Products, Coproducts, and Tensor Products of Rings

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

We now define products and coproducts of rings.

Products of Rings

The product of rings can be defined without difficulty. Let \((A_i)_{i\in I}\) be a family of rings. Then the product of abelian groups \(\prod_{i\in I}A_i\) is well-defined. The multiplication structure \(\mu_i: A_i\otimes A_i \rightarrow A_i\) on each \(A_i\) is identified with a bilinear map \(A_i\times A_i \rightarrow A_i\), and via this identification we obtain a function of 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 therefore induces an abelian group homomorphism \(\left(\prod A_i\right)\otimes\left(\prod A_i\right) \rightarrow \prod A_i\).

Proof

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

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

which defines the multiplication. In other words, the function multiplies two elements componentwise. Bilinearity can now also be verified componentwise.

Thus \(\prod A_i\) acquires a ring structure. Here, the additive identity is the element all of whose components are \(0\), and the multiplicative identity is the element all of whose components are \(1\). Now, 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 \(\alpha,\beta\in\Eq(\phi,\psi)\),

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

hence \(\alpha\beta\in\Eq(\phi,\psi)\). Thus \(\Eq(\phi,\psi)\) is a subring of \(A\), and it defines the equalizer of \(\phi\) and \(\psi\) in \(\Ring\). From this we obtain the following.

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

Coproducts of Rings

Defining coproducts of rings requires some effort. This is essentially because the multiplication operation in a ring is noncommutative, and a similar problem arose in defining coproducts in \(\Grp\). To overcome this, we had to define free products in a rather cumbersome manner in §Free Products. For rings, coproducts can be defined in the same way, but since they will not be used in what follows, we merely record it as a proposition.

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

Now let \(\phi,\psi:A \rightarrow B\) be two ring homomorphisms. Let \(\mathfrak{b}\) be the two-sided ideal of \(B\) generated by the elements \(\phi(\alpha)-\psi(\alpha)\); then \(B/\mathfrak{b}\) is well-defined. Then the same proof as in §Group Isomorphisms, ⁋Proposition 8 yields the following.

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 \(\Ring\). To do so, 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

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

It is an interesting fact that \(\otimes\) coincides with the coproduct in \(\cRing\). To verify this, it suffices to show that for any commutative rings \(A,B\), the map

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

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

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

so its uniqueness is clear. Meanwhile, since the map \((\alpha,\beta)\mapsto \phi_A(\alpha)\phi_B(\beta)\) from \(A\times B\) to \(C\) is bilinear, the universal property of the tensor product yields a ring homomorphism \(A\otimes B \rightarrow C\) satisfying \(\alpha\otimes \beta\mapsto \phi_A(\alpha)\phi_B(\beta)\), and this is precisely \(\phi\).