Direct Product, Direct Sum, and Tensor Product of Algebras

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 on an \(A\)-module (§Algebras, ⁋Definition 1), the subject of this post is to check whether multiplication carries over well to 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 of \(A\)-algebras \((E_i)_{i\in I}\) be given. Then we can first consider the \(A\)-module direct product \(\prod_{i\in I}E_i\), and it is natural to define componentwise multiplication on it.

Proposition 1 For a family of \(A\)-algebras \((E_i)_{i\in I}\), if multiplication is defined 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 the multiplication of \(E_i\) is \(A\)-bilinear in each component, 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 an identity element \(1_{E_i}\), 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 the direct product defined in this way is the product in the category of \(A\)-algebras; that is, the following universal property holds.

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

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 \(u:F \rightarrow\prod E_i\), that is, \(u(x)=(u_i(x))_{i\in I}\), satisfying the condition. This preserves multiplication because each \(u_i\) preserves multiplication, from which

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

follows.

Meanwhile, as we did for other algebraic structures, we can consider the finitely supported elements in 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 only need to check the absorption condition. For any \(x=(x_i)\in\bigoplus E_i\) and \(\alpha=(\alpha_i)\in\prod E_i\), in the componentwise product \(\alpha x=(\alpha_ix_i)\), the \(i\)-th component is not \(0\) only if \(x_i\neq 0\), so the support of \(\alpha x\) is contained in the support of \(x\). Since this is a finite set, \(\alpha x\in\bigoplus E_i\), 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 \(E_i\).

Note that this is not the coproduct in the category of \(A\)-algebras. First, even in the general sense of §Algebras, ⁋Definition 1, the canonical injections \(\iota_j:E_j\hookrightarrow\bigoplus E_i\) preserve addition, scalar multiplication, and multiplication, but these data do not satisfy the universal property. For example, consider the situation where \(E_1=E_2=A\), and each

\[f_i: E_i\rightarrow A\]

is given by \(\id_A\). For \(E_1\oplus E_2\) to be a coproduct, the diagram

must commute for some \(f: E_1\oplus E_2\rightarrow A\). However, for any \((a,b)\in E_1\oplus E_2\), we must have

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

but except when \(ad+bc\) is \(0\), in general the two calculations

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

do not coincide. Therefore, in general, \(f\) does not preserve multiplication.

On the other hand, in the category of associative unital \(A\)-algebras and unital homomorphisms, \(\Alg{A}\), there is also an issue with the canonical injections. If \(I\) is a finite set and each \(E_i\) is unital, the identity element of \(\bigoplus E_i=\prod E_i\) is \((1_{E_i})_{i\in I}\), so in general \(\iota_j\) does not preserve the identity element. Moreover, if \(I\) is an infinite set and every \(E_i\) is nonzero, then no element of \(\bigoplus E_i\) can act as an identity on all components, so \(\bigoplus E_i\) does not have an identity element.

Tensor Product of Algebras

In the category of commutative associative unital \(A\)-algebras, it is the tensor product that gives the correct notion of coproduct. Fundamentally, this is obtained by suitably defining a multiplication on the \(A\)-module \(E\otimes_AE'\) (§Direct Products, Direct Sums, and Tensor Products of Modules, ⁋Proposition 8) to form an \(A\)-algebra, and the multiplication we desire is given by the formula

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

However, since an element of \(E\otimes_AE'\) is not in general uniquely expressed as a sum of elements of the form \(x\otimes x'\), we must first check whether this formula gives a well-defined \(A\)-bilinear map.

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

Proof

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

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

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

Now consider the assignment \((y,y')\mapsto m_{(y,y')}\); this is a function from \(E\times E'\) to \(\End_{\lMod{A}}(E\otimes_AE')\), which is again \(A\)-bilinear by the bilinearity of multiplication. For instance, since

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

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

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

then \(\mu\) is \(A\)-linear in each variable, and satisfies equation \((1)\) on the 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 direct products, the tensor product inherits the properties of both algebras. For instance, if \(E,E'\) are both associative, then since

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

holds on the generators, \(E\otimes_AE'\) is also associative, and in the same way, 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 unital, two unital \(A\)-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 initially introduced, the tensor product is the coproduct in \(\cAlg{A}\). The following theorem explains this.

Theorem 8 Let commutative associative unital \(A\)-algebras \(E,E',F\) and unital \(A\)-algebra homomorphisms \(u:E \rightarrow F\) and \(u':E' \rightarrow F\) be given. Then, satisfying \(w\circ\iota=u\) and \(w\circ\iota'=u'\), there exists a unique unital \(A\)-algebra homomorphism \(w:E\otimes_AE' \rightarrow F\).

Proof

If we define a function \(E\times E' \rightarrow F\) by \((x,x')\mapsto u(x)u'(x')\), this is \(A\)-bilinear, so satisfying \(w(x\otimes x')=u(x)u'(x')\), there exists a unique \(A\)-linear map \(w:E\otimes_AE' \rightarrow F\). To see that \(w\) preserves multiplication, it 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 associative and commutative. Moreover, \(w(1_E\otimes 1_{E'})=u(1_E)u'(1_{E'})=1_F\), and \(w\circ\iota=u\) and \(w\circ\iota'=u'\) are clear from the definition.

We show uniqueness. If \(w'\) satisfies the same condition, then for any generator,

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

and thus \(w'=w\).

That is, \(E\otimes_AE'\) is, in \(\cAlg{A}\), the coproduct of \(E\) and \(E'\). However, inspecting the proof of the theorem above, the commutativity of \(F\) is used only to swap the order of \(u(x)\) and \(u'(x')\), so the condition actually needed instead of commutativity is merely that \(u(E)\) and \(u'(E')\) commute with each other. Thus we obtain the following more general claim.

Let \(E,E'\) be unital \(A\)-algebras, and let \(F\) be an associative unital \(A\)-algebra. If the images of the two unital \(A\)-algebra homomorphisms \(u:E\rightarrow F\) and \(u':E'\rightarrow F\) commute with each other, then, satisfying \(w\circ\iota=u\) and \(w\circ\iota'=u'\), there exists a unique unital \(A\)-algebra homomorphism \(w:E\otimes_AE'\rightarrow F\).

Example 9 The tensor product of polynomial algebras is the polynomial algebra obtained by combining the variables. That is,

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

holds. This follows from the fact that the functor \(A[-]:\Set \rightarrow \cAlg{A}\) we examined in §Algebras, ⁋Proposition 8 is a left adjoint. Since a left adjoint preserves colimits, it sends the coproduct of one-point sets (in sets) \(\{\x\}\sqcup\{\y\}=\{\x,\y\}\) to the coproduct in \(\cAlg{A}\), which by Theorem 8 is precisely the tensor product. Of course, that the above isomorphism is defined by the two formulas

\[\x\otimes 1\mapsto \x,\qquad 1\otimes \y\mapsto \y\]

can also be directly verified.

---

References

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

---