Tag Archives: direct product

The quotient of a product is module isomorphic to the product of quotients

Let R be a ring, let \{A_i\}_{i=1}^n be a finite family of (left, unital) R-modules, and let B_i \subseteq A_i be a submodule for each i. Prove that (\prod A_i)/(\prod B_i) \cong_R \prod A_i/B_i.


We did this previously. D&F, why repeat an exercise?

If V is an infinite dimensional vector space, then its dual space has strictly larger dimension

Let V be an infinite dimensional vector space over a field F, say with basis A. Prove that the dual space \widehat{V} = \mathsf{Hom}_F(V,F) has strictly larger dimension than does V.


We claim that \widehat{V} \cong_F \prod_A F. To prove this, for each a \in A let F_a be a copy of F. Now define \varepsilon_a : \widehat{V} \rightarrow F_a by \varepsilon_a(\widehat{v}) = \widehat{v}(a). By the universal property of direct products, there exists a unique F-linear transformation \theta : \widehat{V} \rightarrow \prod_A F_a such that \pi_a \circ \theta = \varepsilon_a for all a \in A. We claim that \theta is an isomorphism. To see surjectivity, let (v_a) \in \prod_A F_a. Now define \varphi \in \widehat{V} by letting \varphi(a) = v_a and extending linearly; certianly \theta(\varphi) = (v_a). To see injectivity, suppose \varphi \in \mathsf{ker}\ \theta. Then \theta(\varphi) = 0, so that (\pi_a \circ \theta)(\varphi) = 0, and thus \varepsilon_a(\varphi) = 0 for all a. Thus \varepsilon(a) = 0 for all a \in A. Since A is a basis of V, we have \varphi = 0. Thus \theta is an isomorphism, and we have \widehat{V} \cong_F \prod_A F.

By this previous exercise, \widehat{V} has strictly larger dimension than does V.

As an F-vector space, an infinite direct sum of F has strictly smaller dimension than an infinite direct power of F over the same index set

Let F be a field. Prove that a vector space V over F having basis B (regardless of the cardinality of B) is isomorphic as a vector space to \bigoplus_B F. Prove that \prod_B F is also an F-vector space which has strictly larger dimension than that of \bigoplus_B F.


(So a free module on any set is isomorphic to a direct sum. We’ve never gotten around to proving this in the best possible generality, though, so we’ll just prove it for vector spaces here.)

Note that, by the universal property of free modules, the natural injection B \rightarrow \bigoplus_B F which sends b to the tuple which has 1 in the bth component and 0 elsewhere induces a vector space homomorphism \varphi : V \rightarrow \bigoplus_B F. This mapping is clearly surjective, and also clearly injective. So V \cong_F \bigoplus_B F.

Certainly \prod_B F is an F-vector space which contains \bigoplus_B F, so that \mathsf{dim}\ \bigoplus_B F \leq \mathsf{dim}\ \prod_B F. Suppose these two dimensions are in fact equal. Identify B with the usual basis of \bigoplus_B F. By this previous exercise, there is a basis D of \prod_B F which contains B, and as argued above, \prod_B F \cong_F \bigoplus_D F. By our hypothesis, in fact B and D have the same cardinality, and so there exists a bijection \theta : B \rightarrow D. Now \theta induces a vector space isomorphism \Theta : \bigoplus_B F \rightarrow \prod_B F.

However, note that |\prod_B F| = |F|^{|B|}, while |\bigoplus_B F| = |\bigcup_{T \subseteq \mathcal{P}(X), T\ \mathrm{finite}} \prod_T F| = \sum_{|B|} |F|^{|T|} = \sum_{|B|} |F| = |B| \cdot |F|. Since |B| \cdot |F| < |F|^{|B|}, we have a contradiction. Thus the dimension of \bigoplus_B F is strictly smaller than that of \prod_B F.

The interaction of Hom with direct sums and direct products

Let R be a ring with 1. Let A be a left unital R-module, and let \{B_i\}_I be a nonempty family of left unital R-modules. Prove that, as abelian groups, \mathsf{Hom}_R(\bigoplus_I B_i, A) \cong \prod_I \mathsf{Hom}_R(B_i,A) and \mathsf{Hom}_R(A, \prod_I B_i) \cong \prod_I \mathsf{Hom}_R(A,B_i). Prove also that if R is commutative, these pairs are R-module isomorphic.


First we show that \mathsf{Hom}_R(\bigoplus_I B_i, A) \cong \prod_I \mathsf{Hom}_R(B_i,A) as abelian groups. Recall that for each i, we have the canonical injection \iota_i : B_i \rightarrow \bigoplus_I B_i. Define for each i \in I the map \varphi_i : \mathsf{Hom}_R( \bigoplus_I B_i, A) \rightarrow \mathsf{Hom}_R(B_i,A) by \varphi_i(\alpha) = \alpha \circ \iota_i; certainly each \varphi_i is well defined. By the universal property of direct products of abelian groups, there exists a unique group homomorphism \Phi : \mathsf{Hom}_R(\bigoplus_I B_i, A) \rightarrow \prod_I \mathsf{Hom}_R(B_i, A) such that \pi_i \circ \Phi = \varphi_i, where \pi_i denotes the ith natural projection from a direct product. We claim that this \Phi is a group isomorphism.

Suppose \alpha \in \mathsf{ker}\ \Phi, so \Phi(\alpha) = 0. Then (\pi_i \circ \Phi)(\alpha) = 0 for each i, so that \varphi_i(\alpha) = 0 for each i. Thus \alpha \circ \iota_i = 0 for all i. That is, \alpha applied to the ith component of any element in \bigoplus_I B_i is zero, for all i. So \alpha = 0, and thus \mathsf{ker}\ \Phi = 0. So \Phi is injective.

Now suppose \psi = (\psi_i) \in \prod_I \mathsf{Hom}_R(B_i,A). Define \alpha_\psi : \bigoplus_I B_i \rightarrow A by \alpha_\psi(b_i) = \sum \psi_i(b_i); this map is well defined since only finitely many terms of (b_i) are nonzero. Moreover, it is clear that \alpha_\psi is an R-module homomorphism, so in fact \alpha_\psi \in \mathsf{Hom}_R(\bigoplus_I B_i, A). Note that for all i \in I, \Phi(\alpha_\psi)_i(b) = \varphi_i(\alpha_\psi)(b) = (\alpha_\psi \circ \iota_i)(b) = \alpha_\psi(\iota_i(b)) = \psi_i(b). Thus \Phi(\alpha_\psi)_i = \psi_i for all i, and so \Phi(\alpha_\psi) = \psi. Thus \Phi is surjective.

So \Phi is an isomorphism of abelian groups. Suppose now that R is commutative, so that both \mathsf{Hom}_R(\bigoplus_I B_i,A) and \prod_I \mathsf{Hom}_R(B_i,A) are naturally left R-modules. For all r \in R and all module homomorphisms \alpha : \bigoplus_I B_i \rightarrow A, we have \Phi(ra) = ((r \alpha) \circ \iota_i) = (r(\alpha \circ \iota_i)) = r(\alpha \circ \iota_i) = r\Phi(\alpha). Thus \Phi is an isomorphism of left R-modules.

Next we show that \mathsf{Hom}_R(A, \prod_I B_i) \cong \prod_I \mathsf{Hom}_R(A,B_i) as abelian groups. For each i \in I, define \varphi_i : \mathsf{Hom}_R(A, \prod_I B_i) \rightarrow \mathsf{Hom}_R(A,B_i) by \varphi_i(\alpha) = \pi_i \circ \alpha, where \pi_i denotes the ith canonical projection from a direct product. By the universal property of direct products, we have a unique group homomorphism \Phi : \mathsf{Hom}_R(A,\prod_I B_i) \rightarrow \prod_I \mathsf{Hom}_R(A, B_i) such that \pi_i \circ \Phi = \varphi_i for all i. We claim that this \Phi is an isomorphism.

Suppose \alpha \in \mathsf{ker}\ \Phi. So \Phi(\alpha) = 0, and thus (\pi_i \circ \Phi)(\alpha) = 0 for all i. Thus \varphi_i(\alpha) = 0 for all i, so \pi_i \circ \alpha = 0 for all i. That is, every coordinate of any element in the image of \alpha is 0. So \alpha = 0. Thus \mathsf{ker}\ \Phi = 0, and so \Phi is injective.

Now let \psi = (\psi_i) \in \prod_I \mathsf{Hom}_R(A,B_i). Define \alpha_\psi : A \rightarrow \prod_I B_i by \alpha_\psi(a)_i = \psi_i(a). Certainly \alpha_\psi is a module homomorphism. Note that, for all i, \Phi(\alpha_\psi)_i(a) = \varphi_i(\alpha_\psi)(a) = (\pi_i \circ \alpha_\psi)(a) = \psi_i(a). So \Phi(\alpha_\psi)_i = \psi_i for all i, and we have \Phi(\alpha_\psi) = \psi. So \Phi is surjective, and thus an isomorphism of groups.

Finally, suppose R is commutative. If r \in R and \alpha : A \rightarrow \prod_I B_i is a module homomorphism, then \Phi(r\alpha) = (\pi_i \circ r\alpha) = (r(\pi_i \circ \alpha)) = r(\pi_i \circ \alpha) = r\Phi(\alpha). Thus in this case \Phi is an isomorphism of R-modules.

As a ring, the RR-tensor square of CC is isomorphic to the direct square of CC

Let \mathbb{C} be an \mathbb{R}-algebra via the inclusion map and consider \mathbb{C} \otimes_\mathbb{R} \mathbb{C}. Let e_1 = 1 \otimes 1, e_2 = 1 \otimes i, e_3 = i \otimes 1, and e_4 = i \otimes i. Note that every simple tensor (hence every element) of \mathbb{C} \otimes_\mathbb{R} \mathbb{C} can be written in the form a_1e_1+a_2e_2+a_3e_3+a_4e_4 for some real numbers a_i.

  1. Given two elements \alpha = a_1e_1+a_2e_2+a_3e_3+a_4e_4 and \beta = b_1e_1+b_2e_2+b_3e_3+b_4e_4 in \mathbb{C} \otimes_\mathbb{R} \mathbb{C}, compute their product \alpha\beta as a linear combination of the e_i.
  2. Let \varepsilon_1 = \frac{1}{2}(e_1 + e_4) and \varepsilon_2 = \frac{1}{2}(e_1 - e_4). Show that \varepsilon_1\varepsilon_2 = 0, \varepsilon_1 + \varepsilon_2 = 1, \varepsilon_1^2 = \varepsilon_1, and \varepsilon_2^2 = \varepsilon_2. Deduce that as rings, \mathbb{C} \otimes_\mathbb{R} \mathbb{C} \cong (\varepsilon_1) \times (\varepsilon_2).
  3. Prove that \varphi : \mathbb{C} \times \mathbb{C} \rightarrow \mathbb{C} \times \mathbb{C} given by \varphi(z,w) = (zw, z \overline{w}) is \mathbb{R}-bilinear, where overbar denotes complex conjugation.
  4. Show that the \mathbb{R}-module homomorphism \Phi : \mathbb{C} \otimes_\mathbb{R} \mathbb{C} \rightarrow \mathbb{C} \times \mathbb{C} induced by \varphi is also \mathbb{C}-linear, taking care to note how \mathbb{C} acts on \mathbb{C} \otimes_\mathbb{R} \mathbb{C} and \mathbb{C} \times \mathbb{C}. Show that \Phi is bijective.

We can give the following multiplication table for the e_1 (with the left factor on the top row and the right factor on the left column):

e_1 e_2 e_3 e_4
e_1 e_1 e_2 e_3 e_4
e_2 e_2 -e_1 e_4 -e_3
e_3 e_3 e_4 -e_1 -e_2
e_4 e_4 -e_3 -e_2 e_1

Evidently, then, \alpha\beta = (a_1b_1-a_2b_2-a_3b_3+a_4b_4)e_1 + (a_1b_2+a_2b_1-a_3b_4-a_4b_3)e_2 + (a_1b_3-a_2b_4+a_3b_1-a_4b_3)e_3 + (a_1b_4+a_2b_3+a_3b_2+a_4b_1)e_4.

Evidently, we have \varepsilon_1^2 = \varepsilon_1, \varepsilon_2^2 = \varepsilon_2, and \varepsilon_2 = 1 - \varepsilon_1. By this previous exercise, we have a ring isomorphism \mathbb{C} \otimes_\mathbb{R} \mathbb{C} \rightarrow (\varepsilon_1) \times (\varepsilon_2).

Now \varphi(z,w) = (zw,z\overline{w}) is clearly \mathbb{R} bilinear (since real numbers are fixed by complex conjugation). Letting \Phi be the induced \mathbb{R}-module homomorphism \mathbb{C} \otimes_\mathbb{R} \mathbb{C} \rightarrow \mathbb{C} \times \mathbb{C}, we in fact have \Phi(v(z \otimes w)) = \Phi(vz \otimes w) = (vzw, vz\overline{w}) = v(zw,z\overline{w}) = v \Phi(z \otimes w), so that \Phi is a \mathbb{C}-module homomorphism. Evidently, \Phi(\varepsilon_1) = (0,1) and \Phi(\varepsilon_2) = (1,0). In particular, \Phi is surjective, since (0,1) and (1,0) generate \mathbb{C} \times \mathbb{C} as a \mathbb{C}-module. Now \Phi((z_1 \otimes w_1)(z_2 \otimes w_2)) = \Phi(z_1z_2 \otimes w_1w_2) = (z_1z_2w_1w_2, z_1z_2\overline{w_1w_2}) = (z_1w_1z_2w_2, z_1\overline{w_1}z_2\overline{w_2}) = (z_1w_1,z_1\overline{w_1})(z_2w_2,z_2\overline{w_2}) = \Phi(z_1 \otimes w_1) \Phi(z_2 \otimes w_2). Thus \Phi is a ring homomorphism.

Finally, suppose \alpha = a_1e_1 + a_2e_2 + a_3e_3 + a_4e_4 is in \mathsf{ker}\ \Phi. Evidently, (0,0) = \Phi(\alpha) = ((a_1-a_4)+(a_2+a_3)i, (a_1+a_4)+(a_2-a_3)i). Thus a_1 = a_4, and so 2a_1 = 0, and thus a_1 = 0. Likewise, a_2, a_3, and a_4 are all zero. Thus \alpha = 0, and so \mathsf{ker}\ \Phi = 0. So \Phi is injective.

Thus, as rings, we have \mathbb{C} \otimes_\mathbb{R} \mathbb{C} \cong \mathbb{C} \times \mathbb{C}.

Over a commutative ring, the tensor product of a module by a free module of finite rank is a direct power

Let R be a commutative ring and let N be a free R-module of finite rank n; say B = \{e_i\}_{i=1}^n is a basis for N.

  1. Let M be any nonzero R-module. (Since R is commutative, M is naturally an (R,R)-bimodule.) Show that M \otimes_R N \cong_R M^n. In particular, every element of M \otimes_R N can be written uniquely in the form \sum_{i=1}^n m_i \otimes e_i.
  2. Show that if \sum m_i \otimes n_i = 0, where the n_i are merely assumed to be R-linearly independent in N, it need not be the case that the m_i are all zero. (That is, it is crucial that the n_i generate N as an R-module.)

Define a mapping \varphi_B : M \times N \rightarrow M^n by \varphi_B(m, \sum r_ie_i) = (r_i \cdot m). (This is well defined since B is a basis for N.) Moreover, \varphi_B(m_1 + m_2, \sum r_ie_i) = (r_i(m_1 + m_2)) = (r_i m_1) + (r_i m_2) = \varphi_b(m_1, \sum r_ie_i) + \varphi_b(m_2, \sum r_ie_i), \varphi(m, (\sum r_i e_i) + (\sum s_i e_i)) = \varphi_B(m, \sum (r_i+s_i)e_i) = ((r_i+s_i)m) = (r_im) + (s_im) = \varphi_B(m, \sum r_ie_i) + \varphi_B(m, \sum s_i e_i), and \varphi_B(m \cdot a, \sum r_i e_i) = (r_i (ma)) = (ar_i m) = \varphi_B(m, \sum ar_ie_i) = \varphi_B(m, a \sum r_ie_i), so that \varphi_B is bilinear. By the universal property of tensor products, \varphi_B induces a unique group homomorphism (indeed, R-module homomorphism) \Phi_B : M \otimes_R N \rightarrow M^n such that \Phi_B(m \otimes \sum r_i e_i) = (r_im).

Now define \Psi_B : M^n \rightarrow M \otimes_R N by (m_i) \mapsto \sum m_i \otimes e_i. Clearly \Psi_B is an R-module homomorphism.

Moreover, note that (\Psi_B \circ \Phi_B)(m \otimes \sum r_i e_i) = \Psi_B(r_im)) = \sum r_im \otimes e_i = \sum m \otimes r_ie_i = m \otimes \sum r_ie_i. Similarly, (\Phi_B \circ \Psi_b)(m_i) = \Phi_B(\sum m_i \otimes e_i) = \sum \Phi_B(m_i \otimes e_i) = (m_i). Thus \Phi_B and \Psi_B are mutual inverses, and we have M \otimes_R N \cong_R M^n.

In particular, \sum m_i \otimes e_i = 0 if and only if m_i = 0 for all i.

Now for the counterexample, let R = \mathbb{Z}, let N = \mathbb{Z}^1 be the free \mathbb{Z}-module of rank 1, and let M = \mathbb{Z}/(2) be a \mathbb{Z}-module in the usual way. Consider the simple tensor \overline{1} \otimes 2 in M \otimes_R N; note that 1 \neq 0, but that \overline{1} \otimes 2 = \overline{1} \otimes 2 \cdot 1 = \overline{1} \cdot 2 \otimes 1 = \overline{2} \otimes 1 = 0 \otimes 1 = 0.

Direct products of free modules need not be free

Let M = \prod_{\mathbb{N}} \mathbb{Z} and let N = \bigoplus_\mathbb{N} \mathbb{Z}. Consider M to be a left R = \mathbb{Z}-module in the usual way (so M is the direct product of free modules), and consider N \subseteq M as a submodule. Suppose M is free as a \mathbb{Z}-module with basis \mathcal{B}.

  1. Show that N is countable.
  2. Show that there is a countable subset \mathcal{B}_1 \subseteq \mathcal{B} such that N is contained in N_1 = \sum_{\mathcal{B}_1} Rb. Prove also that N_1 is countable.
  3. Show that M/N_1 is free as a \mathbb{Z}-module. Deduce that if x+N_1 \in M/N_1 is nonzero, then there exist at most finitely many k \in \mathbb{Z} such that x = km + N_1 for some m+N_1.
  4. Let \mathcal{S} = \{ (b_i) \ |\ b_i = \pm i! \}. Prove that \mathcal{S} is uncountable. Deduce that there exists some s \in \mathcal{S} such that s \notin N_1.
  5. Show that for every k \in \mathbb{Z}, there exists m + N_1 \in M/N_1 such that s+N_1 = km+N_1. Deduce that M is not free as a \mathbb{Z} module.

  1. For each n \in \mathbb{N}^+, we have a natural inclusion \iota_n : \mathbb{Z}^n \rightarrow N given by \iota_n(a)_i = a_{i+1} if 0 \leq i < n and 0 otherwise. Certainly each element of N is in the image of some \iota_n. Thus N = \bigcup_{n \in \mathbb{N}^+} \mathsf{im}\ \iota_n. In particular, N is a countable union of countable sets, and hence is countable.
  2. For each x \in N, there is a finite subset B_x \subseteq \mathcal{B} such that x = \sum_{b \in B_x} r_b \cdot b. Let \mathcal{B}_1 = \bigcup_{x \in N} B_x; as the countable union of finite sets, \mathcal{B}_1 is countable. Certainly N \subseteq N_1 = \sum_{b \in \mathcal{B}_1} Rb. Certainly N_1 is also countable.
  3. We wish to show that M/N_1 is free. To that end, note that we have the natural inclusion \iota : \mathcal{B}\setminus\mathcal{B}_1 \rightarrow M/N_1 given by b \mapsto b+N_1. By the universal property of free modules, there is a unique R-module homomorphism \Phi : F(\mathcal{B}\setminus \mathcal{B}_1) \rightarrow M/N_1 such that \Phi(b) = b+N_1 for all b \in \mathcal{B} \setminus \mathcal{B}_1. We claim that \Phi is an isomorphism. To that end, let x \in \mathsf{ker}\ Phi. Say x = \sum r_ib_i. Now 0 = \sum r_ib_i +N_1, so that \sum r_ib_i \in N_1. Thus r_i = 0, and we have x = 0. So \Phi is injective. Now let x+N_1 \in M/N_1; since M is free on \mathcal{B}, we have x+N_1 = \sum r_ib_i + N_1 for some r_i and b_i \in \mathcal{B} \setminus \mathcal{B}_1. Certainly then \Phi(\sum r_ib_i) = x, so that \Phi is surjective. Thus \Phi is an R-module isomorphism, and M/N_1 is free as a \mathbb{Z}-module.
  4. We will now prove some lemmas.

    Lemma 1: If n \in \mathbb{Z} is divisible by infinitely many integers, then n = 0. Proof: Certainly 0 is divisible by infinitely many integers, since 0 = 0 \cdot k for all k. Suppose now that n is nonzero and that for some infinite set K \subseteq \mathbb{Z}, k|n for all k \in K. There must exist some k \in K with k > n, a contradiction. \square

    Lemma 2: Let F be a free \mathbb{Z}-module with free basis \mathcal{B} = \{b_i\}_I. If x \in F is nonzero, then there are only finitely many k \in \mathbb{Z} such that x = k \cdot m for some m \in F. Proof: Suppose to the contrary that a nonzero x \in F exists such that, for infinitely many k, there exists m \in F such that x = k \cdot m. Say x = \sum r_ib_i. Since x \neq 0, some a_i is nonzero. Looking at the b_i coefficient in the (unique) expansion of km, we have a_i = km_i for infinitely many integers k. By the lemma, a_i = 0, a contradiction. \square.

    We certainly have a bijection \psi : \mathcal{S} \rightarrow \prod_\mathbb{N} \{1,-1\} = T. A diagonalization argument shows that T is uncountable as follows. Suppose \theta : \mathbb{N} \rightarrow T is a bijection. Construct \epsilon = (e_i) \in T as follows: e_i = 1 if \theta(i)_i = -1 and -1 otherwise. Since \epsilon differs from every element in \mathsf{im}\ \theta = T in some component, \epsilon is not in \mathsf{im}\ T, a contradiction. Thus \mathcal{S} is uncountable. Since N_1 is countable, there must exist s \in \mathcal{S} such that s \notin N_1.

  5. Let k \in \mathbb{Z} be positive. Define m \in M by m_i = i! if i! < k and \sigma_ii!/k if i \geq k, where \sigma_i = 1 if s_i > 0 and -1 if s_i < 0. Now (s-km)_i = 0 for all i > k, so that s-km \in N \subseteq N_1. Thus s + N_1 = k(m+N_1). This is a contradiction of part (4) above. Thus M is not free as a \mathbb{Z}-module.

Arbitrary direct products and direct sums of modules are modules

Let R be a ring and let \{M_i\}_{i \in I} be a family of left R-modules indexed by a (nonempty) set I. Prove that \prod_I M_i is a left R-module via the action r \cdot (m_i) = (r \cdot m_i) (Called the direct product of the modules M_i). Moreover, prove that the subset \bigoplus_I M_i = \{ (a_i) \in \prod_I M_i \ |\ a_i = 0\ \mathrm{for\ all\ but\ finitely\ many}\ i \} is a submodule of \prod_I M_i (Called the direct sum of the modules M_i). Moreover, show that \prod_{n \in \mathbb{N}} \mathbb{Z}/(n) and \bigoplus_{n \in \mathbb{N}} \mathbb{Z}/(n) are not isomorphic as \mathbb{Z}-modules.


We saw in this previous exercise that \prod_I M_i is an abelian group under pointwise addition and multiplication. Thus it suffices to show that the three left-module axioms are satisfied. To that end, let r,s \in R and let (m_i), (n_i) \in \prod_I M_i. We have (r+s) \cdot (m_i) = ((r+s) \cdot m_i) = (r \cdot m_i + s \cdot m_i) = (r \cdot m_i) + (s \cdot m_i) = r \cdot (m_i) + s \cdot (m_i), (rs) \cdot (m_i) = ((rs) \cdot m_i) = (r \cdot (s \cdot m_i)) = r \cdot (s \cdot m_i) = r \cdot (s \cdot (m_i)), and r \cdot ((m_i) + (n_i)) = r \cdot (m_i + n_i) = (r \cdot (m_i + n_i)) = (r \cdot m_i + r \cdot n_i) = (r \cdot m_i) + (r \cdot n_i) = r \cdot (m_i) + r \cdot (n_i). Thus \prod_I M_i is a left R-module. Suppose further that R has a 1 and that each M_i is unital; then 1 \cdot (m_i) = (1 \cdot m_i) = (m_i), for all (m_i), so that \prod_I M_i is unital.

Now we will use the submodule criterion to show that \bigoplus_I M_i is a submodule of \prod_I M_i. To that end, note first that \bigoplus_I M_i is nonempty since 0 \in \bigoplus_I M_i. (The set of indices such that 0 = (0_i) is nonzero is empty, which is certainly finite). Now let (x_i), (y_i) \in \bigoplus_I M_i and let r \in R. Say J,K \subseteq I such that x_j = 0 for j \notin J and y_k = 0 for k \notin K. Consider (x_i) + r \cdot (y_i) = (x_i + r \cdot y_i). Note that if i \notin J \cup K, then x_i + r \cdot y_i = 0. Since J \cup K \subseteq I is finite, (x_i) + r \cdot (y_i) \in \bigoplus_I M_i. By the submodule criterion, \bigoplus_I M_i \subseteq \prod_I M_i is an R-submodule.

Now consider M = \bigoplus_\mathbb{N} \mathbb{Z}/(n) and N = \prod_\mathbb{N} \mathbb{Z}/(n) as \mathbb{Z}-modules in the natural way. We intend to show that these two modules are not isomorphic by showing that one is torsion while the other is not. To that end, let (a_i) \in M. There is a natural number N such that, for all k \geq N, a_k = 0. Now let t = \prod_{n=1}^N n (where this product is taken over the natural numbers). Certainly then t \cdot (a_i) = 0; that is, every element of M is torsion, so that M is torsion. However, note that t \cdot (1) = (t) is nonzero for all integers t since (for instance) the |t|-1 component of (t) is 1. So N is not torsion. Thus M and N are not isomorphic as \mathbb{Z}-modules.

Hom commutes with finite direct products in both slots

Let R be a ring with 1 and let A, B, and M be unital left R-modules. Prove that \mathsf{Hom}_R(A \times B, M) \cong_R \mathsf{Hom}_R(A,M) \times \mathsf{Hom}_R(B,M) and \mathsf{Hom}_R(M,A \times B) \cong_R \mathsf{Hom}_R(M,A) \times \mathsf{Hom}_R(M,B). (Where \cong_R means “isomorphic as R-modules”.)


Given R-module homomorphisms \alpha : A \rightarrow M and \beta : B \rightarrow M, define \Phi(\alpha,\beta) : A \times B \rightarrow M by \Phi(\alpha,\beta)(a,b) = \alpha(a) + \beta(b). Note that, for all a_1,a_2 \in A, b_1,b_2 \in B, and r \in R, we have

\Phi(\alpha,\beta)((a_1,b_1) + r \cdot (a_2,b_2))  =  \Phi(\alpha,\beta)((a_1 + r \cdot a_2, b_1 + r \cdot b_2))
 =  \alpha(a_1 + r \cdot a_2) + \beta(b_1 + r \cdot b_2)
 =  \alpha(a_1) + r \cdot \alpha(a_2) + \beta(b_1) + r \cdot \beta(b_2)
 =  \Phi(\alpha,\beta)(a_1,b_1) + r \cdot \Phi(\alpha,\beta)(a_2,b_2).

So \Phi(\alpha,\beta) : A \times B \rightarrow M is an R-module homomorphism. Consider the map \Phi : \mathsf{Hom}_R(A,M) \times \mathsf{Hom}_R(B,M) \rightarrow \mathsf{Hom}_R(A \times B, M). We claim that \Phi is an R-module homomorphism. To see this, let \alpha_1,\alpha_2 : A \rightarrow M and \beta_1,\beta_2 : B \rightarrow M be R-module homomorphisms and let r \in R. Now

\Phi((a_1,b_1) + r \cdot (a_2,b_2))(a,b)  =  \Phi(\alpha_1 + r \cdot \alpha_2, \beta_1 + r \cdot \beta_2)(a,b)
 =  (\alpha_1 + r \cdot \alpha_2)(a) + (\beta_1 + r \cdot \beta_2)(b)
 =  \alpha_1(a) + r \cdot \alpha_2(a) + \beta_1(b) + r \cdot \beta_2(b)
 =  \Phi(\alpha_1,\beta_1)(a,b) + r \cdot \Phi(\alpha_2,\beta_2)(a,b)
 =  (\Phi(\alpha_1,\beta_1) + r \cdot \Phi(\alpha_2,\beta_2))(a,b).

So \Phi is an R-module homomorphism. Now suppose (\alpha,\beta) \in \mathsf{ker}\ \Phi. In particular, 0 = \Phi(\alpha,\beta)(a,0) = \alpha(a) for all a \in A, so that \alpha = 0. Similarly, \beta = 0. Thus (\alpha,\beta) = 0, and in fact \mathsf{ker}\ \Phi = 0. Hence \Phi is injective. Finally, let \theta : A \times B \rightarrow M be an R-module homomorphism, and define \alpha_\theta : A \rightarrow M and \beta_\theta : B \rightarrow M by \alpha_\theta(a) = \theta(a,0) and \beta_\theta(b) = \theta(0,b). Note that \alpha_\theta(a_1 + r \cdot a_2) = \theta(a_1 + r \cdot a_2,0 = \theta(a_1,0) + r \cdot \theta(a_2,0) = \alpha_\theta(a_1) + r \cdot \alpha_\theta(a_2), so that \alpha_\theta is an R-module homomorphism. Similarly, \beta_\theta is an R-module homomorphism. Since \Phi(\alpha_\theta, \beta_\theta)(a,b) = \alpha_\theta(a) + \beta_\theta(b) = \theta(a,0) + \theta(0,b) = \theta(a,b), we have \Phi(\alpha_\theta,\beta_\theta) = \theta. Hence \Phi is surjective. Thus, we have \mathsf{Hom}_R(A \times B, M) \cong_R \mathsf{Hom}_R(A,M) \times \mathsf{Hom}_R(B,M).

Now let \pi_A : A \times B \rightarrow A and \pi_B : A \times B \rightarrow B be the left and right coordinate projections; certainly these are R-module homomorphisms. Define \Psi : \mathsf{Hom}_R(M,A \times B) \rightarrow \mathsf{Hom}_R(M,A) \times \mathsf{Hom}_R(M,B) by \Psi(\varphi) = (\pi_A \circ \varphi, \pi_B \circ \varphi). If \varphi, \psi \in \mathsf{Hom}_R(M,A \times B) and r \in R, then

\Psi(\varphi + r \cdot \psi)  =  (\pi_A \circ (\varphi + r \cdot \psi), \pi_B \circ (\varphi + r \cdot \psi))
 =  (\pi_A \circ \varphi + r \cdot (\pi_A \circ \psi), \pi_B \circ \varphi + r \cdot (\pi_B \circ \psi))
 =  (\pi_A \circ \varphi, \pi_A \circ \psi) + r \cdot (\pi_B \circ \varphi, \pi_B \circ \psi)
 =  \Psi(\varphi) + r \cdot \Psi(\psi).

Thus \Psi is an R-module homomorphism. Now suppose \varphi \in \mathsf{ker}\ \Psi. In particular, if m \in M, then 0 = \Psi(\varphi)(m,m) = ((\pi_A \circ \varphi)(m), (\pi_b \circ \psi)(m)) = (\pi_A(\varphi(m)), \pi_B(\varphi(m)) = \varphi(m). So \varphi = 0, hence \mathsf{ker}\ \Psi = 0, and thus \Psi is injective. Finally, suppose (\alpha,\beta) \in \mathsf{Hom}_R(M,A) \times \mathsf{Hom}_R(M,B). Define \varphi_{\alpha,\beta} : M \rightarrow A \times B by \varphi_{\alpha,\beta}(m) = (\alpha(m), \beta(m). Certainly \varphi_{\alpha,\beta} is an R-module homomorphism. Moreover, we have \Psi(\varphi_{\alpha,\beta})(m) = ((\pi_A \circ \varphi_{\alpha,\beta})(m), (\pi_B \circ \varphi_{\alpha,\beta})(m)) = (\alpha(m), \beta(m)). Thus \Psi(\varphi_{\alpha,\beta}) = (\alpha,\beta), and so \Psi is surjective. Hence \mathsf{Hom}_R(M, A \times B) \cong_R \mathsf{Hom}_R(M,A) \times \mathsf{Hom}_R(M,B).

Prove that a subset is a submodule

Let R be a ring with 1, let n \geq 1, and let M = R^n. Recall that M is a left R-module via the action r \cdot (a_i) = (ra_i). Let \{I_k\}_{k=1}^n be a family of ideals of R. Prove that the following subsets of R^n are submodules.

  1. N = \prod_{k=1}^n I_k
  2. N = \{(a_i) \in R^n \ |\ \sum a_i = 0\}

We use the submodule criterion for both examples.

  1. Note that 0 \in I_k for each k. Then (0) \in \prod I_k; hence \prod I_k is nonempty. Now suppose (a_i), (b_i) \in \prod I_k, and let r \in R. Because each I_k is an ideal, a_k + rb_k \in I_k for each k. Then (a_i) + r \cdot (b_i) = (a_i + rb_i) \in \prod I_k. Thus \prod I_k is a submodule of M.
  2. Certainly \sum_{i=1}^n 0 = 0, so that (0) \in N. Now suppose (a_i), (b_i) \in N; then \sum a_i = \sum b_i = 0. Let r \in R. Then \sum a_i+rb_i = (\sum a_i) + r(\sum b_i) = 0, and hence (a_i) + r \cdot (b_i) = (a_i+rb_i) \in N. Thus N is a submodule of M.