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\title{Math 210B\\ Fields III}
\author{Brett Hemenway}

\begin{document}
	\maketitle
	\section*{Multiple roots etc}
	\begin{itemize}
		\item [19.]
			Let $f \in F[t]$ with $\Char F = 0$, and $f^\prime = 0$.  If $f = f_n t^n + \ldots + f_0$ with each $f_i \in F$, we have that 
			$f^\prime = n f_n t^{n-1} + (n-1)f_{n-1} t^{n-1} + f_1 = 0$.  Equating powers of $t$, we get $kf_k = 0$ for each $1 \le k \le n$.  
			Since $\Char F = 0$, we have $k \ne 0$, so $f_k = 0$ for $1 \le k \le n$.  Thus $f = f_0 \in F$.
		\item [20.]
			If $\Char F = p \ne 0$ and $f \in F[t]$ with $f^\prime = 0$.  Then $kf_k = 0$ for all $1 \le k \le n$.  So either $p | k$ or $f_k = 0$.
			So if $f_k \ne 0$, $p | k$, so all the monomials in $f$ are of the form $f_k t^{p\ell}$, so $f = g(t^p)$ for some $g$.
		\item [21.]
			Let $x$ be transcendental over $F$.  Let $f =  t^2 - x \in  F(x)[t]$.  Since $f$ has degree 2, $f$ is irreducible iff $f$ has a root in $F(x)$.  Suppose 
			\[
				\left( \frac{ g_m x^m + \ldots + g_0 } { h_n x^n + \ldots + h_0 } \right)^2 = x
			\]
			With $g_m,h_n \ne 0$.  Then cross multiplying we have
			\[
				(g_m x^m + \ldots + g_0 )^2 = x(h_n x^n + \ldots + h_0)^2
			\]
			But the left hand side has degree $2m$, and the right hand side has degree $2n+1$, so they cannot be equal.  Thus $f$ has no roots in $F(x)$.
		\item [22.]
			Suppose $\Char F = p \ne 0$.
			\begin{itemize}
				\item [(a)]
					Let 
					\begin{align*}
						\phi : F	&\rightarrow F\\
							 	x	&\mapsto	x^p
					\end{align*}
					Then we have
					\begin{align*}
						\phi( x + y ) 	&= (x + y)^p\\
										&= \sum_{i=0}^p {p \choose i} x^{p-i}y^i\\
										&= x^p + y^p\\
										&= \phi(x) + \phi(y)
					\end{align*}
					because $p | { p \choose i}$ for $i \ne 0,p$.  Clearly $\phi(xy) = (xy)^p = x^py^p = \phi(x)\phi(y)$, so $\phi$ is a homomorphism.\\
					To show that $\phi$ is injective, suppose $\phi(x) = x^p = 0$.  Since $F$ is a field, it has no zero divisors, so we must have that $x=0$.  Thus $\phi$ is injective.
				\item [(b)]
					Let $K/F$ be algebraic, and $\alpha \in K$ separable over $F(\alpha^p)$.  We know that $\alpha$ is a root of $t^p - \alpha^p \in F(\alpha^p)[t]$, so $m_{F(\alpha^p)}(\alpha) | t^p-\alpha^p$, 
					but in $F(\alpha)$, we have $t^p - \alpha^p = (t - \alpha)^p$ because when $p=2$, $\alpha=-\alpha$, and when $p > 2$, $(-1)^p = -1$.  So $m_{F(\alpha^p)}(\alpha) = (t-\alpha)^k$.  Thus if 
					$\alpha$ is separable over $F(\alpha^p)$, we must have that $m_{F(\alpha^p)}(\alpha) = t - \alpha$, so $\alpha \in F(\alpha^p)$.
				\item [(c)]
					Let $F$ be a finite field, $|F| = p^e$, and $K/F$ algebraic, with $\alpha \in K$.  Suppose $[F(\alpha):F] = n$.  Then $|F(\alpha)| = (p^e)^n = p^{en}$.  Thus $\alpha$ satisfies $t^{p^{en}} - t$, 
					so $m_F(\alpha) | t^{p^{en}} - t$.  But we know that $t^{p^{en}} - t$ has no repeated factors (every element of $F(\alpha)$ is a root of $t^{p^{en}} - t$).  So no divisor of $t^{p^{en}}-t$ has 
					any repeated roots.
			\end{itemize}
		\item [23.]
			Suppose that $\Char F = p \ne 0$.
			\begin{itemize}
				\item [(a)]
					Suppose $K/F$ is separable.  It is clear that $F(K^p) \subset K$.  Now suppose $\alpha \in K$.  Then $\alpha$ is separable over $F$, so $\alpha$ is separable over $F(\alpha^p)$, 
					so by problem 22b, $\alpha \in F(\alpha^p) \subset F(K^p)$.
				\item [(b)]
					Suppose that $[K:F] = n$, and $K = F(K^p)$.  Let $\{ \alpha_1, \ldots, \alpha_n \}$ be a basis of $K$ over $F$.  Then since $K = F(K^p)$, if $\beta \in K$, we can write 
					$\beta = f_1k_1^p + \ldots + f_nk_n^p$.  Because each $k_i \in K$, we can write $k_i = a_{i1}\alpha_1 + \ldots + a_{in}\alpha_n$, which means $k_i^p = a_{i1}^p\alpha_1^p + \ldots + a_{in}^p\alpha_n^p$.  
					thus $\beta$ can be written as a linear combination of $\alpha_1^p,\ldots,\alpha_n^p$, so $\alpha_1^p,\ldots,\alpha_n^p$ forms a basis for $K$, in particular $\alpha_1^p,\ldots,\alpha_n^p$ are linearly 
					independent.  So if $x_1,\ldots,x_m$ are linearly independent, we can extend this to a basis for $K/F$, $x_1,\ldots,x_m,y_1,\ldots,y_{n-m}$.  By the above argument, $x_1^p,\ldots,x_m^p,y_1^p,\ldots,y_{n-m}^p$ 
					are independent, so $x_1^p,\ldots,x_m^p$ are independent.
			\end{itemize}
		\item [24.]
			Let $K/F$.
			\begin{itemize}
				\item [(a)]
					Let $\alpha \in K$ separable over $F$.\\
					\begin{itemize}
						\item [(i)]
							If $\Char F = 0$.  Then let $f(t) = m_F(\alpha)$, so $f$ is irreducible.  Suppose $(t-u)^m | f$ in the splitting field of $f$.  Then if 
							$m > 1$, we have $t - u | \gcd(f,f^\prime)$.  Since $\degree(f^\prime) < \degree(f)$, we have that $\gcd(f,f^\prime) \ne f$, so 
							$\gcd(f,f^\prime)$ is a nontrivial factor of $f$ in $F[t]$, but this contradicts the irreducibility of $f$.  So $f$ has no repeated roots, and hence is separable.
							This means that every characteristic zero field is perfect.  In particular $F(\alpha)/F$ is separable.
						\item [(ii)]
							If $\Char F = p \ne 0$.  Then by problem 22b, $\alpha \in F(\alpha^p)$.  So we have that $F(F(\alpha)^p) = F(\alpha)$.  Now let $\beta \in F(\alpha) \setminus F$, 
							$f = m_F(\beta)$, $[F(\alpha):F] = n$, and let $1=\beta^0,\beta,\ldots,\beta^{n-1}$ be a basis for $F(\alpha)/F$.  So by problem 23b, $\beta^0,\beta^p,\ldots,\beta^{(n-1)p}$ is a basis for 
							$F(\alpha)/F$.  Suppose $(t - u)^m | f$ in a splitting field of $f$ with $m > 1$.  Then $\gcd( f, f^\prime ) \ne 1$.  If $f^\prime \ne 0$, then $\gcd(f,f^\prime) \ne f$, which contradicts the fact 
							that $f$ is irreducible.  On the other hand, if $f^\prime = 0$, by problem 20, there is a $g$ such that $f(t) = g(t^p)$.  We know that $f(\beta) = 0 = g(\beta^p)$, so 
							$0 = g_0 \beta^0 + g_1 \beta^p + \ldots + g_m \beta^{pm}$.  But $m < n$, since $pm = n$, and we know that $\beta^0,\ldots,\beta^{pm}$ are independent, so we have that each 
							$g_i = 0$.  Thus $g \equiv 0$, so $f \equiv 0$.  Which contradicts the fact that $f = m_F(\beta)$.  So we conclude that $m_f(\beta)$ has no repeated roots, and hence $\beta$ is separable over $F$.
							Since $\beta$ was arbitrary, we get that $F(\alpha)/F$ is separable.
					\end{itemize}
				\item [(b)]
					Suppose $\alpha_1,\ldots,\alpha_n$ are separable over $F$.  Then $F(\alpha_1)/F$ is separable by part (a).  Suppose $F(\alpha_1,\ldots,\alpha_{n-1})/F$ is separable.  Then by part (a), $F(\alpha_1,\ldots,\alpha_n)$ is 
					separable over $F(\alpha_1,\ldots,\alpha_{n-1})$.  Thus $F(\alpha_1,\ldots,\alpha_n)$ is separable over $F$.  
				\item [(c)]
					Let $F_{\txt{sep}} = \{ \alpha \in K | \mbox{ $\alpha$ separable over $F$ } \}$.  Then if $\alpha_1,\alpha_2 \in F_{\txt{sep}}$, by part (b), $F(\alpha_1,\alpha_2) \in F_{\txt{sep}}$, in particular, their sum, 
					product, and inverses are in $F_{\txt{sep}}$, so $F_{\txt{sep}}$ is a field.
			\end{itemize}
		\item [25.]
			Suppose $F$ is perfect, and $K/F$ is algebraic.  Then if $L/K$ is algebraic, we have that $L/F$ is algebraic, hence $L$ is separable over $F$ since $F$ is perfect, so $L$ is separable over $K$.  Thus $K$ is perfect.  
			So any algebraic extension of a perfect field is perfect.
		\item [26.]
			Let $F_0$ be a field of characteristic $p > 0$.  Let $F = F_0(t_1^p,t_2^p)$, and $L = F_0(t_1,t_2)$.
			\begin{itemize}
				\item [(a)]
					Let $\theta \in L \setminus F$.  Then $\theta$ is a root of $t^p - \theta^p \in F[t]$.  So $m_F(\theta) | t^p - \theta^p$.  We know that $t^p - \theta^p = (t-\theta)^p \in L[t]$, so 
					$m_F(\theta) = (t-\theta)^k$ for some $1 \le k \le p$.  The constant term of $(t-\theta)^k = \theta^k$.  If $k < p$, then $\gcd(k,p) = 1$ and $\theta^k \in F$, but $\theta^p \in F$, but
					this would imply that $\theta \in F$, so we conclude that $k = p$.  This means that $m_F(\theta) = t^p - \theta^p$, so $[F(\theta):F] = p$.
				\item [(b)]
					Let $K_n = F(t_1 + t_1t_2^n)$.  Clearly $t_1 + t_1t_2^n \not \in F$, so by part (a), $[K_n:F] = p$.  We need to show that $K_n,K_m$ are distinct, for $n \ne m$.  If $K_n = K_m$, then 
					\[
						t_1+t_1t_2^n = \sum_{i=1}^p a_i (t_1+t_1t_2^m)^i
					\]
					For some $a_i \in F_0(t_1^p,t_2^p)$.  Suppose $a_i = \frac{b_i}{c_i}$ where $b_i,c_i \in F_0[t_1^p,t_2^p]$.  Then clearing denominators we have 
					\[
						(t_1 + t_1t_2^n) f = \sum_{i=1}^p g_i (t_1 + t_1t_2^m)^i
					\]
					Where $f,g_i \in F_0[t_1^p,t_2^p]$.  Now we can equate powers of $t_1$.  On the left, every term has a power of $t_1$ equivalent to $1 \mod p$.  On the right, the only term 
					with exponent on $t_1$ equal to $1 \mod p$ is the $i=1$ term, so we conclude that $g_i = 0$ for $i > 1$.  Thus
					\[
						(t_1 + t_1t_2^n) f = g_1 (t_1 + t_1t_2^m)
					\]
					Suppose $n \ge m$, then dividing by $t_1$ and subtracting, we have
					\[
						f - g_1  = t_2^m g_1 - t_2^n f = t_2^m( g_1 - t_2^{n-m}f )
					\]
					If $p \nmid m$, then if $f-g_1 \ne 0$, $t_2^m \nmid f-g_1$, so we get $f - g_1 = 0$.  But then $g_1 - t_2^{n-m}f \ne 0$, and we don't have equality.\\
					Thus if $p \nmid n,m$, $K_n,K_m$ are distinct, so we have created an infinite family of $K_n$ such that $F < K_n < L$.
			\end{itemize}
	\end{itemize}

\end{document}