Tutorial 3 : Separable Extensions and Finite Fields

$(\Rightarrow)$ Let

be irreducible in

. If

for some $b \in F$ , then $x^{p^n}-a = x^{p^n}-b^p = (x^{p^{n-1}}-b)^p$ . This is a contradiction to the fact that

is irreducible. $(\Leftarrow)$ Let $a \notin F^p$ and suppose

is reducible and

where $g(x),h(x) \in F[x]$ are monic of degree

and

respectively. Let

be the splitting field of

over

and

be a root of

. Then $a=b^{p^n}$ and $f(x)=(x-b)^{p^n}$ . Thus

is the only root of

. Since

$b^d \in F$ . Let

. Let $x,y \in \mathbb{Z}$ so that

. Hence

$\displaystyle b^{p^e}=b^{dx}b^{p^ny}=(b^d)^xa^y \in F$

But $a=b^{p^n}=(b^{p^e})^{p^{n-e}} \in F^p$ which is a contradiction.

2. Show that $\bigcap_{i=0}^{\infty} F^{p^i}$ is the largest perfect subfield of

Let

be a perfect subfield of

. Then

$\displaystyle L=L^p=L^{p^2}=\cdots =L^{p^n}.$

Therefore $\cap_{n=0}^\infty L^{p^n} = L \subseteq \cap_{n=0}^{\infty} F^{p^n}$ . Let $K=\cap_{n=1}^{\infty} F^{p^n}$ . To show

is perfect, we show

. Let $x \in K$ . Then

$\displaystyle x=a_1^p=a_2^{p^2}=\cdots=a_n^{p^n}$ for some $\displaystyle a_n \in F\;$ for all $\displaystyle n.$

Now

. Thus $a_1=a_2^p,a_2=a_3^p,a_3=a_4^p=\cdots$ . Therefore $a_1=a_2^p=a_3^{p^2}=a_4^{p^3}=\cdots.$ Hence $a_1 \in \cap_{n=1}^{\infty} F^{p^n} =K$ and

So $x \in K^p$ . Hence

is perfect.

3. Identify the finite fields $\mathbb{Z}[i]/(1+i)$ and $\mathbb{Z}[i]/(2+i).$

More generally we show that if $a \in \mathbb{Z}$ ,then

$\displaystyle \mathbb{Z}/(a^2+1)\mathbb{Z}\simeq \mathbb{Z}[i]/(a+i).$

Since $\overline{a+i}=\overline{0},\overline{i}=-\overline{a}$ . Hence

$\displaystyle \overline{x+iy}=\overline{x}+\overline{i}\:\overline{y}= \overline{x}-\overline{ ay}=\overline{x-ay}.$

Hence $\mathbb{Z}\longrightarrow \mathbb{Z}[i]/(a+i)$ is surjective. If $n=(a+i)(c+id)\;$ then $\; n=(ac-d)+i(c+ad)$ . Thus

$\displaystyle c=-ad \;$ and $\displaystyle \; n=a(-ad)-d=-d(a^2+1) \in (a^2+1)\mathbb{Z}.$

Thus $\mathbb{Z}/(a^2+1)\mathbb{Z}\simeq \mathbb{Z}[i]/(a+i)$ . Now put

and

to see that $\mathbb{Z}[i]/(1+i)\simeq \mathbb{F}_2$ and $\mathbb{Z}[i]/(2+i)\simeq \mathbb{F}_5$

4. Find a necessary and sufficient condition on

and

so that $\mathbb{F}_{p^n}$ is a subfield of $\mathbb{F}_{p^m}.$

We need to show $n \mid m$ if and only if $\mathbb{F}_{p^m} \subset \mathbb{F}_{p^n}$ . Since $[\mathbb{F}_{p^m}:\mathbb{F}_p]=[\mathbb{F}_{p^m}:\mathbb{F}_{p^n}][\mathbb{F}_{p^n};\mathbb{F}_p],$ it follows that $n \mid m$ . Conversely if $n \mid m$ ,that is

then $p^n-1 \mid (p^r)^n-1$ . Hence $x^{p^n-1}-1$ divides $x^{p^m-1}-1$ . Thus roots of $x^{p^n}-x$ are roots of $x^{p^m}-x$ . Hence $\mathbb{F}_{p^m} \subset \mathbb{F}_{p^n}$ .

5. Draw the poset of subfields of $\mathbb{F}_{3^{18}}$

Divisors of

are

Thus the poset of lattices of the subfields are:

$\displaystyle \xymatrix{ &&\mathbb{F}_{3^{18}} \ar@{-}[ld] \ar@{-}[rd] \\ &\ma... ...mathbb{F}_{3^2} \ar@{-}[rd] & & \mathbb{F}_{3^3} \ar@{-}[ld]\\ &\mathbb{F}_3 }$

6. Show that the order of the Frobenius automorphism $\phi : K:=\mathbb{F}_{p^n} \rightarrow \mathbb{F}_{p^n}$ is

Note that $\phi^n(a)=a^{p^n}=a$ for all $a \in \mathbb{F}_{p^n}.$ Let $\circ$ denote the order in the group of all $\mathbb{F}_p$ -automorphisms of

Then $\circ(\phi) \mid n$ . Suppose $\circ(\phi)=r<n$ .Then $\phi^r(a)=a=a^{p^r}\;$ for all $\;a \in \mathbb{F}_{p^n}$ . But $x^{p^r}-x$ has only

roots. This is a contradiction. Hence $\circ(\phi)=n$ .

7. Let

be irreducible over $\mathbb{F}_p$ and let it have a root $r \in \mathbb{F}_{p^n}.$ Show that the roots of

are precisely $r, r^p, r^{p^2}, \ldots, r^{p^{m-1}}.$

is a splitting field of $g(x)=x^{p^n}-x$ over $\mathbb{F}_p,$ any embedding $\sigma : K \rightarrow \mathbb{F}_p^a$ is an $\mathbb{F}_p$ -automorphism of

But $\mathbb{F}_p$ is perfect and so

is a separable extension of $\mathbb{F}_p.$ Therefore the number of $\mathbb{F}_p$ -automorphisms of

But $\circ(\phi)=n.$ Hence all the roots of

are obtained by applying $\phi$ to

successively. Hence the roots are $r^{p^i}$ where $i=0,1,2,\ldots, m-1.$

8. Let

and

be subfields of $\mathbb{F}_{p^n}$ having

and

elements respectively. How many elements does the field $K \cap L$ have ?

As $\mathbb{F}_{p^t} \cap \mathbb{F}_{p^s}=\mathbb{F}_{p^r}$ for some

$r \mid t \;$ and $\;r \mid s$ . Thus $r \mid \gcd(t,s)=d.$ On the other hand $\mathbb{F}_{p^d} \subset \mathbb{F}_{p^t} \cap \mathbb{F}_{p^s}$ . Hence $\mathbb{F}_{p^d} = \mathbb{F}_{p^t} \cap \mathbb{F}_{p^s}.$ Hence we have the following diagram of subfields.

$\displaystyle \xymatrix{ & \mathbb{F}_{p^n} \ar@{-}[ld] \ar@{-}[rd]& \\ \math... ...t} \cap \mathbb{F}_{p^s}=\mathbb{F}_{p^r} \ar@{-}[d]&\\ & \mathbb{F}_p & \\ }$