the other hand, for any two ring homomorphisms \(\phi,\psi:A \rightarrow B\), if we define

\[\Eq(\phi,\psi)=\{\alpha\in A\mid \phi(\alpha)=\psi(\alpha)\}\]

then by §Group Homomorphisms, ⁋Proposition 2, this is a subgroup of \(A\), and moreover, for any \(\alpha,\beta\in\Eq(\phi,\psi)\),

\[\phi(\alpha\beta)=\phi(\alpha)\phi(\beta)=\psi(\alpha)\psi(\beta)=\psi(\alpha\beta)\]

so \(\alpha\beta\in\Eq(\phi,\psi)\). That is, \(\Eq(\phi,\psi)\) is a subring of \(A\), and this defines the equalizer of \(\phi\) and \(\psi\) in \(\Ring\). From this, the following holds.

Theorem 2 The category \(\Ring\) is complete.

Coproducts of Rings

On the other hand, defining coproducts of rings requires some effort. This is essentially because the multiplication operation of a ring is not commutative, and we encountered a similar problem when defining coproducts in \(\Grp\). To overcome this, in §Free Products, we had to define free products in a rather cumbersome way. In the case of rings, coproducts can be defined in the same way, but since this will not be used in subsequent discussions, we simply state it as a proposition.

Proposition 3 For any family \((A_i)_{i\in I}\) of rings, their coproduct exists.

On the other hand, let any two ring homomorphisms \(\phi,\psi:A \rightarrow B\) be given. If we define the ideal \(\mathfrak{b}\) of \(B\) as the two-sided ideal generated by the elements \(\phi(\alpha)-\psi(\alpha)\), then \(B/\mathfrak{b}\) is well-defined. Then by the same proof as §Group Isomorphisms, ⁋Proposition 8, the following holds.

Proposition 4 In the above situation, \(\CoEq(\phi,\psi)=B/\mathfrak{b}\) defines the coequalizer of \(f,g\).

Therefore, the following holds.

Theorem 5 The category \(\Ring\) is a bicomplete category.

Tensor Products of Rings

Finally, we define the tensor product \(\otimes\) in \(\Ring\). For this, it suffices to define, for any two rings \(A,B\), a multiplication structure on the abelian group \(A\otimes B\), that is, the abelian group homomorphism

\[(A\otimes B)\otimes(A\otimes B) \rightarrow A\otimes B\]

By the associativity and commutativity of the tensor product,

\[(A\otimes B)\otimes(A\otimes B)\cong (A\otimes A)\otimes (B\otimes B)\]

holds, and thus \(\mu_A:A\otimes A \rightarrow A\) and \(\mu_B: B\otimes B\) define the multiplication on \(A\otimes B\):

\[(A\otimes B)\otimes(A\otimes B)\cong (A\otimes A)\otimes (B\otimes B)\overset{\mu_A\otimes\mu_B}{\longrightarrow} A\otimes B\]

Definition 6 For any rings \(A,B\), the ring \(A\otimes B\) defined as above is called their tensor product.

Through this, we can verify that the category \(\Ring\) forms a symmetric monoidal category \((\Ring,\otimes, \mathbb{Z})\). Explicitly, the multiplication on \(A\otimes B\) is defined by

\[(\alpha\otimes \beta)(\alpha'\otimes \beta')=\alpha\alpha'\otimes \beta\beta'\]

An interesting fact is that \(\otimes\) coincides with the coproduct in \(\cRing\). To verify this, we need to show that for any commutative rings \(A,B\) and

\[\iota_A: A \hookrightarrow A\otimes B;\quad \alpha\mapsto \alpha\otimes 1\]

and \(\iota_B\) defined similarly, the universal property of coproducts is satisfied. Let any \(\phi_A: A \rightarrow C\) and \(\phi_B: B \rightarrow C\) be given. If a \(\phi: A\otimes B \rightarrow C\) satisfying the universal property of coproducts exists, then it must satisfy

\[\phi(\alpha\otimes \beta)=\phi((\alpha\otimes 1)(1\otimes \beta))=\cdots=\phi_A(\alpha)\phi_B(\beta)\]

so uniqueness is clear. On the other hand, since the function \((\alpha,\beta)\mapsto \phi_A(\alpha)\phi_B(\beta)\) from \(A\times B\) to \(C\) is bilinear, by the universal property of tensor products, there exists a ring homomorphism \(A\otimes B \rightarrow C\) satisfying \(\alpha\otimes \beta\mapsto \phi_A(\alpha)\phi_B(\beta)\), and this is precisely \(\phi\).