Products, Coproducts, and Tensor Products of Rings

Now we define products and coproducts of rings.

Products of Rings

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

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

can be defined.

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

If we express the above function directly in terms of elements, the elements of \(\prod A_i\) are of the form of 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 above function to these two gives multiplication via

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

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

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

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

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

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

we have \(\alpha\beta\in\Eq(\phi,\psi)\), and also since \(\phi(1)=1=\psi(1)\), we have \(1\in\Eq(\phi,\psi)\). 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.

On the other hand, one useful notion in \(\Rng\) is the direct sum. Consider a family of rings \((A_i)_{i\in I}\). Among the elements of the direct product \(\prod_{i\in I}A_i\), the collection of those that are \(0\) except for finitely many components, that is, finitely supported families, is closed under addition and multiplication and forms a subobject of \(\prod A_i\) in \(\Rng\). (The sum and product of two finitely supported families are still finitely supported.) We call this the following.

Definition 3 For a family of rings \((A_i)_{i\in I}\), the collection of finitely supported elements of the direct product \(\prod_{i\in I}A_i\) is called the direct sum of rings and is written \(\bigoplus_{i\in I} A_i\).

This is in exactly the same spirit as the direct sum defined for abelian groups in §Abelian Groups, ⁋Definition 2. However, if all \(A_i\) are nonzero and the index set \(I\) is infinite, then the direct sum \(\bigoplus A_i\) does not contain the identity element \((1)_{i\in I}\), so it is not a (unital) ring; because of this, this notion is mainly used in \(\Rng\). Conversely, when \(I\) is a finite set, the direct sum coincides exactly with the direct product by definition.

In the case of abelian groups the direct sum is the coproduct (§Abelian Groups, ⁋Theorem 1), but this is not the case for rings. Therefore, the coproduct of rings must be defined separately.

Coproducts of Rings

On the other hand, defining the coproduct of rings requires a little effort. This is essentially because, as pointed out above, the multiplication operation of a ring is not commutative, and there was a similar problem when defining the coproduct in \(\Grp\). To overcome this, we had to define the free product in a rather tedious way in §Free Product. In rings, the coproduct can be defined in the same manner, but since its construction repeats that of the free product verbatim, we omit it here.

Proposition 4 For any family of rings \((A_i)_{i\in I}\), their coproduct exists.

On the other hand, let any two ring homomorphisms \(\phi,\psi:A \rightarrow B\) be given. 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. The same proof as in §Group Isomorphisms, ⁋Proposition 8 then shows that the following holds.

Proposition 5 In the above situation, \(\CoEq(\phi,\psi)=B/\mathfrak{b}\) defines the coequalizer of \(\phi,\psi\).

By Proposition 4 and Proposition 5, \(\Ring\) has arbitrary coproducts and coequalizers, so it is cocomplete, and adding Theorem 2 to this we obtain the following.

Theorem 6 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 a multiplication structure on the abelian group \(A\otimes B\) for any two rings \(A,B\), that is, the following abelian group homomorphism:

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

However, 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 therefore \(\mu_A:A\otimes A \rightarrow A\) and \(\mu_B: B\otimes B \rightarrow B\) define a 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 7 For any rings \(A,B\), the ring \(A\otimes B\) defined as above is called their tensor product.

Through this, one 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

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

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

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

and \(\iota_B\) defined in a similar way satisfy the universal property of the coproduct. Let any commutative ring \(C\) and ring homomorphisms \(\phi_A: A \rightarrow C\), \(\phi_B: B \rightarrow C\) be given. If there exists \(\phi: A\otimes B \rightarrow C\) satisfying the universal property of the coproduct, then it must necessarily satisfy

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

so we see that it is unique. On the other hand, since the function from \(A\times B\) to \(C\) given by \((\alpha,\beta)\mapsto \phi_A(\alpha)\phi_B(\beta)\) is bilinear, by the universal property of the tensor product, there exists, satisfying \(\alpha\otimes \beta\mapsto \phi_A(\alpha)\phi_B(\beta)\), an abelian group homomorphism \(A\otimes B \rightarrow C\); since \(C\) is commutative, \(\phi_B(\beta)\) and \(\phi_A(\alpha')\) commute, so this function preserves multiplication and also sends \(1\otimes 1\) to \(1\), hence it is a ring homomorphism and is exactly \(\phi\).

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.