Direct Product, Direct Sum, and Tensor Product of Algebras

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

We examined operations on modules in §Direct Products, Direct Sums, and Tensor Products of Modules, and operations on rings in §Products, Coproducts, and Tensor Products of Rings. Since an \(A\)-algebra is a structure obtained by adding a bilinear multiplication to an \(A\)-module (§Algebras, ⁋Definition 1), the present post verifies that multiplication is compatible with the operations defined at the module level. As in §Algebras, \(A\) is always a commutative ring.

Direct Products and Direct Sums of Algebras

Let a family \((E_i)_{i\in I}\) of \(A\)-algebras be given. We may first consider the direct product \(\prod_{i\in I}E_i\) as an \(A\)-module, and it is natural to equip it with componentwise multiplication.

Proposition 1 For a family \((E_i)_{i\in I}\) of \(A\)-algebras, define multiplication on the \(A\)-module \(\prod_{i\in I}E_i\) by the formula

\[(x_i)_{i\in I}(y_i)_{i\in I}=(x_iy_i)_{i\in I}\]

Then \(\prod_{i\in I}E_i\) becomes an \(A\)-algebra. Moreover, if all \(E_i\) are associative (resp. commutative, unital), then so is \(\prod_{i\in I}E_i\).

Proof

We must show that the above multiplication is \(A\)-bilinear. For arbitrary \(\alpha\in A\) and \(x=(x_i),y=(y_i),z=(z_i)\in\prod E_i\), since multiplication on each \(E_i\) is \(A\)-bilinear, we have

\[\bigl((\alpha x+y)z\bigr)_i=(\alpha x_i+y_i)z_i=\alpha(x_iz_i)+y_iz_i=\bigl(\alpha(xz)+yz\bigr)_i\]

and similarly for the second variable. Associativity and commutativity are checked componentwise, and if all \(E_i\) have identity elements \(1_{E_i}\), then \((1_{E_i})_{i\in I}\) is the identity element of \(\prod E_i\).

We name this as follows.

Definition 2 The \(A\)-algebra \(\prod_{i\in I}E_i\) equipped with the multiplication defined in Proposition 1 is called the direct product of the \(E_i\). The canonical projections \(\pr_i:\prod E_i \rightarrow E_i\) are all \(A\)-algebra homomorphisms.

Then this direct product is the product in the category of \(A\)-algebras; that is, the following universal property holds.

Proposition 3 Let an arbitrary \(A\)-algebra \(F\) and \(A\)-algebra homomorphisms \(u_i:F \rightarrow E_i\) be given. Then there exists a unique \(A\)-algebra homomorphism \(u:F \rightarrow \prod_{i\in I}E_i\) such that \(\pr_i\circ u=u_i\) for all \(i\).

Proof

By the universal property of the product at the level of \(A\)-modules (§Direct Products, Direct Sums, and Tensor Products of Modules, ⁋Theorem 1), there exists a unique \(A\)-linear map \(u:F \rightarrow\prod E_i\), namely \(u(x)=(u_i(x))_{i\in I}\), satisfying the condition. That this preserves multiplication follows from the fact that each \(u_i\) preserves multiplication:

\[u(xy)=(u_i(xy))_{i\in I}=(u_i(x)u_i(y))_{i\in I}=u(x)u(y)\]

On the other hand, as with other algebraic structures, we can consider finitely supported elements inside the direct product.

Proposition 4 The \(A\)-module direct sum \(\bigoplus_{i\in I}E_i\) is a two-sided ideal of the direct product \(\prod_{i\in I}E_i\). In particular, if \(I\) is finite, then \(\bigoplus_{i\in I}E_i=\prod_{i\in I}E_i\).

Proof

That \(\bigoplus E_i\) is a submodule of \(\prod E_i\) holds by definition, so we need only check the absorption condition. For arbitrary \(x=(x_i)\in\bigoplus E_i\) and \(\alpha=(\alpha_i)\in\prod E_i\), the \(i\)-th component of the componentwise product \(\alpha x=(\alpha_ix_i)\) can be nonzero only if \(x_i\neq 0\), so the support of \(\alpha x\) is contained in the support of \(x\). The latter is a finite set, so \(\alpha x\in\bigoplus E_i\), and the same holds for \(x\alpha\). Finally, if \(I\) is finite, the finitely supported condition holds automatically, so the two sets coincide.

Definition 5 The (possibly non-unital) \(A\)-algebra \(\bigoplus_{i\in I}E_i\) obtained by restricting the multiplication of the direct product is called the direct sum of the \(E_i\).

What must be noted is that this is not the coproduct in the category of \(A\)-algebras. First, even in the general sense of §Algebras, ⁋Definition 1, the canonical injection \(\iota_j:E_j\hookrightarrow\bigoplus E_i\) preserves addition, scalar multiplication, and multiplication, yet these data do not satisfy the universal property. For example, consider the case \(E_1=E_2=A\) with

\[f_i: E_i\rightarrow A\]

each given as \(\id_A\). For \(E_1\oplus E_2\) to be a coproduct, there must exist \(f: E_1\oplus E_2\rightarrow A\) making the following diagram

commute. However, for arbitrary \((a,b)\in E_1\oplus E_2\), we must have

\[f\bigl((a,b)\bigr)=f\bigl((a,0)+(0,b)\bigr)=f\bigl((a,0)\bigr)+f\bigl((0,b)\bigr)=(f\circ\iota_1)(a)+(f\circ\iota_2)(b)=a+b\]

yet the following two computations

\[f\bigl((a,b)(c,d)\bigr)=ac+bd\neq (a+b)(c+d)=f(a,b)f(c,d)\]

show that \(f\) cannot preserve multiplication.

Moreover, under our convention that \(A\)-algebras and their homomorphisms are always unital, the situation is even worse. The maps \(\iota_j\) do not send \(1_{E_j}\) to the identity element of \(\bigoplus E_i\), so they are not \(A\)-algebra homomorphisms to begin with, and if \(I\) is infinite and all \(E_i\) are nonzero, then by Proposition 4, \(\bigoplus E_i\) is a proper ideal of \(\prod E_i\) and hence has no identity element.

Tensor Products of Algebras

The notion that yields the correct coproduct in the category of commutative \(A\)-algebras is the tensor product. Essentially, this is the \(A\)-algebra obtained by suitably defining multiplication on the \(A\)-module \(E\otimes_AE'\) (§Direct Products, Direct Sums, and Tensor Products of Modules, ⁋Proposition 8); the desired multiplication is given by the formula

\[(x\otimes x')(y\otimes y')=xy\otimes x'y'\tag{1}\]

However, since elements of \(E\otimes_AE'\) are not in general uniquely expressed as sums of elements of the form \(x\otimes x'\), we must first verify that this formula yields a well-defined \(A\)-bilinear map.

Proposition 6 For two \(A\)-algebras \(E,E'\), there exists a unique \(A\)-bilinear map \(\mu:(E\otimes_AE')\times(E\otimes_AE') \rightarrow E\otimes_AE'\) satisfying equation \((1)\).

Proof

First, fix \((y,y')\in E\times E'\). Then the map

\[E\times E' \rightarrow E\otimes_AE';\qquad (x,x')\mapsto xy\otimes x'y'\]

is \(A\)-bilinear since multiplication on \(E\) and \(E'\) is \(A\)-linear in each variable; therefore, by the universal property of §Direct Products, Direct Sums, and Tensor Products of Modules, ⁋Proposition 8, it induces a unique \(A\)-linear map \(m_{(y,y')}:E\otimes_AE' \rightarrow E\otimes_AE'\) such that \(x\otimes x'\mapsto xy\otimes x'y'\).

Now consider the correspondence \((y,y')\mapsto m_{(y,y')}\). This is a map from \(E\times E'\) to \(\End_{\lMod{A}}(E\otimes_AE')\), and again by the bilinearity of multiplication, it is \(A\)-bilinear. For example,

\[m_{(\alpha y+z,y')}(x\otimes x')=x(\alpha y+z)\otimes x'y'=\alpha(xy\otimes x'y')+xz\otimes x'y'=\bigl(\alpha m_{(y,y')}+m_{(z,y')}\bigr)(x\otimes x')\]

holds on generators, so \(m_{(\alpha y+z,y')}=\alpha m_{(y,y')}+m_{(z,y')}\). Thus, applying the universal property once more, we obtain an \(A\)-linear map \(\tilde{m}:E\otimes_AE' \rightarrow \End_{\lMod{A}}(E\otimes_AE')\) such that \(y\otimes y'\mapsto m_{(y,y')}\). Defining

\[\mu(s,t)=\tilde{m}(t)(s)\]

we see that \(\mu\) is \(A\)-linear in each variable and satisfies equation \((1)\) on generators. Uniqueness is clear from the fact that \(E\otimes_AE'\) is generated by elements of the form \(x\otimes x'\).

Definition 7 For two \(A\)-algebras \(E,E'\), the \(A\)-algebra \(E\otimes_AE'\) equipped with the multiplication from Proposition 6 is called the tensor product of \(E\) and \(E'\).

As with the direct product, the tensor product inherits the properties of the two algebras. For example, if \(E,E'\) are both associative, then on generators we have

\[\bigl((x\otimes x')(y\otimes y')\bigr)(z\otimes z')=(xy)z\otimes (x'y')z'=x(yz)\otimes x'(y'z')=(x\otimes x')\bigl((y\otimes y')(z\otimes z')\bigr)\]

so \(E\otimes_AE'\) is also associative, and similarly if \(E,E'\) are commutative then \(E\otimes_AE'\) is also commutative. Also, if \(E,E'\) are unital, then \(1_E\otimes 1_{E'}\) is the identity element of \(E\otimes_AE'\). In particular, if \(E,E'\) are associative and unital, then the two \(A\)-algebra homomorphisms

\[\iota:E \rightarrow E\otimes_AE';\quad x\mapsto x\otimes 1_{E'},\qquad \iota':E' \rightarrow E\otimes_AE';\quad x'\mapsto 1_E\otimes x'\]

are defined, and their images commute with each other. That is, \((x\otimes 1)(1\otimes x')=x\otimes x'=(1\otimes x')(x\otimes 1)\).

As we introduced at the outset, the tensor product is the coproduct in the category of commutative \(A\)-algebras. The following theorem explains this.

Theorem 8 Let commutative \(A\)-algebras \(E,E'\) and an arbitrary commutative \(A\)-algebra \(F\) be given, together with \(A\)-algebra homomorphisms \(u:E \rightarrow F\) and \(u':E' \rightarrow F\) preserving identity elements. Then there exists a unique \(A\)-algebra homomorphism \(w:E\otimes_AE' \rightarrow F\) satisfying \(w\circ\iota=u\) and \(w\circ\iota'=u'\).

Proof

Define the map \(E\times E' \rightarrow F\) by \((x,x')\mapsto u(x)u'(x')\); this is \(A\)-bilinear, so there exists a unique \(A\)-linear map \(w:E\otimes_AE' \rightarrow F\) such that \(w(x\otimes x')=u(x)u'(x')\). That \(w\) preserves multiplication suffices to check on generators:

\[w\bigl((x\otimes x')(y\otimes y')\bigr)=w(xy\otimes x'y')=u(xy)u'(x'y')=u(x)u(y)u'(x')u'(y')=u(x)u'(x')u(y)u'(y')=w(x\otimes x')w(y\otimes y')\]

where the fourth equality uses the assumption that \(F\) is commutative. Also \(w(1_E\otimes 1_{E'})=u(1_E)u'(1_{E'})=1_F\), and \(w\circ\iota=u\) and \(w\circ\iota'=u'\) are immediate from the definition.

We now show uniqueness. If \(w'\) satisfies the same conditions, then for arbitrary generators

\[w'(x\otimes x')=w'\bigl((x\otimes 1_{E'})(1_E\otimes x')\bigr)=w'(\iota(x))w'(\iota'(x'))=u(x)u'(x')=w(x\otimes x')\]

so \(w'=w\).

Thus, \(E\otimes_AE'\) is the coproduct of \(E\) and \(E'\) in the category of commutative \(A\)-algebras.

Example 9 The tensor product of polynomial algebras is the polynomial algebra with the variables combined. That is,

\[A[\x]\otimes_AA[\y]\cong A[\x,\y]\]

This follows from the fact that the functor \(A[-]:\Set \rightarrow \cAlg{A}\) examined in §Algebras, ⁋Proposition 8 is a left adjoint. Since left adjoints preserve colimits, it sends the coproduct \(\{\x\}\sqcup\{\y\}=\{\x,\y\}\) of one-point sets (in \(\Set\)) to the coproduct in \(\cAlg{A}\), and by Theorem 8, this is precisely the tensor product. Of course, one can also verify the two isomorphisms \(\x\otimes 1\mapsto \x\) and \(1\otimes \y\mapsto \y\) directly.

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References

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

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