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}


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}


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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