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#+TITLE: Assignment Eleven
#+AUTHOR: Lizzy Hunt
#+STARTUP: entitiespretty fold inlineimages
#+LATEX_HEADER: \notindent \notag \usepackage{ dsfont } \usepackage{amsmath} \usepackage[a4paper,margin=1in,portrait]{geometry} \usepackage{polynom} \usepackage{wasysym}
#+LATEX: \setlength\parindent{0pt}
#+OPTIONS: toc:nil
* Section 6.1
** Question Twenty-One
The ideal $(a) + (b)$ is generated by all linear combinations of $a$ and $b$, $x \in (a) + (b) \Rightarrow x = a c_1 + b c_2$
By definition, $d | a$ and $d | b$, so $a = d a_1$ and $b = d b_1$. Then, $x \in (a) + (b) \Rightarrow x = d a_1 c_1 + d b_1 c_2 = d(a_1 c_1 + b_1 c_2)$,
so $(a) + (b) \sube (c)$.
By Theorem 1.2, $d = au + bv$ which is a linear combination of $a$ and $b$, so $d \in (a) + (b)$. Since
every multiple of $d$ is also a multiple of $a$ and $b$, $(d) \sube (a) + (b)$.
So, $(d) = (a) + (b)$.
* Section 6.2
** Question Two
Let $f : F \rightarrow R$ with $R$ being the image, so the kernel of $f$ is an ideal in the field $F$ by Theorem 6.10.
Then assuming from the hint, that question ten in 6.1 is true, the only ideals in $F$ are either $(0_F)$ or $F$
itself. So, $\text{ker}(f) = (0_F)$ or $F$.
When $\text{ker}(f) = (0_F)$ then $f$ is injective by Theorem 6.11, and is thus an isomorphism.
When $\text{ker}(f) = F$ then $R = \{ 0_F \}$.
** Question Four
*** a
$f$ is consistent, as when
$[a]_{12}_{} = [b]_{12}_{}_{} (a \equiv b \text{ mod } 12 \Rightarrow a - b = 12n)$, then $[a]_4 = [b]_4$ as
$a - b \equiv 20n \text{ mod } 4 = a - b \equiv 0 \text{ mod } 4 \Rightarrow a \equiv b \text{ mod } 4$.
*** b
The kernel of $f$ is the ideal $(4)$ as $f(x) \ni x \in (4) \Rightarrow f(x) = [4n]_{4} = [0]_4$.
** Question Six
The kernel of \phi are polynomials in $\mathds{R}[x]$ with a root of 2. By Theorem 4.16, $x - 2$ must be a factor, so
$\text{ker}(\phi)$ is the set of all multiples of $x-2$ in $\mathds{R}[x] = (x - 2)$.
** Question Nine
*** a
There is surjectivity as $x \in \mathds{Z}$, then the matrix $\big(\begin{smallmatrix}
x & 0\\
0 & 0
\end{smallmatrix}\big)$ maps to $x$.
It is a homomorphism since $f(x + y) = \big(\begin{smallmatrix}
a & 0\\
c & d
\end{smallmatrix}\big) + \big(\begin{smallmatrix}
e & 0\\
g & h
\end{smallmatrix}\big) = \big(\begin{smallmatrix}
a + e & 0 \\
c + g & d + h
\end{smallmatrix}\big) = (a + e) = f(x) + f(y)$,
$f(xy) = \big(\begin{smallmatrix}
a & 0\\
c & d
\end{smallmatrix}\big) \cdot \big(\begin{smallmatrix}
e & 0\\
g & h
3\end{smallmatrix}\big) = \big(\begin{smallmatrix}
ae & 0 \\
ce + dc & dh
\end{smallmatrix}\big) = ae = f(x)f(y)$, and the identity matrix is mapped to $1$.
*** b
$\{ \big(\begin{smallmatrix}
0 & 0 \\
c & d
\end{smallmatrix}\big) | c, d \in \mathds{Z} \}$
** Question Ten
*** a
By definition, $x, y \in f(I) \Rightarrow \exists z, t \in I \ni f(z) = x, f(t) = b$, and $f(I)$ is also a subset of $S$.
By hormomorphism, $f(z) - f(t) = x - y \Rightarrow f(z - t) = x - y \in I$.
For $s \in S \Rightarrow \exists r \in R \ni f(r) = s$ by surjection. Then, $x f(r) = s x = f(z) f(r) = f(zr) \in f(I)$
*** b
Let $R = \mathds{R}$, $S = \mathds{C}$, and $f : R \rightarrow S \ni f(x) = x + 0i$. $f$ is not a surjective
homomorphism ($i$ has no associated element in $\mathds{R}$), and $\mathds{R}$ is an ideal in $\mathds{R}$.
However, $f(\mathds{R})$ is not an ideal by the trivial example that
$f(1)(i) = (1 + 0i)(0 + i) = i \notin \mathds{R}$.
** Question Seventeen
*** a
$f(a + b) = ((a + b) + I, (a + b) + J) = ((a + I) + (b + I), (a + J) + (b + J)) = (a + I, a + J) + (b + I, b + J) = f(a) + f(b)$.
Similarly, $f(a)f(b) = f(ab)$: $f(ab) = (ab + I, ab + J)$ and $f(a)f(b) = (a + I, a + J)(b + I, b + J) = ((a + I)(b + I), (a + J)(b + J)) = (ab + I, ab + J)$.
*** b
If we consider $R = \mathds{Z}, I = (2)$, and $J = (4)$, the elements in \mathds{Z}/(2) x \mathds{Z}/(4) are:
{(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3)}
There's at least one domain element that is impossible to get to: $(1, 0)$ - an integer
can't be equivalent to $1$ mod $2$ and also $0$ mod $4$. So, $f$ is not necessarily surjective.
** Question Twenty-One
Let $f : \mathds{Z}_{20} \rightarrow \mathds{Z}_5$ such that $f([a]_{20}) = [a]_5$. $f$ is consistent because when
$[a]_{20} = [b]_{20}_{} (a \equiv b \text{ mod } 20 \Rightarrow a - b = 20n)$, then $[a]_5 = [b]_5$ as
$a - b \equiv 20n \text{ mod } 5 = a - b \equiv 0 \text{ mod } 5 \Rightarrow a \equiv b \text{ mod } 5$.
Also, $f$ is a surjective homomorphism:
+ $f([a]_{20} + [b]_{20}) = f([a + b]_{20}) = [(a + b)]_5 = [a]_5 + [b]_5 = f([a]_{20}) + f([b]_{20})$
+ $f([a]_{20} * [b]_{20}) = f([a * b]_{20}) = [(a * b)]_5 = [a]_5 * [b]_5 = f([a]_{20}) * f([b]_{20})$
+ For each $[x]_5$ there exists $[y]_{20}$ such that $f([y]_{20}) = [x]_5$. The trivial solution is
that $x = y$.
Finally, the kernel of $f$ is the ideal $(5)$ as $x \in (5) \Rightarrow f(x) = [5n]_5 = [0]_5$. Thus,
$\mathds{Z}_{20} / (5)$ is isomorphic to $\mathds{Z}_5$ by the First Isomorphism Theorem.
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