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, ${R/I=\{[a]_\sim:a\in R\},}$ where ${[a]_\sim=\{b:a\sim b\}}$ for ${\sim}$ the equivalence relation defined by ${a\sim b}$ iff ${a-b\in I.}$

Since ${\sim}$ is an equivalence relation, we have that ${[a]_\sim=[b]_\sim}$ if ${a\sim b}$ and ${[a]_\sim\cap[b]_\sim=\emptyset}$ if ${a\not\sim b.}$ 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 ${p\in{\mathbb F}[x],}$ i.e., given any polynomial ${q\in{\mathbb F}[x],}$ ${[q]_\sim=0}$ iff ${p\mid q}$ and, more generally, ${[q]_\sim=[r]_\sim}$ (or, equivalently, ${q\sim r}$ or, equivalently, ${r\in[q]_\sim}$ ) iff ${p\mid (q-r).}$

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 ${p\mid q}$ then ${[q]_\sim=0.}$ Otherwise, ${{\rm gcd}(p,q)=1}$ since ${p}$ is irreducible. So there are polynomials ${r,s\in{\mathbb F}[x]}$ 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 ${p\in{\mathbb F}[x]}$ is irreducible. ${\Box}$

Suppose that ${a,b\in {\mathbb F}.}$ If ${a\sim b}$ then ${p\mid (a-b).}$ 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 ${a\in{\mathbb F}}$ with the corresponding element ${[a]_\sim}$ of ${{\mathbb F}[x]/(p),}$ and we have:

Lemma 2 Suppose that ${p\in{\mathbb F}[x]}$ is irreducible. Under the identification ${a\mapsto[a]_\sim}$ for ${a\in{\mathbb F},}$ 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 ${a\in{\mathbb F}.}$

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 ${p\in{\mathbb F}[x]}$ 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 ${a_0,\dots,a_n\in{\mathbb F}}$ with ${a_0\ne0}$ (since ${p}$ is irreducible) and ${a_n\ne0}$ (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 ${0=a_0+a_1\alpha+\dots+a_n\alpha^n=p(\alpha).}$ This shows that ${\alpha}$ is algebraic over ${{\mathbb F}.}$

Since any element of ${{\mathbb H}}$ is ${[q]_\sim}$ for some ${q\in{\mathbb F}[x],}$ 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 ${p_\alpha\mid p.}$ 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 ${\beta\in{\mathbb C}}$ 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 ${a_0+a_1\alpha+\dots+a_{n-1}\alpha^{n-1}}$ where the numbers ${a_0,\dots,a_{n-1}}$ are in ${{\mathbb F}.}$ Set

$\displaystyle \pi(a_0+a_1\alpha+\dots+a_{n-1}\alpha^{n-1})\!=\!a_0+a_1\beta+\dots+a_{n-1}\beta^{n-1}.$

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 ${q\in{\mathbb F}[x]}$ 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 ${u,v\in{\mathbb F}(\alpha)}$ 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 ${s,t\in{\mathbb F}[x]}$ 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 ${a\in{\mathbb K}.}$ If ${S}$ is a subset of ${{\mathbb F}}$ linearly independent over ${{\mathbb K},}$ then ${\{h(s):s\in S\}}$ is a subset of ${{\mathbb H}}$ linearly independent over ${{\mathbb K}.}$

Proof: Suppose ${s_1,\dots,s_n\in S.}$ Let ${a_1,\dots,a_n\in{\mathbb K}}$ 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 ${s_i\in S}$ are linearly independent over ${{\mathbb K},}$ then ${a_1=\dots=a_n=0.}$ This shows that ${\{h(s):s\in S\}}$ 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 ${\{h(s):s\in S\}}$ is linearly independent over ${{\mathbb K}.}$ Since ${|S|=|\{h(s):s\in S\}|,}$ 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 ${p_\alpha\in{\mathbb F}[x]}$ 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., ${\pi(a_0+a_1\alpha+\dots+a_{n-1}\alpha^{n-1})=}$

$\displaystyle a_0+a_1\pi(\alpha)+\dots+a_{n-1}\pi(\alpha)^{n-1}.$

Moreover, ${p_\alpha(\pi(\alpha))=\pi(p_\alpha(\alpha))=\pi(0)=0,}$ 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 ${{\mathbb Q}({\root3\of2})}$ is isomorphic to precisely three subfields of ${{\mathbb C},}$ namely:

$\displaystyle {\mathbb Q}({\root 3\of2}), {\mathbb Q}({\root3\of2}\zeta_3),\mbox{ and }{\mathbb Q}({\root3\of2}\zeta_3^2)$

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

To see this, note that Theorems 4 and 7 give us that if ${{\mathbb C}:{\mathbb H},}$ then ${{\mathbb Q}({\root 3\of2})}$ is isomorphic to ${{\mathbb H}}$ iff ${{\mathbb H}={\mathbb Q}(\xi)}$ for some root ${\xi}$ of ${p_{\root 3\of2}(x)=x^3-2.}$ This gives that the only subfields of ${{\mathbb C}}$ isomorphic to ${{\mathbb Q}({\root 3\of2})}$ 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, ${{\mathbb Q}({\root 3\of2})\ne{\mathbb Q}({\root 3\of2}\zeta_3)}$ and ${{\mathbb Q}({\root 3\of2})\ne{\mathbb Q}({\root 3\of2}\zeta_3^2),}$ since ${{\mathbb Q}({\root 3\of2})\subseteq{\mathbb R}}$ while the other fields contain complex nonreal numbers. Second, ${{\mathbb Q}({\root 3\of2}\zeta_3)\ne{\mathbb Q}({\root 3\of2}\zeta_3^2),}$ since otherwise

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

Now, ${p_{\zeta_3}(x)=x^2+x+1}$ has degree 2, so ${2=[{\mathbb Q}(\zeta_3):{\mathbb Q}]}$ would have to divide ${[{\mathbb Q}({\root3\of2}\zeta_3):{\mathbb Q}],}$ 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 ${[{\mathbb Q}({\root 3\of2}\zeta_3):{\mathbb Q}]=3.}$

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 ${{\mathbb Q}^{p(x)}={\mathbb Q}({\root3\of2},\zeta_3).}$ Since ${\zeta_3\notin{\mathbb Q}({\root 3\of2}),}$ by the tower law it follows that

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

and that any (strictly) intermediate field between ${{\mathbb Q}({\root 3\of2},\zeta_3)}$ 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

$\displaystyle {\mathbb Q}({\root 3\of2},\zeta_3):{\mathbb F}({\root3\of2}):{\mathbb F}:{\mathbb Q}$

and

$\displaystyle [{\mathbb F}({\root3\of2}):{\mathbb F}]=3$

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

$\displaystyle {\mathbb Q}({\root 3\of2}), {\mathbb Q}({\root3\of2}\zeta_3),\mbox{ and }{\mathbb Q}({\root3\of2}\zeta_3^2).$

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

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

and

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

which again leads to a contradiction with the tower law, since ${4\nmid 6.}$

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 ${a_i\in{\mathbb Z}_p}$ 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 ${p\in{\mathbb Z}_2[x]}$ 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 ${\alpha=[x]_\sim\in{\mathbb F},}$ 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 ${p\in{\mathbb Z}_2[x]}$ 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, ${\alpha^2(\alpha^2+\alpha+1)=[x^4+x^3+x^2]_\sim}$ ${=[(x^3+x+1)(x+1)+1]_\sim=1.}$ Similarly, we find:

${\alpha\times\alpha=\alpha^2,}$ ${\alpha\times(\alpha+1)=\alpha^2+\alpha,}$ ${\alpha\times\alpha^2=\alpha+1,}$ ${\alpha\times(\alpha^2+1)=1,}$ ${\alpha\times(\alpha^2+\alpha)=\alpha^2+\alpha+1,}$ ${\alpha\times(\alpha^2+\alpha+1)=\alpha^2+1.}$
${(\alpha+1)\times(\alpha+1)=\alpha^2+1,}$ ${(\alpha+1)\times\alpha^2=\alpha^2+\alpha+1,}$ ${(\alpha+1)\times(\alpha^2+1)=\alpha^2,}$ ${(\alpha+1)\times(\alpha^2+\alpha)=1,}$ ${(\alpha+1)\times(\alpha^2+\alpha+1)=\alpha.}$
${\alpha^2\times\alpha^2=\alpha^2+\alpha,}$ ${\alpha^2\times(\alpha^2+1)=\alpha,}$ ${\alpha^2\times(\alpha^2+\alpha)=\alpha^2+1,}$ ${\alpha^2\times(\alpha^2+\alpha+1)=1.}$
${(\alpha^2+1)\times(\alpha^2+1)=\alpha^2+\alpha+1,}$ ${(\alpha^2+1)\times(\alpha^2+\alpha)=\alpha+1,}$ ${(\alpha^2+1)\times(\alpha^2+\alpha+1)=\alpha^2+\alpha.}$
${(\alpha^2+\alpha)\times(\alpha^2+\alpha)=\alpha,}$ ${(\alpha^2+\alpha)\times(\alpha^2+\alpha+1)=\alpha^2.}$
${(\alpha^2+\alpha+1)\times(\alpha^2+\alpha+1)=\alpha+1.}$

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:

${\beta\times\beta=\beta^2,}$ ${\beta\times(\beta+1)=\beta^2+\beta,}$ ${\beta\times\beta^2=\beta^2+1,}$ ${\beta\times\beta^2+1)=\beta^2+\beta+1,}$ ${\beta\times(\beta^2+\beta)=1,}$ ${\beta\times(\beta^2+\beta+1)=\beta+1.}$
${(\beta+1)\times(\beta+1)=\beta^2+1,}$ ${(\beta+1)\times\beta^2=1,}$ ${(\beta+1)\times(\beta^2+1)=\beta,}$ ${(\beta+1)\times(\beta^2+\beta)=\beta^2+\beta+1,}$ ${(\beta+1)\times(\beta^2+\beta+1)=\beta^2.}$
${\beta^2\times\beta^2=\beta^2+\beta+1,}$ ${\beta^2\times(\beta^2+1)=\beta+1,}$ ${\beta^2\times(\beta^2+\beta)=\beta,}$ ${\beta^2\times(\beta^2+\beta+1)=\beta^2+\beta.}$
${(\beta^2+1)\times(\beta^2=1)=\beta^2+\beta,}$ ${(\beta^2+1)\times(\beta^2+\beta)=\beta^2,}$ ${(\beta^2+1)\times(\beta^2+\beta+1)=1.}$
${(\beta^2+\beta)\times(\beta^2+\beta)=\beta+1,}$ ${(\beta^2+\beta)\times(\beta^2+\beta+1)=\beta^2+1.}$
${(\beta^2+\beta+1)\times(\beta^2+\beta+1)=\beta.}$

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 ${p(x)=x^3+x+1\in{\mathbb Z}_2[x],}$ 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 ${h(a_0+a_1\alpha+a_2\alpha^2)=}$

$\displaystyle a_0+a_1(\beta+1)+a_2(\beta+1)^2=(a_0+a_1+a_2)+a_1\beta+a_2\beta^2$

for ${a_0,a_1,a_2\in{\mathbb Z}_2,}$ 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 ${x^4-x\in{\mathbb Z}_2[x].}$ In general, if ${{\mathbb F}}$ is a finite field of size ${p^n,}$ then every element of ${{\mathbb F}}$ is a root of ${x^{p^n}-x\in{\mathbb Z}_p[x],}$ 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 ${p\in{\mathbb Z}_3[x]}$ 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 ${(x^2+1)(x^2+x+2)(x^2+2x+2)x(x-1)(x-2)}$ ${=x^9-x\in{\mathbb Z}_3[x].}$

See also this answer in math.stackexchange.

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