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\bec {\Large\bf Three Examples} \ece
In all our examples the ground field shall be $Q$ and the extension field will be a
subfield of the complex numbers $C$.
We take as basic that the {\em only} nonzero rational numbers $c$ for which $\sqrt{c}\in Q$
are those positive $c$ for which each prime factor $p$ appears an even number of times.
In particular, $\sqrt{2},\sqrt{3},\sqrt{3/2}$ are all irrational.
{\tt Example I:} $K=Q(\sqrt{2},\sqrt{3})$.
As $\sqrt{2}$ has minimal polynomial $x^2-2$, $Q(\sqrt{2})$ has basis $1,\sqrt{2}$ over $Q$.
Now we need a simple result:
\begin{theorem}\label{sqrt3}
$\sqrt{3}\not\in Q(\sqrt{2})$
\end{theorem}
{\tt Proof:} If it were we would have
\[ \sqrt{3} = a+b\sqrt{2} \]
with $a,b\in Q$. Squaring both sides
\[ 3 = a^2+2b^2+2ab\sqrt{2} \]
As $1,\sqrt{2}$ is a basis the coefficient of $\sqrt{2}$ would need be zero.
That is, $2ab=0$. So either $a=0$ or $b=0$.
\ben
\item $b=0$: Then $\sqrt{3}=a\in Q$, contradiction.
\item $a=0$. Then $\sqrt{3}=b\sqrt{2}$ so $\sqrt{3/2}=b\in Q$, contradiction.
\een
From Theorem \ref{sqrt3} and that $\sqrt{3}$ satisfies a quadratic
(namely, $x^2-3$) over $Q(\sqrt{2})$, $1,\sqrt{3}$ is a basis for $Q(\sqrt{2},\sqrt{3})$
over $Q(\sqrt{2})$ and so $1,\sqrt{2},\sqrt{3},\sqrt{6}$ is a basis for $K$ over $Q$.
That is, we may write
\beq\label{a11}
K= Q(\sqrt{2},\sqrt{3}) = \{ a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}: a,b,c,d\in Q\}
\eeq
and every $\ah\in K$ has a {\em unique} expression in this form.
Now we turn to the Galois Group $\Gamma(K:Q)$. Any $\sig\in K$ has
$\sig(\sqrt{2})=\pm \sqrt{2}$ and
$\sig(\sqrt{3})=\pm \sqrt{3}$ which gives four ({\tt Caution:} this is $2$
{\em times} $2$) possibilities. The value of $\sig$ on $\sqrt{2},\sqrt{3}$
determines the value on all of $K$.
The four elements of the Galois Group
are $e,\sig_1,\sig_2,\sig_3$ where
\[ e(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \]
\[ \sig_1(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6} \]
\[ \sig_2(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6} \]
\[ \sig_3(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6} \]
{\tt Exercise:} Check that these really are automorphisms, that they are
bijections (easy!) that send sums to sums (easy!) and products to products
(a bit messier!). This will actually come out of more general stuff later.
What does the group $\Gamma(K:Q)$ look like. The identity is the identity,
no problem. Any $\sig$ when squared gives the identity. For example
\[ \sig_1^2(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= \sig_1(a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6})
=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}) \]
We can also see this by notices that either $\sig(\sqrt{2})=\sqrt{2}$ or
$\sig(\sqrt{2})=-\sqrt{2}$ but in either case $\sig^2(\sqrt{2})=\sqrt{2}$ and
similarly $\sig^2(\sqrt{3})=\sqrt{3}$ so that $\sig^2=e$. We also calculate
that if you multiply any two of $\sig_1,\sig_2,\sig_3$ in either directgion
you get the other third one. For example
\[ \sig_2(\sig_1(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})= \sig_2(a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6})
=a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6}) \]
We have the Klein Viergruppe, aka the Fourgroup, which is isomorphic to $Z_2\times Z_2$.
Actually, there are only two groups with four elements (up to isomorphism, of course), the
cyclic group $Z_4$ and the Vierergruppe $Z_2\times Z_2$ so once it isn't the first it
{\em must} be the second!
{\tt Example II:} $K=Q(\eps)$ with $\eps = e^{2\pi i/5}$.
$\eps$ satisfies $x^5-1=0$ and, as $\eps\neq 1$, it satisfies $p(x)=0$ with
\beq\label{minpolyeps}
p(x) = \frac{x^5-1}{x-1} = x^4+x^3+x^2+x+1
\eeq
This is irreducible (one can show this by replacing $x$ by $x+1$ giving
$x^4+5x^3+10x^2+5x+5$ and using Eisenstein's criterion) so $[K:Q]=4$ and we write
\beq \label{basiseps}
K = \{ a + b\eps + c\eps^2+d\eps^3: a,b,c,d\in Q \}
\eeq
The minimal polynomial (\ref{minpolyeps}) has roots $\eps,\eps^2,\eps^3,\eps^4$.
From our general results (earlier notes) the Galois Group $\Gamma(K:Q)$ consists
of four automorphism which we shall label $\sig_1,\sig_2,\sig_3,\sig_4$. They
are determined by their values on $\eps$ and we shall let $\sig_j$ be that
automorphism with $\sig_j(\eps)=\eps^j$. Note that $\sig_1$ is the identity
and $\sig_4$ is our old friend, complex conjugation.
What is the product $\sig_j\sig_k$. Lets see what it does to $\eps$.
\beq \label{prod1}
(\sig_j\sig_k)(\eps)= \sig_j(\sig_k(\eps))= \sig_j(\eps^k)= \sig_j(\eps)^k = (\eps^j)^k = \eps^{jk} \
\eeq
Hmmmm, so it looks like $\sig_j\sig_k = \sig_{jk}$. But we only have four automorphisms. What does
it mean to say $\sig_3\sig_3=\sig_9$. The key is that $\eps^5=1$ so that we can reduce $\eps^{jk}$
by reducing $jk$ modulo $5$. As $\eps^9=\eps^4$ we have $\sig_3\sig_3=\sig_4$. So we {\em can} and
{\em do} say
$\sig_j\sig_k = \sig_{jk}$ with the understanding that $jk$ is computed modulo $5$. With this we
have
\beq\label{iso1} \Gamma(K:Q) \cong Z_5^* \eeq
where we associate $\sig_j$ with $j$. Finally $Z_5^*\cong (Z_4,+)$ (the cyclic group, not the
Vierergruppe) be associating $1,2,3,4$ with $0,1,3,2$ respectively.
{\tt Example III:} $K=Q(2^{1/3},\omega)$ with $\omega = e^{2\pi i/3}$.
The polynomial
\beq \label{minpoly3} p(x)=x^3-2 \eeq
is irreducible (by Eisenstein's criterion, or simply that, as a cubic, it has
no rational roots) and its roots are $\ah,\beta,\gamma$ where for convenience we
write
\beq\label{ahbetagam}
\ah=2^{1/3},\beta=2^{1/3}\omega,\gam=2^{1/3}\omega^2
\eeq
Any field that contains $2^{1/3},\omega$ contains
$\ah,\beta,\gam$
and any field that contains
$\ah,\beta,\gam$
contains $2^{1/3},\omega$ so we may also write $K=Q(\ah,\beta,\gam)$,
so $K$ is the {\em splitting field} (to be defined later) of $p(x)$ over $Q$.
A basis for $Q(\ah)$ over $Q$ is $1,\ah,\ah^2$.
As all elements of $Q(\ah)$ are real,
$\omega\not\in Q(\ah)$. As $\omega$ satisfies the quadratic $1+x+x^2=0$ over
$Q(\ah)$, a basis for $K=Q(\ah,\omega)$ over $Q(\ah)$ is $1,\omega$. Hence
a basic for $K$ over $Q$ is $1,\ah,\ah^2,\omega,\omega\ah,\omega\ah^2$ and we
can write
\beq\label{3basis}
K= \{a+b\ah+c\ah^2+d\omega+e\omega\ah+f\omega\ah^2:a,b,c,d,e,f\in Q \}
\eeq
and each $\zeta\in K$ has a unique such representation.
What are the automorphisms $\sig\in\Gamma(K:Q)$? As $K=Q(\ah,\beta,\gam)$, $\sig$
is determined by its values on $\ah,\beta,\gam$. Further, as $\ah,\beta,\gam$
satisfy the same irreducible (\ref{minpoly3}), $\sig$ of any of them must be one
of them. Further, as $\sig$ must be a {\em bijection}, $\sig$ cannot send
two of $\ah,\beta,\gam$ to the same value and hence $\sig$ must be a permutation
on $\ah,\beta,\gam$. This gives that there are {\em at most} six automorphisms
and that $\Gamma(K:Q)$ is isomorphic to a subgroup of $S_3$, the full symmetric
group on three elements, here $\ah,\beta,\gam$.
Actually, {\em all} permutations of $\ah,\beta,\gam$ yield automorphisms of
$K$ and so
\beq\label{IIIgal} \Gamma(K:Q) \cong S_3 \eeq
This actually will follow from some general stuff but we can give an idea
here. Two automorphisms are easy, the identity (we {\em always} have the
identity) and complex conjugation $\sig$. We do have to check that
complex conjugation is a bijection from $K$ to itself. As $\sig(\ah)=\ah$ and
$\sig(\omega)=\omega^2\in K$ it sends $K$ to $K$ and since $\sig^2=e$ it
must be a bijection. (That is, $\sig^{-1}(\zeta)=\sig(\zeta)$.) This
$\sig$ corresponds to the permutation that keeps $\ah$ fixed and transposes
$\beta,\gam$.
Here is another $\tau\in \Gamma(K:Q)$. Generate it by setting $\tau(\ah)=\beta$
and $\tau(\omega)=\omega$. Then
\beq\label{whitherbeta}
\tau(\beta)=\tau(\ah\omega)=\tau(\ah)\tau(\omega)=\beta\omega=\gamma
\eeq
and
\beq\label{whithergamma}
\tau(\gam)=\tau(\beta\omega)=\tau(\beta)\tau(\omega)=\gamma\omega=\ah
\eeq
so it cycles $\ah$ to $\beta$ to $\gam$ back to $\ah$. With the representation
of (\ref{3basis})
\beq\label{tauzeta}
\tau(a+b\ah+c\ah^2+d\omega+e\omega\ah+f\omega\ah^2)=
a+b\omega\ah + c\omega^2\ah^2+d\omega+e\omega^2\ah+f\ah^2
\eeq
In this form (but, like I said, there are other approaches) one need show
$\tau$ is bijective (pretty easy), that $\tau(\zeta_1+\zeta_2)=\tau(\zeta_1)+\tau(\zeta_2)$ (quite
easy), and that $\tau(\zeta_1\zeta_2)=\tau(\zeta_1)\tau(\zeta_2)$ (lengthy, unless you use some
tricks).
Indeed, here is another approach to show that $\tau$ is indeed an automorphism from $K$ to $K$.
Consider the intermediate field $L=Q(\omega)$. We first claim that $p(x)$ given by (\ref{minpoly3})
is irreducible over $L$. As it is a cubic, if it reduced it would have a root in $L$. So either
$\ah,\beta=\ah\omega,\gam=\ah\omega^2$ would be in $L$. As $\omega\in L$ if any of $\ah,\beta,\gam$
were in $L$ then all three would be in $L$, in particular $\ah\in L$. But then $L$ would have $\omega$
and $\ah$ and so would be $Q(\ah,\omega)=K$. As $[L:Q]=2\neq 6=[K:Q]$ that cannot happen. Now
$\ah,\beta$ have the same minimal polynomial in $L[x]$ and $K=L(\ah)=L(\beta)$, so, from the previous notes, there
does exist an automorphism $\tau$ of $K$ that preserves $L$ and has $\tau(\ah)=\beta$. As it preserves $L$ we also
have $\tau(\omega)=\omega$. Since $\tau$ preserves $L$ it certainly preserves the smaller $Q$ and so
$\tau\in\Gamma(K:Q)$.
Once we have $\sig,\tau\in \Gamma(K:Q)$ we have that $\Gamma(K:Q)$ is a subgroup of $S_3$ that
contains an element $\tau$ of order three and an element $\sig$ of order two. From $\tau$ it
must have at least three elements, from $\sig$ it can't have exactly three elements, so it
has more than three elements, so it has all six elements, it is all of $S_3$.
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