## 305 -Extension fields revisited (3)

1. Isomorphisms

We return here to the quotient ring construction. Recall that if ${R}$ is a commutative ring with identity and ${I}$ is an ideal of ${R,}$ then ${R/I}$ is also a commutative ring with identity. Here, ${R/I={[a]_sim:ain R},}$ where ${[a]_sim={b:asim b}}$ for ${sim}$ the equivalence relation defined by ${asim b}$ iff ${a-bin I.}$

Since ${sim}$ is an equivalence relation, we have that ${[a]_sim=[b]_sim}$ if ${asim b}$ and ${[a]_simcap[b]_sim=emptyset}$ if ${anotsim b.}$ In particular, any two classes are either the same or else they are disjoint.

In case ${R={mathbb F}[x]}$ for some field ${{mathbb F},}$ then ${I}$ is principal, so ${I=(p)}$ for some ${pin{mathbb F}[x],}$ i.e., given any polynomial ${qin{mathbb F}[x],}$ ${[q]_sim=0}$ iff ${p|q}$ and, more generally, ${[q]_sim=[r]_sim}$ (or, equivalently, ${qsim r}$ or, equivalently, ${rin[q]_sim}$) iff ${p|(q-r).}$

In this case, ${{mathbb F}[x]/(p)}$ contains zero divisors if ${p}$ is nonconstant but not irreducible.

If ${p}$ is 0, ${{mathbb F}[x]/(p)cong{mathbb F}.}$

If ${p}$ is constant but nonzero, then ${{mathbb F}[x]/(p)cong{0}.}$

Finally, we want to examine what happens when ${p}$ is irreducible. From now on suppose that this is the case.

Let ${q}$ be any polynomial in ${{mathbb F}[x].}$ If ${p|q}$ then ${[q]_sim=0.}$ Otherwise, ${{rm gcd}(p,q)=1}$ since ${p}$ is irreducible. So there are polynomials ${r,sin{mathbb F}[x]}$ with ${pr+qs=1.}$ Therefore ${[q]_sim[s]_sim=1.}$ It follows that every nonzero element of ${{mathbb F}[x]/(p)}$ is invertible, i.e, we have the following:

Lemma 1 ${{mathbb F}[x]/(p)}$ is a field whenever ${pin{mathbb F}[x]}$ is irreducible. ${Box}$

Suppose that ${a,bin {mathbb F}.}$ If ${asim b}$ then ${p|(a-b).}$ Since ${{rm deg}(p)>0}$ and ${a-b}$ is a constant, we must have ${a-b=0,}$ i.e., ${a=b.}$ This means that we can identify each ${ain{mathbb F}}$ with the corresponding element ${[a]_sim}$ of ${{mathbb F}[x]/(p),}$ and we have:

Lemma 2 Suppose that ${pin{mathbb F}[x]}$ is irreducible. Under the identification ${amapsto[a]_sim}$ for ${ain{mathbb F},}$ we have that ${{mathbb F}[x]/(p)}$ is a field extension of ${{mathbb F}.}$ ${Box}$

Whenever convenient, we will just write ${a}$ instead of ${[a]_sim}$ for ${ain{mathbb F}.}$

Now we connect these field extensions with algebraic numbers. See last lecture for the definition of the minimal polynomial of an algebraic number and its basic properties.

Theorem 3 Let ${{mathbb H}={mathbb F}[x]/(p),}$ where ${pin{mathbb F}[x]}$ is irreducible and monic. Let ${alpha=[x]_sim.}$ Then ${alpha}$ is algebraic over ${{mathbb F}}$ with minimal polynomial ${p_alpha=p.}$ Moreover, ${{mathbb H}={mathbb F}(alpha).}$

Proof: Let ${p(x)=a_0+a_1x+dots+a_nx^n}$ for some ${a_0,dots,a_nin{mathbb F}}$ with ${a_0ne0}$ (since ${p}$ is irreducible) and ${a_nne0}$ (so ${{rm deg}(p)=n}$).

Note first that ${0=[p]_sim=[a_0+a_1x+dots+a_nx^n]_sim.}$ Using our identification of elements of ${{mathbb F}}$ with the corresponding classes in ${{mathbb H},}$ we can write

$displaystyle [a_0+a_1x+dots+a_nx^n]_sim=a_0+a_1[x]_sim+dots+a_n[x^n]_sim.$

Since ${[x^k]_sim=[x]^k_sim=alpha^k,}$ we have ${0=a_0+a_1alpha+dots+a_nalpha^n=p(alpha).}$ This shows that ${alpha}$ is algebraic over ${{mathbb F}.}$

Since any element of ${{mathbb H}}$ is ${[q]_sim}$ for some ${qin{mathbb F}[x],}$ and the same argument as above shows that ${[q]_sim=q(alpha),}$ we obviously have ${{mathbb H}={mathbb F}(alpha).}$

Let ${p_alpha}$ be the minimal polynomial of ${alpha}$ over ${{mathbb F}.}$ Since ${p(alpha)=0}$ then ${p_alpha|p.}$ Since ${p}$ is irreducible, we must have ${{rm deg}(p)={rm deg}(p_alpha).}$ If ${p}$ is also monic, then we must in fact have ${p=p_alpha.}$ $Box$

Suppose for example that ${betain{mathbb C}}$ is algebraic (over ${{mathbb Q}}$) with minimal polynomial ${p_beta.}$ By the above, both ${{mathbb Q}(beta)}$ and ${{mathbb Q}[x]/(p_beta)}$ are field extensions of ${{mathbb Q}}$ of degree ${{rm deg}(p_beta)}$ where ${p_beta}$ has a root (${beta}$ in one case, ${alpha:=[x]_sim}$ in the other). Although on the surface both extensions look very different, as far as field theory is concerned, they are indistinguishable:

Theorem 4 Suppose that ${{mathbb F}}$ is a field, and that ${{mathbb F}(alpha)}$ and ${{mathbb F}(beta)}$ are two field extensions of ${{mathbb F}}$ (not necessarily subfields of a common larger extension field).

If ${alpha}$ and ${beta}$ are algebraic over ${{mathbb F}}$ and ${p_alpha=p_beta,}$ then ${{mathbb F}(alpha)}$ and ${{mathbb F}(beta)}$ are isomorphic. Moreover, if ${n={rm deg}(p_alpha),}$ then the map ${pi:{mathbb F}(alpha)rightarrow{mathbb F}(beta)}$ defined as follows is an isomorphism: Any element of ${{mathbb F}(alpha)}$ can be written in a unique way as ${a_0+a_1alpha+dots+a_{n-1}alpha^{n-1}}$ where the numbers ${a_0,dots,a_{n-1}}$ are in ${{mathbb F}.}$ Set

$displaystyle pi(a_0+a_1alpha+dots+a_{n-1}alpha^{n-1})=a_0+a_1beta+dots+a_{n-1}beta^{n-1}.$

Proof: As shown in Corollary 8 from last lecture, any element of ${{mathbb F}(alpha)}$ can be written as stated and, similarly, any element of ${{mathbb F}(beta)}$ also has the form ${q(beta)}$ for some ${qin{mathbb F}[x]}$ of degree at most ${n-1.}$ Moreover, such representation is unique. It follows that ${pi}$ is well-defined, injective, and surjective. It is straightforward to verify that ${pi(u+v)=pi(u)+pi(v)}$ for any ${u,vin{mathbb F}(alpha)}$ and that ${pi(0)=0.}$

It takes a little bit more effort to check that ${pi(uv)=pi(u)pi(v).}$ Once this is done, it follows that ${pi}$ is an isomorphism and we are done.

Let ${p,q}$ be polynomials in ${{mathbb F}[x]}$ such that ${p(alpha)=u,}$ ${q(alpha)=v}$ and ${{rm deg}(p),{rm deg}(q)le n-1.}$ We can find polynomials ${s,tin{mathbb F}[x]}$ such that ${pq=sp_alpha+t}$ and ${{rm deg}(t)<{rm deg}(p_alpha).}$ Moreover, such polynomials are unique.

It follows that ${uv=p(alpha)q(alpha)=s(alpha)p_alpha(alpha)+t(alpha)=t(alpha)}$ and therefore

$displaystyle pi(uv)=pi(t(alpha))=t(beta).$

Since ${p(beta)q(beta)=s(beta)p_alpha(beta)+t(beta)}$ and ${p_alpha=p_beta,}$ then

$displaystyle t(beta)=p(beta)q(beta)=pi(u)pi(v),$

as needed. $Box$

The converse of the previous result is also true. Before stating this precisely, we need a general result about injective homomorphisms.

Lemma 5 Let ${{mathbb F}}$ and ${{mathbb H}}$ be two field extensions of ${{mathbb K}}$ (not necessarily subfields of a common larger extension field). Suppose that ${h:{mathbb F}rightarrow{mathbb H}}$ is an injective homomorphism and that ${h}$ fixes all the elements of ${{mathbb K},}$ i.e., ${h(a)=a}$ for all ${ain{mathbb K}.}$ If ${S}$ is a subset of ${{mathbb F}}$ linearly independent over ${{mathbb K},}$ then ${{h(s):sin S}}$ is a subset of ${{mathbb H}}$ linearly independent over ${{mathbb K}.}$

Proof: Suppose ${s_1,dots,s_nin S.}$ Let ${a_1,dots,a_nin{mathbb K}}$ and suppose that

$displaystyle a_1h(s_1)+dots a_nh(s_n)=0.$

Since ${a_i=h(a_i),}$ the above equals ${h(a_1s_1+dots+a_n s_n).}$ Also, ${h(0)=0.}$ Since ${h}$ is 1-1, it follows that

$displaystyle a_1s_1+dots+a_ns_n=0.$

Since the ${s_iin S}$ are linearly independent over ${{mathbb K},}$ then ${a_1=dots=a_n=0.}$ This shows that ${{h(s):sin S}}$ is linearly independent over ${{mathbb K}.}$ $Box$

Corollary 6 If ${{mathbb F}}$ and ${{mathbb H}}$ are field extensions of ${{mathbb K}}$ and ${h:{mathbb F}rightarrow{mathbb H}}$ is an isomorphism that fixes all the elements of ${{mathbb K},}$ then ${[{mathbb F}:{mathbb K}]=[{mathbb H}:{mathbb K}].}$

Proof: The lemma shows that if ${S}$ is a basis of ${{mathbb F}}$ over ${{mathbb K},}$ then ${{h(s):sin S}}$ is linearly independent over ${{mathbb K}.}$ Since ${|S|=|{h(s):sin S}|,}$ then ${[{mathbb F}:{mathbb K}]le[{mathbb H}:{mathbb K}].}$

The map ${h^{-1}}$ is an isomorphism from ${{mathbb H}}$ to ${{mathbb F}}$ and it fixes all the elements of ${{mathbb K}.}$ The same argument as above implies that ${[{mathbb H}:{mathbb K}]le[{mathbb F}:{mathbb K}].}$

The result follows. $Box$

Theorem 7 Suppose that ${alpha}$ is algebraic over ${{mathbb F},}$ that ${{mathbb H}:{mathbb F},}$ and that ${pi:{mathbb F}(alpha)rightarrow{mathbb H}}$ is an isomorphism that fixes all the elements of ${{mathbb F}.}$ Then ${{mathbb H}={mathbb F}(pi(alpha)),}$ and ${pi}$ is as given in Theorem 4. Moreover, ${p_{pi(alpha)}=p_alpha.}$

Proof: By the corollary, ${[{mathbb H}:{mathbb F}]=[{mathbb F}(alpha):{mathbb F}]={rm deg}(p_alpha)=n,}$ say, where ${p_alphain{mathbb F}[x]}$ is the minimal polynomial of ${alpha}$ over ${{mathbb F}.}$

We know that ${S={1,alpha,dots,alpha^{n-1}}}$ is a basis of ${{mathbb F}(alpha)}$ over ${{mathbb F}.}$ The corollary also shows that ${{1,pi(alpha),dots,pi(alpha)^{n-1}}}$ is a basis of ${{mathbb H}}$ over ${{mathbb F}.}$This gives that ${{mathbb H}={mathbb F}(pi(alpha))}$ and that ${pi}$ is as in Theorem 4, i.e.,

$displaystyle pi(a_0+a_1alpha+dots+a_{n-1}alpha^{n-1})=a_0+a_1pi(alpha)+dots+a_{n-1}pi(alpha)^{n-1}.$

Moreover, ${pi_alpha(pi(alpha))=pi(p_alpha(alpha))=pi(0)=0,}$ so ${pi(alpha)}$ is a root of ${p_alpha.}$ Since ${p_alpha}$ is irreducible, ${p_{pi(alpha)}=p_alpha.}$ $Box$

2. Examples

The results above give us a very detailed control over the isomorphisms that we can expect to have between fields.

1. For example, it follows that ${{mathbb Q}({root3of2})}$ is isomorphic to precisely three subfields of ${{mathbb C},}$ namely:

$displaystyle {mathbb Q}({root 3of2}), {mathbb Q}({root3of2}zeta_3),mbox{ and }{mathbb Q}({root3of2}zeta_3^2)$

where, as usual, ${zeta_3=cos(2pi/3)+isin(2pi/3).}$

To see this, note that Theorems 4 and 7 give us that if ${{mathbb C}:{mathbb H},}$ then ${{mathbb Q}({root 3of2})}$ is isomorphic to ${{mathbb H}}$ iff ${{mathbb H}={mathbb Q}(xi)}$ for some root ${xi}$ of ${p_{root 3of2}(x)=x^3-2.}$ This gives that the only subfields of ${{mathbb C}}$ isomorphic to ${{mathbb Q}({root 3of2})}$ are the ones listed above. That there are three of them is the same as saying that they are all different.

This can be checked as follows: First, ${{mathbb Q}({root 3of2})ne{mathbb Q}({root 3of2}zeta_3)}$ and ${{mathbb Q}({root 3of2})ne{mathbb Q}({root 3of2}zeta_3^2),}$ since ${{mathbb Q}({root 3of2})subseteq{mathbb R}}$ while the other fields contain complex nonreal numbers. Second, ${{mathbb Q}({root 3of2}zeta_3)ne{mathbb Q}({root 3of2}zeta_3^2),}$ since otherwise

$displaystyle zeta_3=frac{{root 3of2}zeta_3^2}{{root 3of2}zeta_3}in{mathbb Q}({root 3of2}zeta_3).$

Now, ${p_{zeta_3}(x)=x^2+x+1}$ has degree 2, so ${2=[{mathbb Q}(zeta_3):{mathbb Q}]}$ would have to divide ${[{mathbb Q}({root3of2}zeta_3):{mathbb Q}],}$ by the tower law:

$displaystyle [{mathbb F}:{mathbb K}]=[{mathbb F}:{mathbb H}][{mathbb H}:{mathbb K}]$

whenever ${{mathbb F}:{mathbb H}:{mathbb K}.}$ This is of course impossible since ${[{mathbb Q}({root 3of2}zeta_3):{mathbb Q}]=3.}$

2. Continuing with this example, we can now easily identify all the subfields of ${{mathbb Q}^{p(x)},}$ where ${p(x)=x^3-2.}$ We know that ${{mathbb Q}^{p(x)}={mathbb Q}({root3of2},zeta_3).}$ Since ${zeta_3notin{mathbb Q}({root 3of2}),}$ by the tower law it follows that

$displaystyle [{mathbb Q}({root 3of2},zeta_3):{mathbb Q}]=6,$

and that any (strictly) intermediate field between ${{mathbb Q}({root 3of2},zeta_3)}$ and ${{mathbb Q}}$ is a field extension of ${{mathbb Q}}$ of degree either 2 or 3.

Let ${{mathbb F}}$ be such a field. Suppose first that ${[{mathbb F}:{mathbb Q}]=3.}$ If ${p(x)}$ is irreducible over ${{mathbb F},}$ then

$displaystyle {mathbb Q}({root 3of2},zeta_3):{mathbb F}({root3of2}):{mathbb F}:{mathbb Q}$

and

$displaystyle [{mathbb F}({root3of2}):{mathbb F}]=3$

which, by the tower law, would imply that ${9|6=[{mathbb Q}({root 3of2},zeta_3):{mathbb Q}],}$ contradiction.

It follows that ${p(x)}$ is not irreducible over ${{mathbb F}}$ and therefore it has a root ${alpha}$ in ${{mathbb F},}$ from which it follows that ${{mathbb F}={mathbb Q}(alpha).}$ This is because ${{rm deg}(p)=3,}$ so if it factors, one of its factors is linear, so we must have a root (and by the tower law).

Thus, the only subfields of ${{mathbb Q}^{p(x)}}$ of degree 3 over ${{mathbb Q}}$ are

$displaystyle {mathbb Q}({root 3of2}), {mathbb Q}({root3of2}zeta_3),mbox{ and }{mathbb Q}({root3of2}zeta_3^2).$

Suppose now that ${[{mathbb F}:{mathbb Q}]=2.}$ If ${q(x)=x^2+x+1}$ is irreducible over ${{mathbb F},}$ then

$displaystyle {mathbb Q}({root 3of2},zeta_3):{mathbb F}(zeta_3):{mathbb F}:{mathbb Q}$

and

$displaystyle [{mathbb F}(zeta_3):{mathbb F}]=2$

which again leads to a contradiction with the tower law, since ${4not| 6.}$

It follows that ${q(x)}$ is not irreducible over ${{mathbb F}}$ and therefore it factors over ${{mathbb F}}$ and ${{mathbb F}={mathbb Q}(zeta_3).}$

3. Suppose now that ${p}$ is a prime and ${{mathbb F}}$ is a finite field of characteristic ${p.}$ Then, in particular, ${{mathbb F}:{mathbb Z}_p,}$ so ${{mathbb F}}$ is a finite extension of ${{mathbb Z}_p.}$ If ${[{mathbb F}:{mathbb Z}_p]=n,}$ then it follows that ${|{mathbb F}|=p^n.}$ To see this, suppose that ${v_1,dots,v_n}$ is a basis. Then all the combinations

$displaystyle sum_{i=1}^n a_iv_i$

with ${a_iin{mathbb Z}_p}$ are different, and every element of ${{mathbb F}}$ is expressible in this form. To count the number of these expressions, note that there are ${p}$ possible values for each ${a_i.}$

For example, if ${p=2,}$ to build a field of size 4, it would suffice to find an irreducible polynomial ${pin{mathbb Z}_2[x]}$ of degree 2, and to form ${{mathbb F}={mathbb Z}_2[x]/(p).}$ The only candidates are ${p(x)=x^2,x^2+1,x^2+x,x^2+x+1.}$ The first and third are not irreducible since 0 is a root. The second is not irreducible since ${x^2+1=(x+1)^2.}$ The last one is irreducible, and it is our only choice.

Let now ${alpha=[x]_simin{mathbb F},}$ so ${{mathbb F}={mathbb Z}_2(alpha).}$ The elements of ${{mathbb F}}$ are then ${0,1,alpha,alpha+1.}$ To write the multiplication table, only ${alpha(alpha+1)}$ needs to be determined. For this, note that ${alpha(alpha+1)=[x^2+x]_sim=[(x^2+x+1)+1]_sim=1.}$

4. For a slightly less trivial example, let’s now build a field of 8 elements. Then we need an irreducible polynomial ${pin{mathbb Z}_2[x]}$ of degree 3. The candidates are:

• ${p(x)=x^3,x^3+x,x^3+x^2,x^3+x^2+x,}$ none of which are irreducible since ${0}$ is a root,
• ${p(x)=x^3+1=(x+1)(x^2+x+1),}$ ${p(x)=x^3+x^2+x+1=(x+1)^3,}$ which are not irreducible since ${1}$ is a root,
• ${p(x)=x^3+x+1,}$ or ${p(x)=x^3+x^2+1.}$ Both these polynomials are irreducible.

We have therefore two possible fields of size 8: ${{mathbb F}={mathbb Z}_2[x]/(x^3+x+1)}$ and ${{mathbb H}={mathbb Z}_2[x]/(x^3+x^2+1).}$

Case 1: ${{mathbb F}.}$

Let ${alpha=[x]_sim.}$ Then the elements of ${{mathbb F}}$ are ${0,1,alpha,alpha+1,alpha^2,alpha^2+1,alpha^2+alpha}$ and ${alpha^2+alpha+1.}$ To write the multiplication table, we simply have to “reduce the corresponding polynomials modulo ${x^3+x+1.}$” For example, ${alpha^2(alpha^2+x+1)=[x^4+x^3+x^2]_sim=[(x^3+x+1)(x+1)+1]_sim=1.}$ Similarly, we find:

• ${alphatimesalpha=alpha^2,}$ ${alphatimes(alpha+1)=alpha^2+alpha,}$ ${alphatimesalpha^2=alpha+1,}$ ${alphatimes(alpha^2+1)=1,}$ ${alphatimes(alpha^2+alpha)=alpha^2+alpha+1,}$ ${alphatimes(alpha^2+alpha+1)=alpha^2+1.}$
• ${(alpha+1)times(alpha+1)=alpha^2+1,}$ ${(alpha+1)timesalpha^2=alpha^2+alpha+1,}$ ${(alpha+1)times(alpha^2+1)=alpha^2,}$ ${(alpha+1)times(alpha^2+alpha)=1,}$ ${(alpha+1)times(alpha^2+alpha+1)=alpha.}$
• ${alpha^2timesalpha^2=alpha^2+alpha,}$ ${alpha^2times(alpha^2+1)=alpha,}$ ${alpha^2times(alpha^2+alpha)=alpha^2+1,}$ ${alpha^2times(alpha^2+alpha+1)=1.}$
• ${(alpha^2+1)times(alpha^2+1)=alpha^2+alpha+1,}$ ${(alpha^2+1)times(alpha^2+alpha)=alpha+1,}$ ${(alpha^2+1)times(alpha^2+alpha+1)=alpha^2+alpha.}$
• ${(alpha^2+alpha)times(alpha^2+alpha)=alpha,}$ ${(alpha^2+alpha)times(alpha^2+alpha+1)=alpha^2.}$
• ${(alpha^2+alpha+1)times(alpha^2+alpha+1)=alpha+1.}$

Case 2: ${{mathbb H}}$.

Let ${beta=[x]_sim.}$ Just as before, the elements of ${{mathbb H}}$ are ${0,1,beta,beta+1, beta^2,beta^2+1,beta^2+beta,}$ and ${beta^2+beta+1.}$ To write the multiplication table, now we reduce modulo ${x^3+x^2+1.}$ We have:

• ${betatimesbeta=beta^2,}$ ${betatimes(beta+1)=beta^2+beta,}$ ${betatimesbeta^2=beta^2+1,}$ ${betatimes(beta^2+1)=beta^2+beta+1,}$ ${betatimes(beta^2+beta)=1,}$ ${betatimes(beta^2+beta+1)=beta+1.}$
• ${(beta+1)times(beta+1)=beta^2+1,}$ ${(beta+1)timesbeta^2=1,}$ ${(beta+1)times(beta^2+1)=beta,}$ ${(beta+1)times(beta^2+beta)=beta^2+beta+1,}$ ${(beta+1)times(beta^2+beta+1)=beta^2.}$
• ${beta^2timesbeta^2=beta^2+beta+1,}$ ${beta^2times(beta^2+1)=beta+1,}$ ${beta^2times(beta^2+beta)=beta,}$ ${beta^2times(beta^2+beta+1)=beta^2+beta.}$
• ${(beta^2+1)times(beta^2=1)=beta^2+beta,}$ ${(beta^2+1)times(beta^2+beta)=beta^2,}$ ${(beta^2+1)times(beta^2+beta+1)=1.}$
• ${(beta^2+beta)times(beta^2+beta)=beta+1,}$ ${(beta^2+beta)times(beta^2+beta+1)=beta^2+1.}$
• ${(beta^2+beta+1)times(beta^2+beta+1)=beta.}$

On the surface, ${{mathbb F}={mathbb Z}_2(alpha)}$ and ${{mathbb H}={mathbb Z}_2(beta)}$ look rather different. However, notice that if we set ${p(x)=x^3+x+1in{mathbb Z}_2[x],}$ then ${p=p_alpha:}$ In effect, ${p}$ is irreducible and ${alpha^3+alpha+1=0.}$ But also, ${p=p_{beta+1},}$ since ${(beta+1)^3+(beta+1)+1=(beta^3+beta^2+beta+1)+(beta+1)+1=beta^3+beta^2+1=0.}$

Since ${{mathbb H}={mathbb Z}_2(beta+1),}$ we can apply Theorem 4 and conclude that ${{mathbb F}}$ and ${{mathbb H}}$ are isomorphic, and that the map ${h:{mathbb F}rightarrow{mathbb H}}$ given by

$displaystyle h(a_0+a_1alpha+a_2alpha^2)=a_0+a_1(beta+1)+a_2(beta+1)^2=(a_0+a_1+a_2)+a_1beta+a_2beta^2$

for ${a_0,a_1,a_2in{mathbb Z}_2,}$ is an isomorphism. In other words, ${{mathbb F}}$ and ${{mathbb H}}$ are actually the same field, but presented in two slightly different ways; what we call ${alpha}$ in one case, we call ${beta+1}$ in the other.

Notice also that we have that ${p(x)}$ completely factors in ${{mathbb F}:}$ ${p(alpha)=alpha^3+alpha+1=0.}$ Then ${0=(alpha^3+alpha+1)^2=alpha^6+alpha^2+1=(alpha^3)^2+alpha^2+1=p(alpha^2),}$ and ${alpha^2}$ is also a root of ${p.}$ Similarly, ${alpha^4=alpha^2+alpha}$ is a root of ${p,}$ so ${p(x)=(p-alpha)(p-alpha^2)(p-alpha^2-alpha)}$ in ${{mathbb F}[x].}$

Let ${q(x)=x^3=x^2+1.}$ Then ${q}$ also factors in ${{mathbb F}.}$ Equivalently, ${q}$ factors in ${{mathbb H}.}$ Exactly as above, we find ${q(x)=(x-beta)(x-beta^2)(x-beta^2-beta-1)}$ in ${{mathbb H}[x].}$ Equivalently, ${q(x)=(x-alpha-1)(x-alpha^2-1)(x-alpha^2-alpha-1)}$ in ${{mathbb F}[x].}$ Notice that all the `new’ elements of ${{mathbb F}}$ are roots of either ${p}$ or ${q,}$ and none of them is a double root (we say that they are simple roots) or a root of both ${p}$ and ${q.}$

Finally, note that ${(x^3+x+1)(x^3+x^2+1)=x^6+x^5+x^4+x^3+x^2+x+1}$ in ${{mathbb Z}_2[x],}$ so ${(x^3+x+1)(x^3+x^2+1)x(x-1)=x^8-x,}$ which means that ${{mathbb Z}_2^{x^8-x}={mathbb F},}$ and that every element of ${{mathbb F}}$ is a simple root of the polynomial ${x^8-x.}$ This is a particular case of a general phenomenon present in all finite fields.

5. To illustrate the same phenomenon, note that in the field of 4 elements from example 3, we have that every element is a simple root of the polynomial ${x^4-xin{mathbb Z}_2[x].}$ In general, if ${{mathbb F}}$ is a finite field of size ${p^n,}$ then every element of ${{mathbb F}}$ is a root of ${x^{p^n}-xin{mathbb Z}_p[x],}$ so ${{mathbb F}={mathbb Z}_p^{t(x)}}$ where ${t(x)=x^{p^n}-x.}$

I’ll leave it as an exercise to verify that this is the case with the field of 9 elements, and that, in particular, there is only one such field, up to isomorphism. Note that if ${|{mathbb F}|=9,}$ then ${{mathbb F}}$ is an extension of ${{mathbb Z}_3}$ of degree 2, so one way of building such a field is by considering ${{mathbb Z}_3[x]/(p),}$ where ${pin{mathbb Z}_3[x]}$ is irreducible of degree 3.

There are three such polynomials, namely ${x^2+1,}$ ${x^2+x+2}$ and ${x^2+2x+2.}$ One can check that the three extensions one obtains are actually “the same” (i.e., they are isomorphic), and a direct computation verifies that ${(x^2+1)(x^2+x+2)(x^2+2x+2)x(x-1)(x-2)=x^9-xin{mathbb Z}_3[x].}$

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