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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Solutions for Problem Set 5}
\centerline{February 18, 2008}
\vskip 10pt
\noindent
{\it Section 10.3}
\vskip 10pt
\noindent
3.  Following the example on page 349 of the text,
consider the tower of field extensions
\[
\Q \subset \Q(\sqrt{2}) \subset \Q(\sqrt{2})(\sqrt{3}) \subset \Q(\sqrt{2},\sqrt{3})(\sqrt{5}).
\]
We have $[\Q(\sqrt{2}):\Q] = 2$ with basis $\{1,\sqrt{2}\}$ since the minimal polynomial
of $\sqrt{2}$ over $\Q$ is $x^2 - 2$.  At the next step, we claim that 
$[\Q(\sqrt{2})(\sqrt{3}) : \Q(\sqrt{2})] = 2$ also and a basis is $\{1,\sqrt{3}\}$.
Clearly $x^2 - 3$ is a polynomial in $\Q(\sqrt{2})[x]$ with $\sqrt{3}$ as a root.
Moreover, it is irreducible because no $a + b\sqrt{2}$ is a root of this polynomial.
(If there were such a root, then $(a + b\sqrt{2})^2 = 3$ would imply $\sqrt{2} = \frac{3 - a^2 - 2b^2}{2ab} \in \Q$, which we know is not the case.)  Hence as in the proof of Theorem 10.4 (the multiplicativity
of degree in towers), $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ is a basis for $\Q(\sqrt{2})(\sqrt{3}) = 
\Q(\sqrt{2},\sqrt{3})$ over
$\Q$.  Now we argue the same way for the final extension above.  $\sqrt{5}$ is a root of 
$x^2 - 5 \in \Q(\sqrt{2},\sqrt{3})[x]$, but no $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$
is a root of this polynomial, so $[\Q(\sqrt{2},\sqrt{3})(\sqrt{5}): \Q(\sqrt{2},\sqrt{3})] = 2$
and $\{1,\sqrt{2},\sqrt{3},\sqrt{6},\sqrt{5},\sqrt{10},\sqrt{15},\sqrt{30}\}$ is a 
basis of $\Q(\sqrt{2},\sqrt{3},\sqrt{5})$ over $\Q$.
\vskip 10pt
\noindent
5.  We follow the same procedure as in the previous problem.  Consider
\[
\Q \subset \Q(\sqrt{3}) \subset \Q(\sqrt{3},i).
\]
We have $[\Q(\sqrt{3}):\Q] = 2$ with basis $\{1,\sqrt{3}\}$ since the minimal polynomial
of $\sqrt{3}$ over $\Q$ is $x^2 - 3$.   Next, $i$ is a root of $x^2 + 1 \in \Q(\sqrt{3})[x]$
and this is the minimal polynomial because no $a + b\sqrt{3}$ can be a root of $x^2 + 1$ 
(since $a + b \sqrt{3}$ are all real numbers).  Hence $[Q(\sqrt{3},i) : \Q(\sqrt{3})] = 2$
and $\{1,i\}$ is a basis.  As in the proof of Theorem 10.4, $\{1,\sqrt{3},i,i\sqrt{3}\}$ is a
basis for $\Q(\sqrt{3},i)$ over $\Q$.  Hence $[\Q(\sqrt{3},i): \Q] = 4$. 
\vskip 10pt
\noindent
7.  Assume $[K:F]$ is finite and $u$ is algebraic over $K$.  By Theorem 10.7, $[K(u):K]$
is finite (this equals the degree of the minimal polynomial of $u$ over $K$, $p(x) \in K[x]$.  
Now, we have $F \subseteq K \subseteq K(u)$, so also $F \subseteq F(u) \subseteq K(u)$.
From the first tower $[K(u):F] = [K(u):K][K:F]$ is finite.  This implies $[F(u):F]$ is
also finite from the second tower.  By Theorem 10.7 again, $[F(u):F]$ equals
the degree of the minimal polynomial $q(x) \in F[x]$ 
of $u$ over $F$.  Since this polynomial has $u$ as a root and also lies in $K[x]$, it
follows that $p(x) | q(x)$ in $K[x]$.  Therefore $\deg q(x) = [F(u):F] \ge [K(u) : K] = \deg p(x)$.
\vskip 10pt
\noindent
8.  Let $[K:F]$ be finite and assume $u$ is algebraic over $K$.  By Theorem 10.7, 
every element $v$ of $K(u)$ is of the form 
\begin{equation}\label{one}
v = a_0 + a_1 u + \cdots a_{n-1} u^{n-1},
\end{equation}
with $a \in K$
(where $n$ is the degree of the minimal polynomial of $u$ over $K$).  Now let $\{b_1,\ldots,b_k\}$
be a basis of $K$ over $F$.  We can expand each coefficient 
\[
a_i = c_{i1}b_1 + \cdots + c_{ik}b_k,
\]
where $c_{ij} \in F$.  Substituting these into (\ref{one}), and regrouping terms we see that 
\[
v = (c_{01} + c_{11}u + \cdots + c_{n-1,1}u^{n-1})b_1 + \cdots + (c_{0k} + c_{1k}u + \cdots + c_{n-1,k}u^{n-1})b_k.
\]
This shows that $\{b_1,\ldots,b_k\}$ also spans $K(u)$ over $F(u)$.  Hence
$[K(u):F(u)] \le [K:F]$ (with equality if and only if $\{b_1,\ldots,b_k\}$ is linearly
independent over $F(u)$, which is \emph{not automatic}).
\vskip 10pt
\noindent
11.  (a)  Consider the towers of field extensions $F \subseteq F(u) \subseteq F(u)(v) = F(u,v)$
and $F \subseteq F(v) \subseteq F(v)(u) = F(u,v)$.  
By hypothesis and Theorem 10.7, $[F(u):F] = \deg p(x) = m$.  So then by Theorem 10.4, 
$m | [F(u,v):F]$.  Similarly, 
$[F(v):F] = \deg q(x) = n$.  So then by Theorem 10.4, 
$n | [F(u,v):F]$.  So this says $[F(u,v) : F]$ is a \emph{common multiple} of $m,n$.
Since $(m,n) = 1$, then least common multiple is of $m,n$ is the product $mn$.
However, from problem 7 above with $K = F(v)$, we also know $[F(u,v):F(v)] \le [F(u):F] =m$.
Therefore $[F(u,v):F] = [F(u,v):F(v)][F(v):F] \le mn$.  Therefore $[F(u,v): F] = mn$.  

(b) For a simple example, consider $F = \Q$, $u = \sqrt{2}$ and $v = 1 + \sqrt{2}$.  We have
$[\Q(u):\Q] = [\Q(v):\Q] = 2$ (look at the minimal polynomials).  So the degrees
are not relatively prime.  Moreover 
$v \in \Q(u)$, so in fact $\Q(u,v) = \Q(u)$ and $[\Q(u,v):\Q] = 2$, not $2\cdot 2 = 4$.

(c) $[\Q(\sqrt{2},{\root 3\of 2}):\Q] = 6$ by part (a) since 
$[\Q(\sqrt{2}):\Q] = 2$ and $[\Q({\root 3\of 2}):\Q] = 3$ are relatively prime.
\vskip 10pt
\noindent
12.  Method 1:  (A ``brute force'' method not following the Hint.)
Since we are given that $D$ is a ring, 
the only thing ``missing'' is that we must show that every $u \in D$ has an 
inverse in $D$.  So let $u \in D$.  Since $K$ is algebraic over $F$, every element
of $K$ is algebraic over $F$, and it follows that $u$ is a root
of a nonzero polynomial $c_n x^n + \cdots + c_1 x + c_0 \in F[x]$, or
\[
c_n u^n + \cdots + c_1 u + c_0 = 0.
\]
We can assume that zero is not also a root of this polynomial (why?).  Hence
we may assume $c_0 \ne 0$.  But then rearranging the last equation, 
\[
(-c_0^{-1}(c_n u^{n-1} + \cdots + c_1)) u = 1.
\]
This shows $u^{-1} = -c_0^{-1}(c_n u^{n-1} + \cdots + c_1)$, which is in $D$ 
since $D$ is closed under sums and products and $u \in D$.  Hence $D$ is 
a field.  

Method 2:  (Following the Hint.)  Let $u \in D$.  Since $u$ is in $K$, 
which is a field, we have the tower $F \subseteq F(u) \subset K$.  Since
$K$ is algebraic over $F$, 
by Theorem 10.7 we know $F(u)$ has a basis $\{1,u,\ldots,u^{n-1}\}$ if 
the minimal polynomial of $u$ over $F$ has degree $n$.  Therefore all 
the elements of $F(u)$ look like $v = a_0 + a_1 u + \cdots + a_{n-1}u^{n-1}$ 
where $a_i \in F$.  But $u \in D$ and $D$ is a ring (hence closed under sums 
and products).  It follows that $F(u) \subseteq D$.  Therefore $D$ contains 
$u^{-1}$, the multiplicative inverse of $u$ in $K$.  Hence $D$ is a field. 

\noindent
\emph{Comment:}
I happen to prefer proofs like the first one, because they give a method
for finding $u^{-1}$ in addition to establishing the existence of $u^{-1}$
in $D$.      

\vfill\eject
\noindent
{\it Section 10.4}
\vskip 10pt 
\noindent
2.  For $x^2 - 3$, we have the roots $x = \pm \sqrt{3}$.  Hence the splitting
field is $\Q(\sqrt{3},-\sqrt{3}) = \Q(\sqrt{3})$.  For $x^2 - 2x - 2$, the 
quadratic formula gives 
\[
x = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2\pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}.
\]
Therefore the splitting field of this polynomial is $\Q(1 + \sqrt{3},1 - \sqrt{3}) = \Q(1 + \sqrt{3})$.
Since $1 + \sqrt{3}$ is an element of $\Q(\sqrt{3})$ and conversely every element of
$\Q(\sqrt{3})$ is contained in $\Q(1 + \sqrt{3})$, it follows that the splitting fields
of these two polynomials are equal.
\vskip 10pt 
\noindent
6.  We have $u \in K$, so $F(u) \subseteq K$.  Since $K$ is a splitting field of 
a polynomial in $F[x]$ it is a finite (algebraic) extension of $F$.
By Theorem 10.4, $[K:F] = [K:F(u)][F(u):F]$.  Since $[K:F] = p$ is prime, but we
are given $u \ne F$, it follows that $[F(u):F] = p$ and $[K:F(u)] = 1$.  Therefore
$K = F(u)$.
\vskip 10pt 
\noindent
7.  Since $F(u)$ is normal over $F$, by definition, if $p(x)$ is any 
irreducible polynomial in $F[x]$ and $p(x)$ has one root in $F(u)$, then
$p(x)$ splits completely over $F(u)$ (has all its roots in $F(u)$).  
Apply this to $p(x) = $ the minimal polynomial of $u$ over $F$.  It
follows that all the roots $u = u_1, u_2, \ldots, u_n$ 
of $p(x)$ are contained in $F(u)$. Therefore
$F(u,u_2,\ldots,u_n) \subseteq F(u)$.  The other inclusion is automatic,
so $F(u,u_2,\ldots,u_n) = F(u)$.  This shows that $F(u)$ is the splitting
field of $p(x)$, which is what we wanted to show.   
\vskip 10pt 
\noindent
8.  We use Theorem 10.15 for all the parts here.  

(a) $\Q(\sqrt{3})$ is normal over $\Q$ since
this is the splitting field of the polynomial $x^2 - 3$ over $\Q$.  

(b) $\Q({\root 3 \of 3})$ is not a normal extension of $\Q$ since this 
number is a root of the irreducible polynomial $x^3 - 3 \in \Q[x]$, but
the other two roots of this polynomial are not contained in $\Q({\root 3 \of 3})$
(they are the nonreal numbers $\rho {\root 3 \of 3}, \rho^2 {\root 3 \of 3}$
where $\rho = \frac{-1}{2} + \frac{i\sqrt{3}}{2}$).

(c) $\Q(\sqrt{5},i)$ is a normal extension of $\Q$, since this is 
the splitting field of $p(x) = (x^2 - 5)(x^2 + 1)$.  (Note that
the polynomial need not be irreducible in $F[x]$ in the theorem.)
\vskip 10pt 
\noindent
18.  Let $u_1,\ldots,u_t$ be the roots of $f(x)\in F[x]$ adjoined 
to obtain $E$, and let $u_1,\ldots, u_n$, $n \ge t$ be all of the 
roots of $f(x)$.  By the definition, $K$ is a splitting field of $f(x)$
over $f$ if and only if $K = F(u_1,\ldots,u_n)$.  But this is true
if and only if 
\begin{eqnarray*}
K &=& F(u_1,\ldots,u_n) \\
  &=& F(u_1,\ldots,u_t)(u_{t+1},\ldots,u_n)\\
  &=& E(u_{t+1},\ldots,u_n)\\
  &=& E(u_1,\ldots,u_n) \text{ since $u_1,\ldots,u_t$ are already in $E$ .}
\end{eqnarray*}
Finally, $K = E(u_1,\ldots,u_n)$ if and only if $K$ is a splitting field
of $f(x)$ over $E$, again by the definition.

%19.  
%(i) $\Rightarrow$ (ii):  By induction on the degree of $f(x) \in K[x]$.  Polynomials
%of degree 1 always have roots in $K$, so there is nothing to prove for the base case.  
%So now assume i holds and that every polynomial of degree $k$ splits completely in $K[x]$.
%Let $f(x) \in K[x]$ have degree $k + 1$.  By (i), $f(x)$ has a root $a$ in $K$.  Therefore
%we know $f(x) = (x - a)q(x)$ for some $q(x) \in K[x]$, which has degree $k$.  By induction,
%$q(x)$ splits completely in $K[x]$.  Therefore the same is true for $f(x)$.
%
%(ii) $\Rightarrow$ (iii): If $f(x)$ has degree $n > 1$, then (ii) implies that
%$f(x) = c(x - a_1)\ldots (x - a_n)$ for some $a_i \in K$.  This implies that
%$f$ is not irreducible in $K[x]$.  Therefore the only irreducible polynomials
%in $K[x]$ have degree 1.
%
%(iii) $\Rightarrow$ (iv):  Let $u$ be algebraic over $K$ and consider the minimal
%polynomial $p(x)$ of $u$ in $K[x]$.  We know $p(x)$ must be irreducible in $K[x]$.
%Therefore (iv) implies that $p(x)$ must have degree 1, so $u \in K$.  It follows
%that the only algebraic extension of $K$ is $K$ itself. 
%
%(iv) $\Rightarrow$ (i):  It suffices to prove this for irreducible polynomials.
%Let $f(x)$ be an irreducible polynomial in $K[x]$. 
%We know there exist algebraic extension fields of $K$ which contain roots of $f(x)$, such
%as $L = K[x]/(f(x))$.  But (iv) implies that $L = K$, which means that $f(x)$ has all of 
%its roots in $K$.  

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