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|q} and, more generally, {[q]_\sim=[r]_\sim} (or, equivalently, {q\sim r} or, equivalently, {r\in[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,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|(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|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.,

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


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

which, by the tower law, would imply that {9|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}


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

which again leads to a contradiction with the tower law, since {4\not| 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+x+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

\displaystyle  h(a_0+a_1\alpha+a_2\alpha^2)=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].}

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