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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 $R$.
+