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\noindent
Math 541 \\
Problem Set 8
\begin{enumerate}


\item[4.5.8.] The order or $S_5$ is $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 2^3 \cdot 3 \cdot 5$, so a Sylow 2-subgroup will have order $2^3 = 8$.  The dihedral group $D_8$ has order 8 and acts naturally on the 4 corners of the square, so it can be made to act on 5 elements by leaving the fifth alone.  The only group element that does nothing in this action is the identity, so the map $D_8 \to S_5$ given by $r \mapsto (1\ 2\ 3\ 4)$, $s \mapsto (2\ 4)$ is injective.  Thus $\langle (1\ 2\ 3\ 4), (2\ 4) \rangle$ is a subgroup of $S_5$ of order 8.  If we conjugate this by $(1\ 2)$, we have $(1\ 2) (1\ 2\ 3\ 4) (1\ 2) = (1\ 3\ 4\ 2)$ and $(1\ 2) (2\ 4) (1\ 2) = (1\ 4)$, so $\langle (1\ 3\ 4\ 2), (1\ 4) \rangle$ is another Sylow 2-subgroup.


\item[7.1.5.]
\begin{enumerate}
\item If $a/b$ and $c/d$ have odd denominators then so too does $a/b - c/d = (ad-bc)/bd$, and this will still be true after we write it in lowest terms; thus the set is closed under subtraction.  Similarly, it is closed under multiplication since $a/b \times c/d = ac/bd$.  It is not empty since $1/3$ is an element.  Thus it is a subring of $\QQ$.
\item This is not closed under addition, since $1/6 + 1/6 = 1/3$.
\item This is does not contain additive inverses, hence is not a group under addition.
\item This is not closed under addition, since $1/4 + 1/4 = 1/2$.
\item This is not closed under addition, since $1 + 1 = 2$.
\item If $a/b$ is written in lowest terms and $a$ is even then $b$ is odd.  If $a/b$ and $c/d$ have even numerators and odd denominators then so too does $a/b - c/d = (ad-bc)/bd$, and this will still be true after we write it in lowest terms, so the set is closed under subtraction.  Similarly, it is closed under multiplication since $a/b \times c/d = ac/bd$.  It is not empty since $2/3$ is an element.  Thus it is a subring of $\QQ$.
\end{enumerate}


\item[7.1.6.] Throughout, let $S$ denote the subset.
\begin{enumerate}
\item Let $f, g \in S$, so $f(q) = g(q) = 0$ for all $q \in \QQ \cap [0,1]$.  $S$ is closed under subtraction since $(f-g)(q) = f(q) - g(q) = 0-0 = 0$, so $f-g \in S$.  $S$ is closed under multiplication since $(fg)(q) = f(q)g(q) = 0 \cdot 0 = 0$.  $S$ is not empty since $0 \in S$.  Thus $S$ is a subring.

\item It is well-known that this is closed under subtraction and multiplication and is not empty.

\item This is not closed under subtraction.  Let
\[ f(x) = \begin{cases}
0 &\text{if $x = \tfrac{1}{3}$} \\
1 &\text{otherwise}
\end{cases} \qquad
g(x) = \begin{cases}
0 &\text{if $x = \tfrac{2}{3}$} \\
1 &\text{otherwise}
\end{cases}. \]
Then $f \in S$ and $g \in S$, but $f-g$ vanishes on an infinite set.

\item This is not closed under subtraction.  Let
\[ f(x) = \begin{cases}
0 &\text{if $x < \tfrac{1}{2}$} \\
1 &\text{otherwise}
\end{cases} \qquad
g(x) = \begin{cases}
0 &\text{if $x > \tfrac{1}{2}$} \\
1 &\text{otherwise}
\end{cases}. \]
Then $f, g \in S$, but $f-g$ vanishes only at $\frac{1}{2}$.

\item Let $f, g \in S$.  $S$ is closed under subtraction since
\[ \lim_{x\to1^-} [f(x)-g(x)] = \lim_{x\to1^-} f(x) - \lim_{x\to1^-} g(x) = 0 - 0 = 0. \]
$S$ is closed under multiplication since
\[ \lim_{x\to1^-} f(x)g(x) = \lim_{x\to1^-} f(x) \cdot \lim_{x\to1^-} g(x) = 0 \cdot 0 = 0. \]
$S$ is not empty since $0 \in S$.  Thus $S$ is a subring.

\item Clearly $S$ is nonempty and closed under subtraction.  To show that it is closed under multiplication, it suffices to observe that
\begin{align*}
\sin mx \cos nx &= \tfrac{1}{2}[\sin(m+n)x + \sin(m-n)x] \\
\cos mx \cos nx &= \tfrac{1}{2}[\cos(m+n)x + \cos(m-n)x] \\
\sin mx \sin nx &= \tfrac{1}{2}[-\cos(m+n)x + \cos(m-n)x].
\end{align*}
If $m-n$ is negative, we can rewrite $\sin(m-n)x$ as $-\sin(n-m)x$ and $\cos(m-n)x$ as $\cos(n-m)x$.  Thus $S$ is a subring.
\end{enumerate}


\item[7.2.2.] Let $p(x) = a_n x^n + \dotsb + a_0 \in R[x]$, where $a_n \ne  0$.  If there is a nonzero $b \in R$ such that $b p(x) = 0$ then $p(x)$ is a zero divisor.  Conversely, suppose that $p(x)$ is a zero divisor and let $g(x) = b_m x^m + \dotsb + b_0$, where $b_m \ne 0$, be of minimal degree such that $g(x) p(x) = 0$.  The leading term of $g(x) p(x)$ is $b_m a_n x^{m+n}$, so $b_m a_n = 0$.  Since $R$ is commutative, $a_n b_m = 0$ as well.  Now
\[ a_n g(x) = a_n b_m x^m + a_n b_{m-1} x^{m-1} + \dotsb = 0 + a_n b_{m-1} x^{m-1} + \dotsb \]
has lower degree than $g(x)$ and $(a_n g(x)) p(x) = a_n (g(x) p(x)) = 0$, but we assumed that $g(x)$ had minimal degree among such polynomials, so $a_n g(x) = 0$.  Now
\begin{multline*}
0 = g(x) p(x) = g(x) (a_n x^n + a_{n-1} x^n + \dotsb) \\ = g(x) a_n x^n + g(x) (a_{n-1} x^{n-1} + \dotsb) = 0 + g(x) (a_{n-1} x^{n-1} + \dotsb)
\end{multline*}
so $b_m a_{n-1} = 0$, so $a_{n-1} g(x) = 0$ by the same argument as above.  Similarly $b_m a_{n-2} = 0$, $b_m a_{n-3} = 0$, and so on, so $b_m g(x) = 0$, as desired.


\item[7.2.3.]
\begin{enumerate}
\item Since addition in $R[[x]]$ is done componentwise, it is an abelian group under $+$ because $R$ is; in fact, $R[[x]] \equiv R \times R \times R \times \dotsb$ as abelian groups.  For the rest, let $a(x) = a_0 + a_1 x + a_2 x^2 + \dotsb$ be an element of $R[[x]]$ and similarly $b(x)$ and $c(x)$.  The $n^\th$ coefficient of $a(x)b(x)$ is 
\[ \sum_{i=0}^n a_i b_{n-i} = \sum_{\substack{0 \le i, j \le n \\ i+j = n}} a_i b_j \]
and since multiplication in $R$ is commutative, this is the same as the $n^\th$ coefficient of $b(x)a(x)$.  The $n^\th$ coefficient of $(a(x)b(x))c(x)$ is
\[ \sum_{\substack{0 \le m, k \le n \\ m+k = n}} \Biggl( \sum_{\substack{0 \le i, j \le m \\ i+j = m}} a_i b_j \Biggr) c_k = \sum_{\substack{0 \le i, j, k \le n \\ i+j+k = n}} (a_i b_j) c_k \]
and since multiplication in $R$ is associative, this is the same as the $n^\th$ coefficient of $a(x)(b(x)c(x))$.  The $n^\th$ coefficient of $(a(x) + b(x))c(x)$ is
\[ \sum_{\substack{0 \le i, j \le n \\ i+j = n}} (a_i + b_i) c_j = \sum_{\substack{0 \le i, j \le n \\ i+j = n}} (a_i c_j + b_i c_j) = \sum_{\substack{0 \le i, j \le n \\ i+j = n}} a_i c_j + \sum_{\substack{0 \le i, j \le n \\ i+j = n}} b_i c_j \]
which is the $n^\th$ coefficient of $a(x)c(x) + b(x)c(x)$.  The proof of distributivity on the other side is similar.  The identity is $1 + 0x + 0x^2 + 0x^3 + \dotsb$, since the $n^\th$ coefficient of $1 a(x)$ is $1 a_n + 0 a_{n-1} + 0 a_{n-2} + \dotsb + 0 a_1 + 0 a_0 = a_n$ and similarly $a(x) 1 = a(x)$.

\item
\begin{align*}
(1-x)(1 + x + x^2 + \dotsb)
&= (1 + x + x^2 + \dotsb) - x(1 + x + x^2 + \dotsb) \\
&= 1 + x + x^2 + \dotsb - x - x^2 - x^3 - \dotsb \\
&= 1.
\end{align*}

\item Suppose that $a(x) = a_0 + a_1 x + a_2 x^2 + \dotsb$ is a unit in $R[[x]]$, and let $b(x) = b_0 + b_1 x + b_2 x^2 + \dotsb$ be its inverse.  Then $1 = a(x) b(x) = a_0 b_0 + (a_0 b_1 + a_1 b_0) x + (a_0 b_2 + a_1 b_1 + a_2 b_0) x^2 + \dotsb$, so $a_0 b_0 = 1$, so $a_0$ is a unit in $R$.  Convsersely, suppose $a(x)$ is given and we wish to find a $b(x)$ such that $a(x) b(x) = 1$; then we need to solve the equations
\begin{align*}
1 &= a_0 b_0 \\
0 &= a_0 b_1 + a_1 b_0 \\
0 &= a_0 b_2 + a_1 b_1 + a_2 b_0 \\
0 &= a_0 b_3 + a_1 b_2 + a_2 b_1 + a_3 b_0 \\
&\dots
\end{align*}
and if $a_0$ is invertible then we can do it:
\begin{align*}
b_0 &= {a_0}^{-1} \\
b_1 &= -{a_0}^{-1} (a_1 b_0) \\
b_2 &= -{a_0}^{-1} (a_1 b_1 + a_2 b_0) \\
b_3 &= -{a_0}^{-1} (a_1 b_2 + a_2 b_1 + a_3 b_0) \\
&\dots.
\end{align*}
\end{enumerate}


\end{enumerate}
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