Maxmin Strategy in Battle of the Sexes

Kevin Leyton-Brown and Yoav Shoham Essentials of Game Theory, page 15:

Given the following payoff table for a typical Battle of the Sexes game, calculate the maxmin strategies and values for each player.

	L	W
L	2 1	0 0
W	0 0	1 2

Proof: Player 2 must minimize $$u_1(s)=2s_1(L)s_2(L)+s_1(W)s_2(W)=$$ $$2s_1(L)s_2(L)+(1-s_1(L))(1-s_2(L))=$$ $$s_2(L)(3s_1(L)-1)+1-s_1(L)$$ When $s_1(L) > 1/3$ and $3s_1(L)-1$ is positive, player 2 minimizes by playing $s_2(L)=0$ and now the expression collapses to $1-s_1(L)$ so that $u_1(s) < 2/3$. If $s_1(L) < 1/3$ and $3s_1(L)-1$ is negative, player 2 minimizes with $s_2(L)=1$ and the expression collapses to $2s_1(L)$ showing $u_1(s) < 2/3$. When $s_1(L)=1/3$ then the expression invariably maximizes as $u_1(s)=2/3$. This shows $$s_1(L)=1/3$$ $$s_1(W)=2/3$$ is player 1's maxmin strategy with maxmin value $2/3$. Symmetrically, similar holds for player 2.$~\square$

Tensor Products and Extensions of Polynomial Ring Quotients (10.4.26)

Dummit and Foote Abstract Algebra, section 10.4, exercise 26:

Let $S$ and $R \subseteq S$ be commutative rings with $1_R = 1_S$, and let $I \subseteq R[x_1,...,x_n]$ be an ideal. Prove $$S \otimes_R (R[x_1,...,x_n]/I) ≅ S[x_1,...,x_n]/IS[x_1,...,x_n]$$ as $S$-algebras.

Proof: We must prove there is an isomorphism of rings and of $S$-modules between the two. Define a mapping $$φ : S \times (R[x_1,...,x_n]/I) → S[x_1,...,x_n]/IS[x_1,...,x_n]$$ $$(s,\overline{r(x)}) \mapsto \overline{\overline{sr(x)}}$$ This map is well defined since $$(s,\overline{r(x)}) = (s,\overline{r(x)+i}) ⇒$$ $$φ(s,\overline{r(x)}) = φ(s,\overline{r(x)+i}) - is⇒$$ $$\overline{\overline{φ(s,\overline{r(x)})}} = \overline{\overline{φ(s,\overline{r(x)+i})}}$$ This map is evidently $R$-balanced, and thus induces a homomorphism $\Phi$ between the targets. For each $s(x)= \overline{\overline{\sum s_ix^i}}∈S[x_1,...,x_n]/IS[x_1,...,x_n]$ we can choose a preimage $\sum s_i \otimes \overline{x^i}∈S \otimes (R[x_1,...,x_n]/I)$ so that this homomorphism is surjective. Introduce the homomorphism $$S[x_1,...,x_n] → S \otimes_R (R[x_1,...,x_n]/I)$$ $$\sum s_ix^i \mapsto \sum s_i \otimes \overline{x^i}$$ which is seen to factor through $IS[x_1,...,x_n]$ so that there is an inverse to $\Phi$ making it injective, as well.

Now we may demonstrate $$s\Phi(\sum s_i \otimes \overline{r_i(x)})=\sum \overline{\overline{ss_ir_i(x)}} = \Phi(s\sum s_i \otimes \overline{r_i(x)})$$ so that $\Phi$ is an $S$-module isomorphism. Finally, to show the ring isomorphism, we have $$\Phi(\sum s_i \otimes \overline{r_i(x)})\Phi(\sum s_j \otimes \overline{r_j(x)})=\sum \sum \overline{\overline{s_is_jr_i(x)r_j(x)}}=$$ $$\Phi(\sum s_i \otimes \overline{r_i(x)} \cdot \sum s_j \otimes \overline{r_j(x)})~\square$$

Nonsimple Tensors (10.4.20)

Dummit and Foote Abstract Algebra, section 10.4, exercise 20:

Let $R=\mathbb{Z}[x]$ and let $I=(2,x)$. Show that the element $2 \otimes 2 + x \otimes x$ in $I \otimes_R I$ is not a simple tensor.

Proof: Define a map $$I \times I → R$$ $$(i,j) \mapsto ij$$ This is clearly seen to be $R$-balanced, and so induces a homomorphism $\Phi$ on $I \otimes_R I$. Assuming $2 \otimes 2 + x \otimes x = a \otimes b$, then also $$\Phi(2 \otimes 2 + x \otimes x)=\Phi(2 \otimes 2) + \Phi(x \otimes x)=x^2+4=\Phi(a \otimes b)=ab$$ Since $x^2+4$ is a monomial quadratic with no roots, it does not factor in $R$, and thus either $a$ or $b$ is $\pm 1$ and now $a \otimes b∉I \otimes I$.$~\square$

Linear Associativity and Tensor Products (10.4.12)

Dummit and Foote Abstract Algebra, section 10.4, exercise 12:

Let $V$ be a vector space over a field $F$ and let $v,v'∈V$ be nonzero. Prove $v \otimes_F v' = v' \otimes_F v ⇔ v=av'$ for some $a∈F$.

Proof: ($\Leftarrow$) $$v' \otimes v = v' \otimes av' = av' \otimes v' = v \otimes v'$$ ($⇒$) We may assume $V$ has a basis $\{e_i\}_{i∈I}$.

Let $v=\sum_{j∈J} f_je_j$ and $v'=\sum_{j∈J} f_j'e_j$ for appropriate, finite $J \subseteq I$. Let $W = \sum_{j∈J}Fe_j$, so that $W≅F^n$ for some finite $n$.

Now, $v \otimes v'=v \otimes \sum_{j∈J} f_j'e_j = \sum_{j∈J} v \otimes f_j'e_j = \sum_{j∈J} f_j'v \otimes e_j$, and similarly $v' \otimes v = \sum_{j∈J} f_jv' \otimes e_j$. We have $v \otimes v',v' \otimes v ∈ W \otimes W≅F^n \otimes F^n$ and by 10.4.10(a) we have $f_j'v=f_jv'$ for all $j∈J$ and since $v,v'≠0$ we may say $v=(f_j'^{-1}f_j)v'$.

Element Collapsing Between Tensor Products (10.4.2)

Dummit and Foote Abstract Algebra, section 10.4, exercise 2:

Show $2 \otimes 1=0∈\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$ but $2 \otimes 1≠0∈\mathbb{2Z} \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$.

Proof: For the first, notice $2 \otimes 1 = 1 \cdot 2 \otimes 1 = 1 \otimes 2 \cdot 1 = 1 \otimes 2 = 1 \otimes 0 = 0$.

Now, if $2 \otimes 1$ is zero in the latter tensor product, then by natural extension as a $\mathbb{Z}$-module, for an arbitrary element we have $2a \otimes b = (ab)(2 \otimes 1) = 0$ so that $\mathbb{2Z} \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z}$ is zero. However, we have $$\mathbb{2Z} \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z} ≅ \mathbb{Z}/2\mathbb{Z} \otimes_\mathbb{Z} \mathbb{2Z} ≅ \mathbb{2Z}/((\mathbb{2Z})(\mathbb{2Z})) ≅ \mathbb{2Z}/\mathbb{4Z} \not ≅ 0~\square$$

Universal Multilinear Object (10.4 Corollary 16)

Dummit and Foote Abstract Algebra, section 10.4, corollary 16:

Show that if $R$ is a commutative ring, $A_i$ is an $R$-module, $L$ is an abelian group, and $φ : A_1 \times ... \times A_n → L$ is a multilinear map, then there exists a unique homomorphism of $R$-modules $\Phi : A_1 \otimes ... \otimes A_n → L$ such that $φ = \Phi \circ \iota$, where $\iota$ is the natural map from $A_1 \times ... \times A_n$ to $A_n \otimes ... \otimes A_n$. Conversely, show that if there is a homomorphism of $R$-modules $\Phi : A_1 \otimes ... \otimes A_n → L$ then $φ = \Phi \circ \iota$ is a multilinear map.

Proof: Lemma 1: We have the natural map on simple tensors $$\iota ' : A_1 \otimes ... \otimes A_{n-1} \times A_n → (A_1 \otimes ... \otimes A_{n-1}) \otimes A_n ≅ A_1 \otimes ... \otimes A_n$$ is well defined, and together with the evidently well-defined natural map $$\iota^* : A_1 \times ... \times A_n → A_1 \otimes ... \otimes A_{n-1} \times A_n$$ we have the relation $\iota = \iota ' \circ \iota^*$. Proof: The lemma will follow as soon as we show $\iota '$ is well defined. If $(a_1 \otimes ... \otimes a_{n-1}, a_n)=(a_1' \otimes ... \otimes a_{n-1}', a_n')$ then by using these two given equivalencies in the coordinates we clearly have $$\iota '(a_1 \otimes ... \otimes a_{n-1}, a_n)=(a_1 \otimes ... \otimes a_{n-1}) \otimes a_n =$$ $$(a_1' \otimes ... \otimes a_{n-1}') \otimes a_n' = \iota '(a_1' \otimes ... \otimes a_{n-1}', a_n')~\square$$ Proceed by induction on $n$. For each $a_n∈A_n$ define $$φ_{a_n} : A_1 \times ... \times A_{n-1} → L$$ $$φ_{a_n}(a_1,...,a_{n-1})=φ(a_1,...,a_n)$$ Note that $φ_{a_n}$ is multilinear $$φ_{a_n}(a_1,...,ra_m+r'a_m',...,a_{n-1})=φ(a_1,...,ra_m+r'a_m',...,a_n)=$$ $$rφ(a_1,...,a_m,...,a_n)+r'φ(a_1,...,a_m',...,a_n)=$$ $$rφ_{a_n}(a_1,...,a_m,...,a_{n-1})+r`φ_{a_n}(a_1,...,a_m',...,a_{n-1})$$ so that by induction there exists a unique homomorphism of $R$-modules $$\Phi_{a_n} : A_1 \otimes ... \otimes A_{n-1} → L$$ such that $φ_{a_n}=\Phi_{a_n} \circ \iota^\Delta$ where $\iota^\Delta$ is the natural map from $A_1 \times ... \times A_{n-1}$ to $A_1 \otimes ... \otimes A_{n-1}$. Now define $$σ' : A_1 \otimes ... \otimes A_{n-1} \times A_n → A_1 \otimes ... \otimes A_{n-1}$$ $$σ'(a_1 \otimes ... \otimes a_{n-1}, a_n)=a_1 \otimes ... \otimes a_{n-1}$$ $$ψ : A_1 \otimes ... \otimes A_{n-1} \times A_n → L$$ $$ψ = \Phi_{a_n} \circ σ'$$ (Where $a_n$ in the definition of $ψ$ depends on the input) Prove $ψ$ is bilinear by $$ψ(r \sum a_{i,1} \otimes ... \otimes a_{i,n-1} + r' \sum a_{i,1}' \otimes ... \otimes a_{i,n-1}',a_n)=$$ $$\Phi_{a_n}(r \sum a_{i,1} \otimes ... \otimes a_{i,n-1} + r' \sum a_{i,1}' \otimes ... \otimes a_{i,n-1}')=$$ $$r\Phi_{a_n}(\sum a_{i,1} \otimes ... \otimes a_{i,n-1})+r'\Phi_{a_n}(\sum a_{i,1}' \otimes ... \otimes a_{i,n-1}')=$$ $$rψ(\sum a_{i,1} \otimes ... \otimes a_{i,n-1},a_n)+r'ψ(\sum a_{i,1}' \otimes ... \otimes a_{i,n-1}',a_n)$$ as well as $$ψ(\sum a_{i,1} \otimes ... \otimes a_{i,n-1},ra_n+r'a_n')=\Phi_{ra_n+r'a_n'}(\sum a_{i,1} \otimes ... \otimes a_{i,n-1})=$$ $$\sum \Phi_{ra_n+r'a_n'}(a_{i,1} \otimes ... \otimes a_{i,n-1})=\sum (\Phi_{ra_n+r'a_n'} \circ \iota^\Delta)(a_{i,1},...,a_{i,n-1})=$$ $$\sum φ_{ra_n+r'a_n'}(a_{i,1},...,a_{i,n-1})=\sum φ(a_{i,1},...,a_{i,n-1},ra_n+r'a_n')=$$ $$\sum [rφ(a_{i,1},...,a_n)+r'φ(a_{i,1},...,a_n')]=$$ $$\sum rφ(a_{i,1},...,a_n) + \sum r'φ(a_{i,1},...,a_n')=$$ $$r\sum φ(a_{i,1},...,a_n) + r'\sum φ(a_{i,1},...,a_n')=$$ $$rψ(\sum a_{i,1} \otimes ... \otimes a_{i,n-1},a_n)+r'ψ(\sum a_{i,1} \otimes ... \otimes a_{i,n-1},a_n')$$ so that there exists a unique homomorphism of $R$-modules $$\Phi : A_1 \otimes ... \otimes A_n → L$$ such that, in part, $ψ = \Phi \circ \iota '$ on simple tensors where $\iota '$ is described in the lemma. Now we can show $$\Phi \circ \iota = \Phi \circ \iota ' \circ \iota^* = ψ \circ \iota^* = \Phi_{a_n} \circ σ' \circ \iota^*$$ Since clearly $σ' \circ \iota^* = \iota^\Delta \circ σ$ where $$σ : A_1 \times ... \times A_n → A_1 \times ... \times A_{n-1}$$ $$σ(a_1,...,a_n)=(a_1,...,a_{n-1})$$ we can continue $$\Phi_{a_n} \circ σ' \circ \iota^* = \Phi_{a_n} \circ \iota^\Delta \circ σ = φ_{a_n} \circ σ = φ$$ This completed $⇒$ argument in reverse also shows $ψ = \Phi \circ \iota ' ⇔ φ = \Phi \circ \iota$ since $\iota^*$ is surjective on simple tensors, which is seen to extend to $\Leftarrow$ by $$ψ(\sum a_{i,1} \otimes ... \otimes a_{i,n-1},a_n)=\sum ψ(a_{i,1} \otimes ... \otimes a_{i,n-1},a_n)=$$ $$\sum (\Phi \circ \iota ')(a_{i,1} \otimes ... \otimes a_{i,n-1},a_n)= \sum \Phi(a_{i,1} \otimes ... \otimes a_n)=$$ $$\Phi(\sum a_{i,1} \otimes ... \otimes a_n) = \Phi((\sum a_{i,1} \otimes ... \otimes a_{i,n-1}) \otimes a_n)=$$ $$(\Phi \circ \iota ')(\sum a_{i,1} \otimes ... \otimes a_{i,n-1},a_n)$$ This establishes the first part. For the converse, observe $$(\Phi \circ \iota)(a_1,...,ra_m+r'a_m',...,a_n)=\Phi(a_1 \otimes ... \otimes (ra_m + r'a_m') \otimes ... \otimes a_n)=$$ $$r\Phi(a_1 \otimes ... \otimes a_m \otimes ... \otimes a_n) + r'\Phi(a_1 \otimes ... \otimes a_m' \otimes ... \otimes a_n) =$$ $$r(\Phi \circ \iota)(a_1,...,a_m,...,a_n)+r'(\Phi \circ \iota)(a_1,...,a_m',...,a_n)~\square$$

Direct Products of Free Modules Need Not be Free (10.3.24)

Dummit and Foote Abstract Algebra, section 10.3, exercise 24:

For each $i \in \mathbb{Z}^+$ let $M_i=\mathbb{Z}$ be free and let $M=\prod_{i \in \mathbb{Z}^+}M_i$. Let $N$ be the restricted direct product of $M$ considered as a submodule of $M$ by componentwise multiplication by $\mathbb{Z}$. Assume $M$ is free with basis $B$.

(a) Prove $N$ is countable.
(b) Prove there exists countable $B_1 \subseteq B$ such that $N \subseteq N_1 = RB_1$. Prove $N_1$ is countable.
(c) Prove $\overline{M}=M/N_1$ is free. Deduce that for any nonzero $\overline{x} \in \overline{M}$ there are only a finite number of integers $k$ such that $\overline{x}=k\overline{m}$ for some $\overline{m} \in \overline{M}$.
(d) Let $S=\{(b_1,...)~|~b_i = \pm i!\}$, and prove $S$ is uncountable. Deduce there exists $s\in S$ with $s \not \in N_1$.
(e) Prove there exists $s \in S$ with $s \not \in N_1$ such that for every positive integer $k$ there exists $\overline{m} \in \overline{M}$ such that $\overline{s}=k\overline{m}$, a contradiction by (c).

Proof: (a) Recall that countable unions of countable sets are countable. We have $$N=\bigcup_{i∈\mathbb{Z}^+}N_i'$$ where $N_i'=\{x~|~x∈N \land \text{last}(x)=i\}$ for $\text{last}(x)$ being the position of the last nonzero coordinate of $x$, is a countable union, so that it suffices to show $N_i'$ is countable. We have $$N_i'=\bigcup_{j∈\mathbb{Z}^++\{0\}}N_{i,j}''$$ where $N_{i,j}''=\{x~|~x∈N_i'' \land \sum_{y∈x} |y| = j\}$ is also a countable union, so we must prove $N_{i,j}''$ is countable; in fact, since for every $y∈x∈N_{i,j}''$ we must have $-j≤y≤j$ implying $|N_{i,j}''| < (2j+1)^i$ is finite, and as such is countable.

In general, this implies that any restricted direct product of copies of $\mathbb{Z}$ will be countable.

(b) Let $$B_1=\bigcup_{x∈N}V_x$$ where $V_x$ is the set of elements from $B$ involved in the nonzero terms of the unique sum representation of $x$. Since each $V_x$ is finite and $N$ is countable, $B_1$ is countable and clearly $N \subseteq N_1 = RB_1$. Since $N_1$ has $B_1=\{a_1,...\}$ as a free basis (uniqueness of sums by the freeness of $M$), we have $N_1 ≅ Ra_1 \oplus ...$ by $n=z_1a_1+... \mapsto (z_1,...)$ and in this fashion $N_1$ is the restricted direct product of a number of copies of $\mathbb{Z}$ and by the note at the end of (a) $N_1$ is thus countable.

(c) We claim $\overline{B \setminus B_1}$ is a free basis for $\overline{M}$. We have $\overline{B}$ generates $\overline{M}$, and the terms involving elements from $B_1$ may be removed as they are within $N_1$, so that $\overline{B \setminus B_1}$ generates $\overline{M}$. Further, if $\overline{z_1c_1+...+z_kc_k}=\overline{z_1'c_1+...+z_k'c_k}$ as sums with $c_i∈B \setminus B_1$, then for some $n∈N-1$ we have $z_1c_1+...+z_kc_k+n=z_1c_1+...+z_kc_k+z_1''a_1+...+z_n''a_n=z_1'c_1+...+z_k'c_k$ and now $z_1'c_1+...+z_k'c_k-z_1c_1-...-z_kc_k=z_1''a_1+...+z_n''a_n$. Since the terms $a_i$ and $c_i$ don't coincide, by the freeness of $M$ we can conclude $z_1c_1+...+z_kc_k=z_1'c_1+...+z_k'c_k$ and again by freeness that $z_i=z_i'$.

Since such $k$ manifest as the common divisors of the coefficients in $\mathbb{Z}$ of the elements from $\overline{B \setminus B_1}$, and there are only a finite number of divisors for any integer, we have there are only a finite number of $k$ per $\overline{x}$.

(d) Since there are two choices per coordinate of an element of $b_i$, this is equivalent to proving $\prod^\infty \{0,1\}$ is uncountable, which is standard by Cantor's diagonal argument. Since $S$ is now uncountable, $S \subseteq N_1$ would allow a natural surjection $\mathbb{Z} → S$ by the countability of $N_1$; therefore, such $s∈S$ and $s∉N_1$ exists.

(e) Let $s=(b_1,...)$ be as in (d) and choose arbitrary $k∈\mathbb{Z}$. We have $$\overline{(b_1,...)}=\overline{(0,...,b_k,...)+(b_1,...,b_{k-1},0,...)}=$$ $$\overline{(0,...,b_k,...)}=k\overline{(0,...,b_k',...)}$$ as $k \mid b_i=\pm i!$ for all $i \geq k$.$~\square$

Generalization of the Primary Decomposition Theorem (10.3.22)

Dummit and Foote Abstract Algebra, section 10.3, exercise 22:

Let $M=Tor(M)$ be an $R$-module for $R$ a PID and let $p∈R$ be prime. The $p$-primary component of $M$ is the set of all elements of $M$ that are annihilated by some positive power of $p$.

(a) Prove the $p$-primary component of $M$ is a submodule of $M$.
(b) Prove this definition of $p$-primary components of $M$ is consistent with the one given in 10.3.18 when $Ann(M)≠0$.
(c) Letting $M_p$ be the $p$-primary component of $M$, prove $$M=\oplus_{p\text{prime}}M_p$$ Proof: (a) Let $x,y∈M_p$ with $p^ax=p^by=0$. We have $p^{a+b}(x+ry)=p^b(p^ax)+p^ar(p^by)=0$ so that $x+ry∈M_p$.

(b) For $Ann(M)=(a)$ for $a=p_1^{\alpha_1}...p_k^{\alpha_k}$ with the previous definition of $M^*_{p_i}=\{m~|~p_i^{\alpha_i}m=0\}$ we clearly have $M^*_{p_i} \subseteq M_{p_i}$. As well, for any $m∈M_{p_i}$ we have $am=0$, so that for $Ann(m)=(d)$ we have $d \mid a$. As well, by $m∈M_{p_i}$ we have $d \mid p_i^a$ for some positive $a$. This implies $d \mid (a,p_i^a)=p_i^b$ for $b ≤ \alpha_i$ so that now $p_i^{\alpha_i}m=p^c(dm)=0$ and $M_{p_i} = M^*_{p_i}$.

(c) Choose arbitrary $x∈M$. Letting $X=Rx$ we have $(a) ⊆ Ann(x) = (b)$ so that $b | a = p_1^{\alpha_1}...p_k^{\alpha_k}$. Say $b = q_1^{β_1}...q_k^{β_k}$ where $β_i ≤ α_i$. By 10.3.18 we have $X=X_{q_1} \oplus ... \oplus X_{q_k} ⊆ M_{p_1} ⊕ ... ⊕ M_{p_k}$ so that $x$ is writable as a sum while in $M_{p_1} + ... + M_{p_k}$ and now $M=\sum_{p prime}M_p$.

Proceed in the fashion of 10.3.22(ii) $⇒$ (i). Let $m∈M_{p_1} \cap (M_{p_2} + ... + M_{p_k})$ for any selection such that $p_j$ are distinct primes. Write $m=n_1=n_2+...+n_k$ in accordance with the first and second halves, with $p_i^{\alpha_i}n_i=0$ for any $i$. For $Ann(m)=(d)$ have $d \mid p_1^{\alpha_1}$ for some positive $a$, and by observing the result of $(\prod_{2 ≤ i ≤ k}p_k^{\alpha_k})m$, we notice $d \mid \prod_{2 ≤ i ≤ k}p_k^{\alpha_k}$ and now $d=1$ and $1m=m=0$.$~\square$

Direct Sum Equivalences (10.3.21)

Dummit and Foote Abstract Algebra, section 10.3, exercise 21:

Let $I$ be an indexing set for submodules $N_i$ of $M$. Prove $$\sum_{i∈I}N_i ≅ \oplus_{i∈I}N_i ⇔$$ $$(\{i_1,...,i_k\} \subseteq I ⇒ N_{i_1} ∩ (N_{i_2}+...+N_{i_k}) = 0) ⇔$$ $$(\{i_1,...,i_k\} \subseteq I ⇒ N_{i_1}+...+N_{i_k} = N_{i_1} \oplus ... \oplus N_{i_k}) ⇔$$ $$(x∈\sum_{i∈I}N_i ⇒ ∃! \prod a_i ∈ \prod N_i~s.t.~(|\{y~|~0≠y∈\prod a_i\}|< \infty \land x=\sum y))$$ Proof: (i) $⇒$ (ii) Let $n∈N_{i_1} ∩ (N_{i_2}+...+N_{i_k}))$. By the first half, we can simply write $n$ as a sum, and by the second half we can write a sum only using elements of $N_{i_2},...,N_{i_k}$, which would necessarily be distinct sums when $n≠0$.

(ii) $⇒$ (iii) This is simply proposition 5 applied to $N_{i_1}+...+N_{i_k}$.

(iii) $⇒$ (iv) Since $x∈\sum_{i∈I}N_i$, we must necessarily have it written as a finite sum of nonzero elements, say $x=\sum_{k∈K} b_k$ for some finite $K \subseteq I$ where $b_k≠0$ for all $k$. Assume further that $x=\sum_{j∈J} c_j$ for some finite $J \subseteq I$ where $b_j≠0$ for all $j$, is another sum. We thus have $x∈\sum_{l∈J \cup K} N_l = \oplus_{l∈J \cup K} N_l$ is writable as two sums, so that these two sums are in fact the same.

(iv) $⇒$ (i) Define $φ : \sum_{i∈I}N_i → \oplus_{i∈I}N_i$ by $x = \sum n_i \mapsto \prod n_i$ where $\sum n_i$ is the unique sum representation of $x$. We easily see this is a well-defined isomorphism of $R$-modules.$~\square$

Primary Decomposition Theorem (10.3.18)

Dummit and Foote Abstract Algebra, section 10.3, exercise 18:

Let $M$ be an $R$-module for $R$ a PID, and let $Ann(M)=(a)≠0$. Let $a=p_1^{\alpha_1}...p_k^{\alpha_k}$ be its decomposition into prime factors. Let $M_i=Ann(p_i^{\alpha_i})$, i.e. the $p_i$-primary component of $M$. Prove $$M=M_1 \oplus~...~\oplus M_k$$ Proof: Let $n_i=a/p_i^{\alpha_i}$. We can see that no prime factor divides all of the $n_i$, so that their greatest common divisor is $1$ and there exist $r_i$ such that $1=r_1n_1+...+r_kn_k$. For any $m∈M$ we have $m=r_1n_1m+...+r_kn_km$, and since $p_i^{\alpha_i}(r_in_im)=r_i(am)=0$, we have $r_in_im∈M_i$ and now $M=M_1+...+M_k$. To observe uniqueness of sums, let $$m∈M_j∩(M_1+...+M_{j-1}+M_{j+1}+...+M_k)$$ with $Ann(m)=(d)$. We have $d \mid p_j^{\alpha_j}$ from the first half as well as $d \mid n_j$ from the second so that $d=1$ and now $1m=m=0$.$~\square$

R-Module Homomorphisms from Free Modules (10.3.13)

Dummit and Foote Abstract Algebra, section 10.3, exercise 13:

Let $R$ be a commutative ring. Prove $Hom_R(R^n,R)≅R^n$ as $R$-modules.

Proof: Define $ψ : Hom_R(R^n,R) → R^n$ by $φ \mapsto (φ(e_1),...,φ(e_n))$. We have $$ψ(φ_1+rφ_2) = ((φ_1+rφ_2)(e_1),...,(φ_1+rφ_2)(e_n)) =$$ $$(φ_1(e_1),...,φ_1(e_n)) + r(φ_2(e_1),...,φ_2(e_n)) = ψ(φ_1)+rψ(φ_2)$$ so $ψ$ is a homomorphism of $R$-modules. If $ψ(φ)=(0,...,0)$, then for arbitrary $r^*=(r_1,...,r_n)=r_1e_1+...+r_ne_n∈R^n$ we have $φ(r^*)=r_1φ(e_1)+...+r_nφ(e_n)=0$ so that $φ=0$ and now $ψ$ is injective. Again for arbitrary $r^*∈R^n$, define $φ∈Hom_R(R^n,R)$ by $φ(e_i)=r_i$ and extending linearly, so that $ψ(φ)=r^*$ and now $ψ$ is surjective and an isomorphism of $R$-modules.$~\square$

Quotients and Finite Generation of Modules (10.3.7)

Dummit and Foote Abstract Algebra, section 10.3, exercise 7:

If $N$ is a submodule of the $R$-module $M$, prove that if $M/N$ and $N$ are finitely generated, then so is $M$.

Proof: Let $A=\{a_1,...,a_n\}$ generate $N$, and let $B=\{b_1,...,b_m\}$ be such that $\overline{B}$ generates $M/N$. We claim $A \cup B$ generates $M$. Note that for an arbitrary element $m∈M$ we have $\overline{m}=\overline{r_1b_1+...+r_mb_m}$ for some $r_i∈R$ so that $m=r_1b_1+...+r_mb_m+n$ for some $n∈N$, so that $m=r_1b_1+...+r_mb_m+s_1a_1+...+s_na_n$ for some $s_i∈R$, and now $M$ is generated by $A \cup B$.$~\square$

R-Module Homomorphisms from R (10.2.9-10)

Dummit and Foote Abstract Algebra, section 10.2, exercises 9-10:

Let $R$ be a commutative ring with 1.
9. Prove $Hom_R(R,M)≅M$ as $R$-modules.
10. Prove $End(R) ≅ R$ as rings.

Proof: (9) Define a mapping $ψ : Hom_R(R,M) → M$ by $φ \mapsto φ(1)$, and prove it is an $R$-module isomorphism. We have $ψ(φ_1+rφ_2)=φ_1(1)+rφ_2(1)=ψ(φ_1)+rψ(φ_2)$ for $R$-module homomorphicity, and if $ψ(φ)=φ(1)=0$ then $φ(r)=rφ(1)=0$ so that $φ=0$ and $ψ$ is thus injective. We show $ψ$ is surjective by constructing the mapping $φ$ defined by $φ(1)=m$ and $φ(r)=r \cdot m$ and showing $φ∈Hom_R(R,M)$: We have this because $φ(r_1+r^* \cdot r_2)=(r_1+r^* \cdot r_2)m=r_1 \cdot m +r^* \cdot (r_2 \cdot m)=φ(r_1)+r^* \cdot φ(r_2)$.

(10) By the above they are already isomorphic as abelian groups, so it suffices to show $ψ$ in this case fulfills $$ψ(φ_1 \circ φ_2)=(φ_1 \circ φ_2)(1)=φ_1(φ_2(1))=φ_1(φ_2(1) \cdot 1))=$$ $$φ_1(1)φ_2(1)=ψ(φ_1)ψ(φ_2)~\square$$