Proposition

If $a_1, a_2$ are ideals of $A$ and if $b_1, b_2$ are ideals of $B$, then

$(a_1 + a_2)^e = a_1^e + a_2^e, (b_1 + b_2)^c \supset b_1^c + b_2^c$.
$(a_1 \cap a_2)^2 \subset a_1^e \cap a_2^e, (b_1 \cap b_2)^c = b_1^c \cap b_2^c$.
$(a_1a_2)^e = a_1^ea_2^e, (b_1b_2)^c \supset b_1^cb_2^c$.
$(a_1:a_2)^e \subset (a_1^e:a_2^e), (b_1:b_2)^c \subset (b_1^c:b_2^c)$.
$r(a)^e \subset r(a^e), r(b)^c = r(b^c)$.

Solution

1

Let $\sum_{i=1}^{n} y_if(x_{i, 1} + x_{i, 2}) \in (a_1 + a_2)^e$ where $y_i \in B$ and $x_{i, 1} \in a_1, x_{i, 2} \in a_2$.

\[\begin{align*} \sum_{i=1}^{n} y_if(x_{i, 1} + x_{i, 2}) &= \sum_{i=1}^{n} y_i(f(x_{i, 1}) + f(x_{i, 2}) ) \\ &= \sum_{i=1}^{n} y_if(x_{i, 1}) + \sum_{i=1}^{n} y_if(x_{i, 2}) \\ &\in a_1^e + a_2^e. \end{align*}\]

Therefore, $(a_1 + a_2)^e \subset a_1^e + a_2^e$.

On the other hand, let $\sum_{i=1}^{n} y_{i, 1} f(x_{i, 1}) \in a_1^e$ and $\sum_{i=1}^{m} y_{i, 2} f(x_{i, 2}) \in a_2^e$. Then $\sum_{i=1}^{n} y_{i, 1} f(x_{i, 1}) + \sum_{i=1}^{m} y_{i, 2} f(x_{i, 2}) = \sum_{i=1}^{n + m} y_i f(x_i)$ where

\[\begin{align*} y_i &= \begin{cases} y_{i, 1} & (i \leq n) \\ y_{i - n, 2} & (i \geq n + 1) \end{cases} \\ x_i &= \begin{cases} x_{i, 1} & (i \leq n) \\ x_{i - n, 2} & (i \geq n + 1). \end{cases} \end{align*}\]

Then each $y_i$ is in $B$ and each $x_i$ is in $a_1 + a_2$. Thus $\sum_{i=1}^{n + m} y_i f(x_i) \in (a_1 + a_2)^e$.

Let $x_1 + x_2 \in b_1^c + b_2^c$ be given. Then $f(x_1) \in b_1$ and $f(x_2) \in b_2$. Thus $f(x_1) + f(x_2) \in b_1 + b_2$, so $x_1 + x_2 \in (b_1 + b_2)^c$. Therefore, $b_1^c + b_2^c \subset (b_1 + b_2)^c$.

2

Since $a_1 \cap a_2 \subset a_1$, $(a_1 \cap a_2)^e \subset a_1^e$. Similarly, $(a_1 \cap a_2)^e \subset a_2^e$. Therefore, $(a_1 \cap a_2)^e \subset a_1^e \cap a_2^e$.

$b_1 \cap b_2 \subset b_1$, so $(b_1 \cap b_2)^c \subset b_1^c$. Similarly, $(b_1 \cap b_2)^c \subset b_2^c$. Therefore, $(b_1 \cap b_2)^c \subset b_1^c \cap b_2^c$.

Let $x \in b_1^c \cap b_2^c$. Then $x \in b_1^c$ and $x \in b_2^c$. Therefore, $f(x) \in b_1$ and $f(x) \in b_2$. Thus $f(x) \in b_1 \cap b_2$, so $x \in (b_1 \cap b_2)^c$.

3

$\begin{align*} (a_1a_2)^e &= Bf(a_1a_2) \\ &= Bf(a_1)f(a_2) \\ &= BBf(a_1)f(a_2) \\ &= Bf(a_1)Bf(a_1) \\ &= a_1^ea_2^e. \end{align*}$

Since $(b_1b_2)^c$ is closed under addition and every element in $b_1^cb_2^c$ is a finite sum of elements of the form $xy$ where $x \in b_1^c$ and $y \in b_2^c$, it suffices to show that each $xy$ is in $(b_1b_2)^c$.

$x \in b_1^c \implies x \in f^{-1}(b_1) \implies f(x) \in b_1$.
$y \in b_2^c \implies y \in f^{-1}(b_2) \implies f(y) \in b_2$.

Therefore, $f(xy) = f(x)f(y) \in b_1b_2$, so $xy \in f^{-1}(b_1b_2) = (b_1b_2)^c$.

4

Let $bf(x) \in (a_1:a_2)^e$ where $b \in B$ and $x \in (a_1:a_2)$.

\[\begin{align*} xa_2 \subset a_1 &\implies f(xa_2) \subset f(a_1) \\ &\implies f(x)f(a_2) \subset f(a_1) \\ &\implies B(f(x)f(a_2)) \subset Bf(a_1) \\ &\implies f(x)(Bf(a_2)) \subset Bf(a_1) \\ &\implies f(x)a_2^e \subset a_1^e \\ &\implies f(x) \in (a_1^e:a_2^e) \\ &\implies bf(x) \in (a_1^e:a_2^e). \end{align*}\] \[\begin{align*} x \in (b_1:b_2)^c &\implies f(x) \in (b_1:b_2) \\ &\implies f(x)b_2 \in b_1 \\ &\implies f^{-1}(f(x)b_2) \subset f^{-1}(b_1) \\ &\implies xf^{-1}(b_2) \subset f^{-1}(f(x)b_2) \subset f^{-1}(b_1) \\ &\implies xf^{-1}(b_2) \subset f^{-1}(b_1) \\ &\implies x \in (f^{-1}(b_1):f^{-1}(b_2)) \\ &\implies x \in (b_1^c:b_2^c). \end{align*}\]

5

\[\begin{align*} bf(x) \in r(a)^e &\implies x \in r(a) \\ &\implies x^m \in a & (m \in \mathbb{N})\\ &\implies f(x^m) \in f(a) \\ &\implies (f(x))^m \in f(a) \subset a^e \\ &\implies (f(x))^m \in a^e \\ &\implies f(x) \in r(a^e) \\ &\implies bf(x) \in r(a^e). \end{align*}\] \[\begin{align*} x \in r(b)^c &\iff f(x) \in r(b) \\ &\iff \exists m \in \mathbb{N}, f(x)^m \in b \\ &\iff \exists m \in \mathbb{N}, f(x^m) \in b \\ &\iff \exists m \in \mathbb{N}, x^m \in f^{-1}(b) \\ &\iff \exists m \in \mathbb{N}, x^m \in b^c \\ &\iff x \in r(b^c). \end{align*}\]