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#+TITLE: Assignment Ten
#+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 Four
No. As a counterexample, suppose $n = \big(\begin{smallmatrix}
0 & 0\\
0 & 1
\end{smallmatrix}\big) \in J$, and $m = \big(\begin{smallmatrix}
1 & 1 \\
1 & 1
\end{smallmatrix}\big) \in M(\mathds{R})$, $mn = \big(\begin{smallmatrix}
0 & 1 \\
0 & 1
\end{smallmatrix}\big) \notin J$
** Question Five
By Theorem 3.6, $K$ is a subring since it is closed under subtraction: $\big(\begin{smallmatrix}
a & b \\
0 & 0
\end{smallmatrix}\big) -
(\begin{smallmatrix}
c & d \\
0 & 0
\end{smallmatrix}\big) =
(\begin{smallmatrix}
a - c & b - d \\
0 & 0
\end{smallmatrix}\big)$, and absorbs products of $M(\mathds{R})$ on the right (which is a superset of $K$, so $K$ is closed under this multiplication):
$(\begin{smallmatrix}
a & b \\
0 & 0
\end{smallmatrix}\big) \cdot
(\begin{smallmatrix}
c & d \\
e & f
\end{smallmatrix}\big)
=
(\begin{smallmatrix}
ac + be & ad + bf \\
0 & 0
\end{smallmatrix}\big)$
But from the left, as a counterexample $n = (\begin{smallmatrix}
1 & 1 \\
0 & 0
\end{smallmatrix}\big) \in K, m =(\begin{smallmatrix}
2 & 1 \\
3 & 0
\end{smallmatrix}\big) \in M(\mathds{R})$, then $mn = (\begin{smallmatrix}
2 & 2 \\
3 & 3
\end{smallmatrix}\big) \notin K$
** Question Eleven
*** a
$(1) = (2) = (3) = (4) = \mathds{Z}_5$ and $(0) =$ { 0 }
*** b
$(1) = (2) = (4) = (5) = (7) = (8) = \mathds{Z}_9$, $(0) =$ { 0 }, and $(3) = (6) =$ { 0, 3, 6 }
*** c
$(1) = (5) = (7) = (11) = \mathds{Z}_{12}$, $(2) = (6) = (10) =$ { 0, 2, 4, 6, 8, 10 }, $(4) = (8) =$ { 0, 4, 8 },
$(3) = (9) =$ { 0, 3, 6, 9 }, and $(6) =$ { 0, 6 }
** Question Thirteen
No, $\mathds{Z}_5$ is a commutative ring, but in question above, $(1) = (2)$ but $1 \neq 2$.
** Question Sixteen
*** a
If we can show that $(4, 6) \sube (2)$ and $(2) \sube (4, 6)$, then we can conclude that $(4, 6) = (2)$
Each element $x \in (4, 6) \Rightarrow x = 4m + 6n$ for $m,n \in \mathds{Z}$.
Thus $x = 2(2m + 3n)$ which shows that $x$ is a multiple of two, and thus, $(4, 6) \sube (2)$.
Each element $y \in (2) \Rightarrow y = 2p$ for $p \in \mathds{Z}$. When $p$ is itself a multiple of two, this can be rewritten as $y = 2(2o) = 4o$ and thus $y \in (4, 6)$.
When $y$ is odd, $y = 2(2(q - 1) + 3) = 4q + 6$ for some $q \in \mathds{Z}$. Therefore, $y$ is in $(4, 6)$.
*** b
We'll take the same approach as a:
Each element $x \in (6, 9, 15) = 6m + 9n + 15o$ with $m, n, o \in \mathds{Z}$.
Thus $x = 3(2m + 3n + 5o)$ is a multiple of three, so $x \in (3)$.
Each element $y \in (3) \Rightarrow y = 3p$ for $p \in \mathds{Z}$. When $p$ is a multiple of two, then $y = 3(2q) = 6q \in (6, 9, 15)$.
When $p$ is odd, $y = 3(2(u - 1) + 7)$ for some $u \in \mathds{Z}$, $y = 6u + 15 \in (6, 9, 15)$.
** Question Seventeen
*** a
If $a \in I \cap J$, and $b \in I \cap J$, then $a \in I$, $b \in I$, $a \in J$, and $b \in J$. Then, $a - b \in I$ and $a - b \in J$, so
$a - b \in I \cap J$.
If $a \in I \cap J$ and $r \in R$, then $ra \in I$, $ra \in J$, $ar \in J$, and $ar \in I$ by definition of I and J.
Therefore, $I \cap J$ is an ideal in $R$.
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