Commutative Algebra/Direct products, direct sums and the tensor product

Direct products and direct sums
Theorem 8.2:

In the category of modules, the direct product constitutes a product.

Proof:

Let $$\mathcal J$$ be any index category, that contains one element $$j_\alpha \in \mathcal J$$ for each $$\alpha \in A$$, no other elements, and only the identity morphisms. Let $$N$$ be any other object such that

Lemma 8.4:

Let $$M_\alpha, \alpha \in A$$ be modules. Their direct sum is a submodule of the direct product.

Proof:

Both have the same elements and the same operations, and the direct product is a subset that is a module with those operations. Therefore we have a submodule.

Lemma 8.5:

For each $$\alpha \in A$$, there is a canonical morphism
 * $$M_\alpha \to \bigoplus_{\alpha \in A} M_\alpha$$.

Proof:


 * $$\iota_\alpha: M_\alpha \to \bigoplus_{\alpha \in A} M_\alpha, \iota_\alpha(m) := (0, \ldots, 0, \overbrace{m}^{\alpha \text{-th place}}, 0, \ldots, 0)$$.

Lemma 8.6:


 * $$\operatorname{Hom}_R \left( \bigoplus_{\alpha \in A} M_\alpha, N \right) \cong \prod_{\alpha \in A} \operatorname{Hom}_R (M_\alpha, N)$$.

Proof:

Consider the morphism
 * $$\varphi: \operatorname{Hom}_R \left( \bigoplus_{\alpha \in A} M_\alpha, N \right) \to \prod_{\alpha \in A} \operatorname{Hom}_R (M_\alpha, N), f \mapsto (f \circ \iota_\alpha)_{\alpha \in A}$$.

We claim that this is an isomorphism, so we check all points.

1. Well-defined:

Both $$\iota_\alpha$$ and $$f$$ are morphisms (with suitable domains and images), so $$f \circ \iota_\alpha$$ is as well.

2. Injective:

Assume $$(f \circ \iota_\alpha)_{\alpha \in A} = (g \circ \iota_\alpha)_{\alpha \in A}$$. Then for any $$(m_\alpha)_{\alpha \in A}$$ contained in $$\bigoplus_{\alpha \in A} M_\alpha$$ we have
 * $$f((m_\alpha)_{\alpha \in A}) = \sum_{\alpha \in A} f((0, \ldots, 0, \overbrace{m}^{\alpha \text{-th place}}, 0, \ldots, 0)) = \sum_{\alpha \in A} f \circ \iota_\alpha ((m_\alpha)_{\alpha \in A}) = \sum_{\alpha \in A} g \circ \iota_\alpha ((m_\alpha)_{\alpha \in A}) = g((m_\alpha)_{\alpha \in A})$$;

note that the sum is finite, since we are in the direct sum; this is necessary since infinite sums are not defined. Hence $$f = g$$.

3. Surjective:

Let $$(h_\alpha)_{\alpha \in A} \in \prod_{\alpha \in A} \operatorname{Hom}_R (M_\alpha, N)$$. Define
 * $$f: \bigoplus_{\alpha \in A} M_\alpha \to N, f((m_\alpha)_{\alpha \in A}) := \sum_{\alpha \in A} h_\alpha(m_\alpha)$$.

The latter sum is finite because $$h_\alpha(0) = 0$$ and all but finitely many $$m_\alpha$$ are nonzero. Thus this is well-defined as a function, and direct computation proves easily that it is $$R$$-linear. Hence we have a morphism, and further
 * $$\varphi(f) = (f \circ \iota_\alpha)_{\alpha \in A} = (h_\alpha)_{\alpha \in A}$$.

Theorem 8.7:

direct sum is coproduct in category of modules

Quotient spaces
To be then used to construct the tensor product.

The tensor product
The following theorem shows that the tensor product has something to do with bilinear maps:

Proof:

Let $$f: M \times N \to K$$ be any $$R$$-bilinear map. Define
 * $$g\left( \left[ \sum_{j=1}^l r_j (m_j, n_j) \right] \right) := \sum_{j=1}^l r_j f(m_j, n_j)$$,

where the square brackets indicate the equivalence class.

Once we proved that this is well-defined, the linearity of $$g$$ easily follows. We thus have to show that $$g$$ maps equivalent vectors to the same element, which after subtracting the right hand side follows from $$g$$ mapping $$S$$ to zero.

Indeed, let
 * $$\sum_{j=1}^l r_j s_j \in S$$,

where all $$s_j$$ are one of the four types of generators of $$S$$. By distinguishing cases, one obtains that each type of generator of $$S$$ is mapped to zero by $$f$$ because of bilinearity. Well-definedness follows, and linearity is clear from the definition and since addition and module operation interchange with equivalence class formation.

Note that from a category theory perspective, this theorem 8.9 states that for any two modules $$M, N$$ over the same ring, the arrow
 * $$M \times N \overset{\otimes}{\rightarrow} M \otimes N$$

is a universal arrow. Hence, we call the result of theorem 8.9 the universal property of the tensor product.

Lemma 8.10:

Let $$R$$ be a ring and $$M$$ be an $$R$$-module. Recall that using canonical operations, $$R$$ is an $$R$$-module over itself. We have
 * $$R \otimes_R M \cong M$$.

Proof:

Define the morphism
 * $$R \times M \to M, (r, m) \to rm$$,

extend it to all formal linear combinations via summation
 * $$\sum_{k=1}^l s_k (r_k, m_k) \mapsto \sum_{k=1}^l s_k r_k m_k$$

and then observe that
 * $$\varphi: R \otimes_R M \to M, \varphi\left( \left[ \sum_{k=1}^l s_k (r_k, m_k) \right] \right) := \sum_{k=1}^l s_k r_k m_k$$

is well-defined; again, by subtracting the right hand side, it's enough to show that $$S$$ is mapped to zero, and this is again done by consideration of each of the four generating types.

This is a morphism as shown by direct computation (using the rules for the module operation), it is clearly surjective (map $$[(1, m)]$$) and it is injective because if
 * $$\sum_{k=1}^l s_k r_k m_k = \sum_{j=1}^r s_j' r_j' m_j'$$, then
 * $$\left[ \sum_{k=1}^l s_k r_k m_k - \sum_{j=1}^r s_j' r_j' m_j' \right] = 0$$

since $$0 \in S$$.

Lemma 8.11:

Let $$M, N, K$$ be $$R$$-modules. Then
 * $$M \otimes (N \otimes K) \cong (M \otimes N) \otimes K$$.

Proof:

For $$m \in M$$ fixed, define the bilinear function
 * $$f_m : N \times K \to (M \otimes N) \otimes K, f_m((n,k)) := [([m, n], k)]$$.

Applying theorem 8.9 yields
 * $$g_m: N \otimes K \to (M \otimes N) \otimes K$$

such that $$g_m([(n, k)]) = [([(m, n)], k)]$$. Then define
 * $$F: M \times (N \otimes K) \to (M \otimes N) \otimes K, F \left( m, \left[ \sum_{j=1}^l r_j (n_j, k_j) \right] \right) := g_m \left(\left[ \sum_{j=1}^l r_j (n_j, k_j) \right] \right)$$.

This function is bilinear (linearity in $$m$$ from
 * $$[([(m + \lambda m', n)], k)] = [([(m,n) + \lambda (m', n)], k)] = [([(m,n)] + \lambda [(m', n)], k)] = \cdots = [([(m, n)], k)] + \lambda [([(m', n)], k)]$$)

and thus theorem 8.9 yields a morphism
 * $$G: M \otimes (N \otimes K) \to (M \otimes N) \otimes K$$

such that
 * $$G([(m, [(n,k)])]) = F(m, [(n,k)]) = g_m([(n,k)]) = [([(m, n)], k)]$$.

An analogous process yields a morphism
 * $$H: (M \otimes N) \otimes K \to M \otimes (N \otimes K)$$

such that
 * $$H([([(m, n)], k)]) = [(m, [(n,k)])]$$.

Since addition within tensor products commutes with equivalence class formation, $$G$$ and $$H$$ are inverses.

Lemma 8.12:

Let $$N_\alpha, \alpha \in A$$ be $$R$$-modules, let $$M$$ be an $$R$$-module. Then
 * $$M \otimes \left( \bigoplus_{\alpha \in A} N_\alpha \right) \cong \bigoplus_{\alpha \in A} (M \otimes N_\alpha)$$.

Proof:

We define
 * $$f: M \times \left( \bigoplus_{\alpha \in A} N_\alpha \right) \to \bigoplus_{\alpha \in A} (M \otimes N_\alpha), (m, (n_\alpha)_{\alpha \in A}) \mapsto \left( [(m, n_\alpha)] \right)_{\alpha \in A}$$.

This is bilinear (since formation of equivalence classes commutes with summation and module operation), and hence theorem 8.9 yields a morphism
 * $$g: M \otimes \left( \bigoplus_{\alpha \in A} N_\alpha \right) \to \bigoplus_{\alpha \in A} (M \otimes N_\alpha)$$

such that
 * $$g([(m, (n_\alpha)_{\alpha \in A})]) = \left( [(m, n_\alpha)] \right)_{\alpha \in A}$$.

This is obviously surjective. It is injective because
 * $$\begin{align}

& g \left( \left[ \sum_{j=1}^l r_j (m_j, (n^j_\alpha)_{\alpha \in A}) \right] \right) = g \left( \left[ \sum_{q=1}^p s_q (x_q, (y^q_\alpha)_{\alpha \in A}) \right] \right) \\ \Leftrightarrow & \left( \left[ \sum_{j=1}^l r_j (m_j, n^j_\alpha) \right] \right)_{\alpha \in A} = \left( \left[ \sum_{q=1}^p s_q (x_q, y^q_\alpha) \right] \right)_{\alpha \in A} \end{align}$$ by the linearity of $$g$$ and component-wise addition in the direct sum, and equality for the direct sum is component-wise. We split the argument up into sums where only one component of the right direct sum matters, and observe equality since we divide out isomorphic spaces.

Lemma 8.13:

$$M \otimes N \cong N \otimes M$$.

Proof:

Linear extension of
 * $$[(m, n)] \mapsto [(n, m)]$$

defines a morphism which is well-defined due to symmetry, linear by definition and bijective because of the obvious inverse.

We have proven:

Note that we have more: From lemma 8.12 even infinite direct sums (uncountably many, as many as you like, ...) distribute over the tensor product. Incidentally, only finite direct sums are identical to the direct product. This may give hints for an infinite distributive law for infinitesimals.

Proof:

Set
 * $$\varphi: \operatorname{Hom} (M \otimes N, K) \to \operatorname{Hom}(M, \operatorname{Hom}(N,K)), \varphi(f)(m) = n \mapsto f([(m, n)])$$.

Due to the equalities holding for elements of the tensor product and the linearity of $$f$$, this is well-defined. Further, we obviously have linearity in $$f$$ since function addition and module operation are defined point-wise.

Further set
 * $$\psi: \operatorname{Hom}(M, \operatorname{Hom}(N,K)) \to \operatorname{Hom} (M \times N, K), \psi(g)(m,n) = g(m)(n)$$.

By theorem 8.9 and thinking outside the box, we get a map
 * $$\theta: \operatorname{Hom}(M, \operatorname{Hom}(N,K)) \to \operatorname{Hom} (M \otimes N, K)$$

such that
 * $$\theta(g)([(m,n)]) = g(m)(n)$$.

Then $$\theta$$ and $$\varphi$$ are inverse morphisms, since $$f: M \otimes N \to K$$ is determined by what it does on elements of the form $$[(m,n)]$$.

Proof:

The map
 * $$\phi: M \times K \to N \otimes K, \phi(m,k) = \theta(m) \otimes k$$

is bilinear, and hence induces a map
 * $$\varphi: M \otimes K \to N \otimes K$$

such that
 * $$\varphi([(m, k)]) = [(\theta(m), k)]$$.

Similarly, the map
 * $$\phi_{-1} N \times K \to M \otimes K, \phi(n,k) = \theta^{-1}(n) \otimes k$$

induces a map
 * $$\varphi^{-1}: N \otimes K \to M \otimes K$$

such that
 * $$\varphi^{-1}([(n, k)]) = [(\theta^{-1}(n), k)]$$.

These maps are obviously inverse on elements of the type $$m \otimes k$$, $$n \otimes k$$, and by their linearity and since addition and equivalence classes commute, they are inverse to each other.