Part 1

In order for $IS$ to be a submodule of $A$, we need to show the following implication: \begin{align*} x\in IS,~a\in A \implies xa, ax \in IS. \end{align*}

Suppose $x\in IS$. Then by definition, $x = \sum_{i=1}^n r_i a_i$ for some $r_i \in R, a_i\in A$.

But then \begin{align*} \begin{align*} xa &= \left( \sum_{i=1}^n r_i a_i \right) a \\ &= \sum_{i=1}^n r_i a_i a \\ &\coloneqq\sum_{i=1}^n r_i a_i', \end{align*} \end{align*}

where $a_i' \coloneqq a_i a$ for each $i$, which is still an element of $A$ since $A$ itself is a module and thus closed under multiplication.

But this expresses $xa$ as an element of $IS$. Similarly, we have \begin{align*} \begin{align*} ax &= a \left( \sum_{i=1}^n r_i a_i \right)\\ &= \sum_{i=1}^n a r_i a_i a \\ &\coloneqq\sum_{i=1}^n r_i a a_i, \\ &\coloneqq\sum_{i=1}^n r_i a_i', \end{align*} \end{align*}

and so $ax \in IS$ as well.

Part 2

Letting $R/I \curvearrowright A/IA$ be the action given by $r+I \curvearrowright+ IA \coloneqq ra + IA$, we need to show the following:

$r.(x + y) = r.x + r.y$,
$(r + r').x = r.x + r'.x$,
$(rs).x = r.(s.x)$, and
$1.x = x$.

Letting $\oplus$ denote the addition defined on cosets, we have \begin{align*} \begin{align*} r \curvearrowright(x + IA \oplus y + IA) &\coloneqq r \curvearrowright x + y + IA \\ &\coloneqq r(x+y) + IA \\ &= rx + ry + IA \\ &\coloneqq rx + IA \oplus ry + IA \\ &\coloneqq(r \curvearrowright x + IA) \oplus (r\curvearrowright y + IA) .\end{align*} \end{align*}

\begin{align*} \begin{align*} (r + s) \curvearrowright x + IA &\coloneqq(r+s)x + IA \\ &\coloneqq rx + sx + IA \\ &\coloneqq rx + IA \oplus sx + IA \\ &\coloneqq(rs \curvearrowright IA) \oplus (sx \curvearrowright IA) .\end{align*} \end{align*}

\begin{align*} \begin{align*} (rs) \curvearrowright x + IA &\coloneqq rsx + IA \\ &= r(sx) + IA \\ &\coloneqq r \curvearrowright(sx + IA) \\ &= r \curvearrowright(s \curvearrowright x + IA) .\end{align*} \end{align*}

\begin{align*} \begin{align*} 1 \curvearrowright x + IA &\coloneqq 1x + IA = x + IA .\end{align*} \end{align*}

Problem 2

Part 1

We want to show that every simple $R{\hbox{-}}$module $M$ is cyclic, i.e. if the only ideals of $M$ are $(0)$ and $M$ itself, that $M = \left\langle{m}\right\rangle$ for some element $m\in M$.

Towards a contradiction, let $M$ be a simple $R{\hbox{-}}$module and suppose $M$ is not cyclic, so $M\neq \left\langle{m}\right\rangle$ for any $m\in M$. But then let $a\in M$ be an arbitrary nontrivial element; then $(a)$ is a non-empty ideal (since it contains $a$), so $(a) \neq 0$. Since $M$ is simple, we must have $(a) = M$, a contradiction.

Part 2

Let $\phi: A \to A$ be a module endomorphism on a simple module $A$. Then $\operatorname{im}\phi \coloneqq\phi(A)$ is a submodule of $A$. Since $A$ is simple, we have either $\operatorname{im}\phi = 0$, in which case $\phi$ is the zero map, or $\operatorname{im}\phi = A$, so $\phi$ is surjective. In this case, we can also consider $\ker \phi$, which is a submodule of $A$. Since $A$ is simple, we can again only have $\ker \phi = A$, which can not happen if $\phi$ is not the zero map, or $\ker \phi = 0$, in which case $\phi$ is both a surjective and an injective map and thus an isomorphism of modules.

Problem 3

Part 1

We want to show that if $A, B$ are $R{\hbox{-}}$modules then $X = (\hom_{R{\hbox{-}}\text{mod}}(A, B), +$ is an abelian group. Let $f, g, h \in X$, we then need to show the following:

Closure: $f + g \in X$
Associativity: $f + (g + h) = (f + g) + h$

Identity: $\operatorname{id}\in X$
Inverses: $f^{-1}\in X$

Commutativity: $f + g = g + f$

Closure: This follows from the definition, because $(f + g) \curvearrowright x \coloneqq f(x) + g(x)$ pointwise, which is well-defined homomorphism $A \to B$.

Associativity: We have \begin{align*} \begin{align*} f + (g + h) \curvearrowright x &\coloneqq f(x) + (g + h)(x) \\ &\coloneqq f(x) + (g(x) + h(x)) \\ &= (f(x) + g(x)) + h(x) \\ &= (f+g) + h \curvearrowright x .\end{align*} \end{align*}

Identity: We can define $\mathbf{0}: A \to B$ by $\mathbf{0}(x) = 0 \in B$. Then \begin{align*}(f + \mathbf{0})\curvearrowright x = f(x) + 0 = f(x) = 0 + f(x) = (\mathbf{0} + f) \curvearrowright x.\end{align*}

Inverses: Given $f\in X$, we can define $-f: A \to B$ as $-f(x) = -x$. Then \begin{align*} \begin{align*} (f + -f) \curvearrowright x = f(x) + -f(x) &= f(x) - f(x) = x - x = 0 = \mathbf{0} \curvearrowright x \\ (-f + f) \curvearrowright x = -f(x) + f(x) &= -f(x) + f(x) = -x + x = 0 = \mathbf{0} \curvearrowright x .\end{align*} \end{align*}

Commutativity: Since $B$ is a module, by definition $(B, +)$ is an abelian group. Thus

\begin{align*} \begin{align*} (f + g) \curvearrowright x &= f(x) + g(x) = g(x) + f(x) = (g+f)\curvearrowright x .\end{align*} \end{align*}

Part 2

By part 1, $(\hom_{R{\hbox{-}}\text{mod}}(A, A), +)$ is an abelian group, We just need to check that $(\hom_R(A, A), \circ)$ is a monoid, i.e.:

Associativity: $f \circ (g\circ h) = (f\circ g) \circ h$
Identity: $\operatorname{id}\circ f = f$
Closure: $f\circ g \in \hom_{R{\hbox{-}}\text{mod}}(A, A)$

Associativity: We have \begin{align*} \begin{align*} f\circ (g\circ h) \curvearrowright x &\coloneqq(f \circ (g \circ h))(x) \\ &= f((g\circ h)(x)) \\ &= f(g(h(x))) \\ &= (f\circ g)(h(x)) \\ &= ((f\circ g) \circ h)(x)\\ &\coloneqq(f \circ g) \circ h \curvearrowright x .\end{align*} \end{align*}

Identity: Take $\operatorname{id}_A: A \to A$ given by $\operatorname{id}_A(x) = x$, then \begin{align*} \begin{align*} f\circ \operatorname{id}_A \curvearrowright x = f(\operatorname{id}_A(x)) = f(x) = \operatorname{id}_A(f(x)) = \operatorname{id}_A \circ f \curvearrowright x .\end{align*} \end{align*}

Closure: If $f: A\to A$ and $g: A\to A$ are homomorphisms, then $f\circ g: A \to A$ as a set map, and is an $R{\hbox{-}}$module homomorphism because \begin{align*} \begin{align*} f\circ g \curvearrowright(r+s)(x+y) &= f(g((r+s)(x+y)))\\ &= f((r+s)(g(x) + g(y))) \\ &= (r+s)(f(g(x)) + f(g(y))) \\ &= (f \curvearrowright(r+s)(x+y)) \circ (g \curvearrowright(r+s)(x+y)) .\end{align*} \end{align*}

Part 3

For arbitrary $x, y \in A$, we need to check the following:

$f\curvearrowright(x+y) = f\curvearrowright x + f \curvearrowright y$
$(f+g)\curvearrowright x = f \curvearrowright x + g \curvearrowright x$

$f\circ g \curvearrowright x = f \curvearrowright(g \curvearrowright x)$
$\operatorname{id}_a \curvearrowright x = x$

For (a): \begin{align*} \begin{align*} f \curvearrowright(x + y) &\coloneqq f(x + y) \\ &= f(x) + f(y)\quad\quad\text{since $f$ is a homomorphism} \\ &= f\curvearrowright x + f \curvearrowright y \\ .\end{align*} \end{align*}

For (b): \begin{align*} \begin{align*} (f+g)\curvearrowright x &= (f+g)(x) \\ &= f(x) + g(x) \\ &= f \curvearrowright x + g \curvearrowright x .\end{align*} \end{align*}

For (c): \begin{align*} \begin{align*} f\circ g \curvearrowright x &= (f\circ g)(x) \\ &= f(g(x)) \\ &= f \curvearrowright g(x) \\ &= f \curvearrowright(g \curvearrowright x) .\end{align*} \end{align*}

For (d): \begin{align*} \begin{align*} \operatorname{id}_A \curvearrowright x &= \operatorname{id}_A(x) = x .\end{align*} \end{align*}

Problem 4

Injectivity: We have the following situation:

\begin{center}
\begin{tikzcd}
a'                                              & a                                          & x                             & 0                                     \\
A_1 \arrow[dd, "\alpha_1", two heads] \arrow[r] & A_2 \arrow[dd, "\alpha_2", hook] \arrow[r] & A_3 \arrow[dd, "f"] \arrow[r] & A_4 \arrow[dd, "\alpha_4", two heads] \\
                                                &                                            &                               &                                       \\
B_1 \arrow[r]                                   & B_2 \arrow[r]                              & B_3 \arrow[r]                 & B_4                                   \\
0                                               & \alpha_2(a)                                & y = f(x) = 0                  & 0                                    
\end{tikzcd}
\end{center}

where we would like to show that $f$ is a monomorphism, i.e. that $\ker f = 0$. So let $x\in \ker f$, so $y \coloneqq f(x) = 0 \in B_3$.

We will show that $x=0 \in A_3$:

Since $y = 0 \in B_3$, applying $B_3 \to B_4$ yields $y \mapsto 0 \in B_4$ since these maps are homomorphisms and always map zero to zero.
Pull back $0 \in B_4$ to $0 \in B_3$ along $\alpha_4$, which can be done since $\alpha_4$ is injective, giving $0 \in A_4$.
Since this is $0$ in $A_4$, it is in the kernel of $A_3 \to A_4$, yielding some $x\in A_3$.
By commutativity of the third square, $x\mapsto f(x)$ under $f: A_3 \to B_3$.
Since $x\in \ker (A_3 \to A_4) = \operatorname{im}(A_2 \to A_3)$ by exactness, there is some $\alpha \in A_2$ such that $\alpha_2(a) = x \in A_3$.
By injectivity of $\alpha_2$, $a$ maps to a unique element $\alpha_2(a) \in B_2$.
By commutativity of the middle square, since $a \in A_2 \mapsto 0 \in B_3$, we must have $\alpha_2(a) \mapsto 0 f(x)$ under $B_2 \to B_3$.
Then $\alpha_2(a) \in \ker(B_2 \to B_3) = \operatorname{im}(B_1 \to B_2)$, so it pulls back to some $b\in B_1$.
By surjectivity of $\alpha_1$, $b$ pulls back to some $a' \in A_1$.
By commutativity of square 1, $a' \mapsto a$ under $A_1 \to A_2$.
So $a \mapsto x$ under $A_1 \to A_3$.
But then $a \in \operatorname{im}(A_1 \to A_2) = \ker(A_2 \to A_3)$, so $a \mapsto 0$ under $A_1 \to A_3$.
So $x=0$ as desired.

\newpage

Surjectivity: We now have this situation:

\begin{center}
\begin{tikzcd}
A_2 \arrow[dd, "\alpha_2", two heads] \arrow[r] & A_3 \arrow[dd, "f"] \arrow[r] & A_4 \arrow[dd, "\alpha_4", two heads] \arrow[r] & A_5 \arrow[dd, "\alpha_5", hook] \\
                                                &                               &                                                 &                                  \\
B_2 \arrow[r]                                   & B_3 \arrow[r]                 & B_4 \arrow[r]                                   & B_5                             
\end{tikzcd}
\end{center}

Let $y \in B_3$; we want to then show that there exists an $x\in A_3$ such that $f(x) = y$.

Apply $B_3 \to B_4$ to $y$ to obtain $y_4 \in B_4$.
By surjectivity of $\alpha_4$, this pulls back to some $a_4 \in A_4$.
Also by exactness of $B_3 \to B_4 \to B_5$, $y_4$ pushes forward to $0 \in B_5$
By injectivity of $\alpha_5$, this pulls back to $0\in A_5$.
By commutativity of the right square, $y_4 \mapsto 0$ under $A_4 \to A_5$.
Since $a_4 \in \ker(A_4 \to A_5)$, it pulls back to some $x\in A_3$ by exactness of $A_3 \to A_4 \to A_5$.
Then $f(x) \in B_3$, and it remains to show that $f(x) = y$.
By commutativity of the middle square, $f(x) \mapsto y_4$ under $B_3 \to B_4$.
Since $a \mapsto y_4$ we as well, we have $z \coloneqq f(x) - y \in B_3$ maps to $0\in B_4$.
Since $z\in \ker(B_3 \to B_4)$, by exactness it pulls back to some $b_2 \in B_2$.
By surjectivity of $\alpha_2$, this pulls back to some $a_2 \in A_2$.
By commutativity of the first square, $a_2 \mapsto z \in B_3$.
$a_2 \mapsto a_3 \in A_3$, where $a_3$ may not equal $x$, but $f(a_3) = z \coloneqq f(a) - y$.
Then $f(a_3) = f(x) - y \implies y = f(x) - f(a_3) = f(x - a_3)$ since $f$ is a homomorphism.
This shows that $x-a_3 \mapsto y$ under $f$, which is the element we wanted to produce.

Problem 5

Part (a)

We want to show that if $(p) {~\trianglelefteq~}R$ is a prime ideal then $R/(p)$ is a field, so we’ll proceed by letting $x + (p) \in R/(p)$ be arbitrary where $x\not \in (p)$ and producing a multiplicative inverse.

Since $R$ is a principal ideal domain, prime ideals are maximal, so $(p)$ is maximal. Then $x\in R \setminus (p)$, so define \begin{align*} I \coloneqq\left\{{p + rx {~\mathrel{\Big\vert}~}p\in (p), r\in R}\right\} {~\trianglelefteq~}R, \end{align*}

which is an ideal in $R$.

In particular, since $x\not\in (p)$, we have a strict containment $(p) < I$, but since $(p)$ was maximal this forces $I = R$.

Then $1 \in I$, so there exists some $p, r$ such that $p + rx = 1$, i.e. $rx - 1 \in (p)$.

But then

\begin{align*} r + (p) \cdot x + (p) = rx + (p) = 1 + (p), \end{align*}

which says that $(x + (p))^{-1}= r + (p)$ in $R/(p)$.

Part (b)

Images and kernels of module homomorphisms are always submodules, so define \begin{align*} \begin{align*} \phi: A \to A \\ x \mapsto px .\end{align*} \end{align*}

This is a module homomorphism, and \begin{align*} \begin{align*} \operatorname{im}\phi &\coloneqq\left\{{px {~\mathrel{\Big\vert}~}x \in A}\right\} \coloneqq pA,\\ \ker \phi &\coloneqq\left\{{a\in A {~\mathrel{\Big\vert}~}pA = 0}\right\} \coloneqq A[p] .\end{align*} \end{align*}

Part (c)

Since $R/(p)$ is a field, we just need to show that $A/pA \curvearrowright R/(p)$ defines a module.

$r\cdot(x + y) = rx + ry$: \begin{align*} \begin{align*} r + (p) \curvearrowright x + pA \oplus y + pA &\coloneqq r + (p) \curvearrowright x + y + pA \\ &\coloneqq r(x+y) + pA \\ &= rx + ry + pA \\ &\coloneqq rx + pA \oplus ry + pA \\ &\coloneqq r\curvearrowright x + pA \oplus r \curvearrowright y + pA .\end{align*} \end{align*}

$(r + s)\cdot x = rx + sx$: \begin{align*} \begin{align*} r + (p) \oplus s + (p) \curvearrowright x + pA &\coloneqq r + s + (p) \curvearrowright x + pA \\ &\coloneqq(r+s)x + pA \\ &= rx + sx + pA \\ &\coloneqq rx + pA \oplus sx + pA \\ &\coloneqq r+(p) \curvearrowright x + pA \oplus s+(p) \curvearrowright x + pA .\end{align*} \end{align*}

$rs\cdot x = r\cdot (s\cdot x)$: \begin{align*} \begin{align*} r+ (p) \cdot s + (p) \curvearrowright x + pA &\coloneqq rs + (p) \curvearrowright x + pA \\ &= rsx + pA \\ &\coloneqq r + (p) \curvearrowright sx + pA \\ &\coloneqq r + (p) \curvearrowright s + (p) \curvearrowright x + pA .\end{align*} \end{align*}

$1\cdot x = x$: \begin{align*} \begin{align*} 1_R + (p) \curvearrowright x + pA &= 1_R x + pA = x + pA .\end{align*} \end{align*}

Part (d)

Similarly, since $R/(p)$ is a field, it suffices to show that $R/(p)\curvearrowright A[p]$ defines a module.

$r\cdot(x + y) = rx + ry$: \begin{align*} \begin{align*} r + (p) \curvearrowright(a + a') &\coloneqq r(a + a') \\ &= ra + ra' \\ &= r\curvearrowright a + r\curvearrowright a' .\end{align*} \end{align*} $(r + s)\cdot x = rx + sx$: \begin{align*} \begin{align*} r + s + (p) \curvearrowright a &= (r+s)a \\ &= ra + sa \\ &= r\curvearrowright a + s\curvearrowright a .\end{align*} \end{align*}

$rs\cdot x = r\cdot (s\cdot x)$: \begin{align*} \begin{align*} rs + (p) \curvearrowright a &= rsa \\ &= r \curvearrowright sa \\ &= r \curvearrowright s \curvearrowright a .\end{align*} \end{align*} $1\cdot x = x$: \begin{align*} \begin{align*} 1_R + (p) \curvearrowright a &= 1a = a .\end{align*} \end{align*}

Problem 6

Supposing that $\dim V = n$, let $\mathcal B \coloneqq\left\{{\mathbf{b}_k \mathrel{\Big|}1 \leq k \leq n}\right\}$ be a basis for $V$, and define \begin{align*} \mathbf{e}_i \coloneqq[0, 0, \cdots, 1, \cdots, 0] \in V^{\oplus m} \end{align*}

where the $1$ occurs in the $i$th position. The claim is that $\mathcal{B}^{m} \coloneqq\left\{{\mathbf{e}_i \mathbf{b}_k \mathrel{\Big|}1 \leq i \leq n,~~1\leq k \leq m}\right\}$ forms a basis for $V^{\oplus m}$.

Elements in $\mathcal{B}^{m}$ are of the form \begin{align*} \begin{align*} [\mathbf{b}_1, 0, 0, \cdots, 0]\\ [\mathbf{b}_2, 0, 0, \cdots, 0]\\ \cdots \\ [0, \mathbf{b}_1, 0, \cdots, 0]\\ [0, \mathbf{b}_2, 0, \cdots, 0]\\ \cdots ,\end{align*} \end{align*}

and by construction, ${\left\lvert {\mathcal B} \right\rvert} = mn = m\dim V$.

To see that this is a spanning set, let $\mathbf{x} \in V^{\oplus m}$, so $\mathbf{x} = [\mathbf{v}_1, \mathbf{v}_2, \cdots, \mathbf{v}_m]$ where each $\mathbf{v}_i \in V$.

Then each $\mathbf{v}_i \in \mathcal B$, so $\mathbf{v}_i = \sum_{k=1}^n \alpha_{k, i} \mathbf{b}_k$. But then \begin{align*} \mathbf{x} = [\sum_{k=1}^n \alpha_{k, 1} \mathbf{b}_k, \sum_{k=1}^n \alpha_{k, 2} \mathbf{b}_k, \cdots, \sum_{k=1}^n \alpha_{k, m} \mathbf{b}_k] \coloneqq\sum_{i=1}^m \sum_{k=1}^n \alpha_{k, i} \mathbf{b}_k \mathbf{e}_i, \end{align*}

which exhibits $\mathbf{x} \in \mathcal{B}^m$.

To see that it is linearly independent, supposing that $\mathbf{x} = \sum_i \sum_k \alpha_{k, i} \mathbf{b}_k \mathbf{e}_i = 0$, this says that $\mathbf{x} = [0, 0, \cdots, 0]$, which forces $\sum_k \alpha_{k, i} \mathbf{b}_k$ to be zero for each $i$.

But for a fixed $i$, since $\left\{{\mathbf{b}_k}\right\}$ was a basis for $V$, this means that $\alpha_{k, i} = 0$ for all $k$. But then $\alpha_{k, i} = 0$ for all pairs $i, k$.

Problem 7

Let $F_1, F_2$ be free, so they have bases $\mathcal B_1 = \left\{{\mathbf{b}_{1, k}}\right\}, \mathcal B_2 = \left\{{\mathbf{b}_{2, k}}\right\}$. Supposing that they have the invariant dimension property, we can assume that ${\sharp}\mathcal B_1 \coloneqq\operatorname{rank}F_1$ and similarly ${\sharp}\mathcal B_2 \coloneqq\operatorname{rank}F_2$.

The claim is that the set \begin{align*}\mathcal B = \left\{{(v, 0) \mathrel{\Big|}v\in \mathcal{B}_1 }\right\} \cup\left\{{(0, w) \mathrel{\Big|}w \in \mathcal{B}_2}\right\}\end{align*} is a basis for $F_1 \oplus F_2$, where ${\sharp}\mathcal B = {\sharp}\mathcal B_1 + {\sharp}\mathcal B_2 = \operatorname{rank}F_1 + \operatorname{rank}F_2$.

So see that $\mathcal B$ spans $F_1 \oplus F_2$, let $x\in F_1 \oplus F_2 = (f_1, f_2)$ be arbitrary. Since $f_1 \in F_1$, we have $f_1 = \sum_i r_i \mathbf{b}_{1, i}$, and similarly $f_2 = \sum_j s_j \mathbf{b}_{2, j}$.

We can then write \begin{align*} x = (f_1, f_2) = (f_1, 0) + (0, f_2) = (\sum_i r_i \mathbf{b}_{1, i}, 0) + (0, \sum_j s_j \mathbf{b}_{2, j}), \end{align*}

which exhibits $x$ as a linear combination of elements in $\mathcal B$.

To see linear independence, we just note that \begin{align*} \begin{align*} x &= (0, 0) \\ &= \sum_i r_i (v_i, 0) + \sum_j s_j (0, w_j) \\& = \sum_i (r_i v_i, 0) + \sum_j (0, s_j w_j) \\ &= (\sum_i r_i v_i, \sum_j s_j w_j) \\ & \implies \sum_i r_i v_i = 0 \quad \& \quad \sum_j s_j w_j = 0 ,\end{align*} \end{align*}

but since the $v_i$ were a basis of $F_1$ and the $w_j$ a basis of $F_2$, this forces $r_i = 0, w_j = 0$ for all $i, j$.

Homework 7

Part 1

Part 2

Problem 2

Part 1

Part 2

Problem 3

Part 1

Part 2

Part 3

Problem 4

Problem 5

Part (a)

Part (b)

Part (c)

Part (d)

Problem 6

Problem 7