## 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 $b$th 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 $i$th 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 $i$th 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 $i$th 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 $M$.