Existence of Finite Fields with p^n Elements

[Math Amateur]

I am reading Beachy and Blair's book: Abstract Algebra (3rd Edition) and am currently studying Theorem 6.5.7.

I need help with the proof of the Theorem.

Theorem 6.5.7 and its proof read as follows:View attachment 2853In the above proof, Beachy and Blair write:

By Lemma 6.5.4, the set of all roots of \(\displaystyle f(x)\) is a subfield of \(\displaystyle F\), and so we conclude that F must consist precisely of the roots of f(x) ... ... "

My question is as follows:

Can someone please explain exactly how the set of all roots of \(\displaystyle f(x)\) being a subfield of \(\displaystyle F \) leads to F consisting of precisely the roots of f(x)?NOTES:

1. I have the nagging suspicion that my question involves issues I have recently asked about on MHB ... if this is actually the case then ... my apologies ...

2, Lemma 6.5.4 is referred to in the above post ... Lemma 6.5.4 reads as follows:

View attachment 2854

 

[Euge]

I am reading Beachy and Blair's book: Abstract Algebra (3rd Edition) and am currently studying Theorem 6.5.7.

I need help with the proof of the Theorem.

Theorem 6.5.7 and its proof read as follows:View attachment 2853In the above proof, Beachy and Blair write:

By Lemma 6.5.4, the set of all roots of \(\displaystyle f(x)\) is a subfield of \(\displaystyle F\), and so we conclude that F must consist precisely of the roots of f(x) ... ... "

My question is as follows:

Can someone please explain exactly how the set of all roots of \(\displaystyle f(x)\) being a subfield of \(\displaystyle F \) leads to F consisting of precisely the roots of f(x)?NOTES:

1. I have the nagging suspicion that my question involves issues I have recently asked about on MHB ... if this is actually the case then ... my apologies ...

2, Lemma 6.5.4 is referred to in the above post ... Lemma 6.5.4 reads as follows:

View attachment 2854

Since the set of all roots of $f$ is a subfield of $F$ containing $\Bbb Z_p$, and since $F$ is by definition the smallest extension field of $\Bbb Z_p$ that contains the roots of $f$, $F$ must equal to this set. This is why $F$ consists of exactly the roots of $f$.

 

[Math Amateur]

Since the set of all roots of $f$ is a subfield of $F$ containing $\Bbb Z_p$, and since $F$ is by definition the smallest extension field of $\Bbb Z_p$ that contains the roots of $f$, $F$ must equal to this set. This is why $F$ consists of exactly the roots of $f$.

Thanks Euge ... most helpful ...

Can anyone help with a further question, as follows:

In the proof of the Theorem - in the second sentence to be precise - Beachy and Blair take some care to show that \(\displaystyle f(x)\) has distinct roots ... ... why is this important in the context of this Theorem?

Peter

 

[Deveno]

It is possible for a polynomial to have" multiple roots", for example the polynomial:

$x^2 - 1 = (x - 1)^2$ in $\Bbb Z_2[x]$. 1 is a "double root" for this polynomial, so the cardinality of the set of its roots is less than its degree.

If a polynomial does not have such multiple roots, then the cardinality of the set of roots equals the degree of the polynomial.

So if $x^{p^n} - x$ has no multiple roots, we may conclude it has exactly $p^n$ roots. This means that every element of a field of cardinality $p^n$, being a root of said polynomial, leaves "no room left over" for any other elements besides the subfield:

$F' = \{a \in F: a^{p^n} = a\}$

consisting of the roots.

This is quite remarkable, because it is not immediately obvious that $x^{p^n} - x$ has no multiple roots.

*******************

Suppose we have $p = 2$ and $n = 3$, for a field of 8 elements. We can look at the polynomial:

$x^8 - x = x(x^7 - 1) = x(x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)$. The "obvious" roots 0 and 1 correspond to the first two factors.

Perhaps you may recall that Euge gave a representation of $\Bbb F_8$ as:

$\{0,1,u,1+u,u^2,1+u^2,u+u^2,1+u+u^2\}$ where $u^3 = u + 1$. In that thread, we saw that $u^7 = 1$.

It follows that since $u \neq 0,1$ and $u^3 + u + 1 = 0$, that $u$ must be a root of:

$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$, and indeed we see that:

$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = (x^3 + x + 1)(x^3 + x^2 + 1)$.

So over $\Bbb Z_2$, we have this factorization:

$x^8 - x = x(x - 1)(x^3 + x + 1)(x^3 + x^2 + 1)$ <--these are irreducible factors.

We know one root of the third factor, $u$, and we can use the division algorithm (in $\Bbb F_8[x]$) to find that:

$x^3 + x + 1 = (x - u)(x^2 + ux + 1 + u^2)$ (the first factor could also be written $x + u$).

By trial-and-error (there are only 5 roots left), we find that:

$(u^2)^2 + u(u^2) + 1 + u^2 = u^4 + u^3 + 1 + u^2 = (1 + u)u^3 + 1 + u^2$

$= (1 + u)^2 + 1 + u^2 = 1 + u^2 + 1 + u^2 = 0$, so $u^2$ is a root of:

$x^2 + ux + 1 + u^2$, and thus of $x^3 + x + 1$. Now we can divide $x^3 + x + 1$ by:

$(x - u)(x - u^2) = x^2 + (u + u^2)x + 1 + u$ to get:

$x^3 + x + 1 = (x - u)(x - u^2)(x - (u+u^2))$.

It should thus be the case that:

$x^3 + x^2 + 1 = (x - (1+u))(x - (1+u^2))(x - (1+u+u^2))$

which can be verified by multiplying out the RHS, or checking:

$(1+u)^3 + (1+u)^2 + 1 = (1+u)(1+u)^2 + (1+u)^2 + 1 = u(1+u)^2 + 1 = u(1+u^2) + 1$

$= u + u^3 + 1 = u + u + 1 + 1 = 0$, and similarly for the other two roots.

In short:

$x^8 - x = x(x - 1)(x - u)(x - u^2)(x - (u+u^2))(x - (1+u))(x - (1+u^2))(x - (1+u+u^2))$

this polynomial does indeed split over $\Bbb F_8$, and 3 is prime, so $\Bbb F_8$ has no other subfields between it and $\Bbb Z_2$. Furthermore, we see, in fact, that all the roots are distinct (we have exactly 8, no more, no less, all different).

 

[Math Amateur]

It is possible for a polynomial to have" multiple roots", for example the polynomial:

$x^2 - 1 = (x - 1)^2$ in $\Bbb Z_2[x]$. 1 is a "double root" for this polynomial, so the cardinality of the set of its roots is less than its degree.

If a polynomial does not have such multiple roots, then the cardinality of the set of roots equals the degree of the polynomial.

So if $x^{p^n} - x$ has no multiple roots, we may conclude it has exactly $p^n$ roots. This means that every element of a field of cardinality $p^n$, being a root of said polynomial, leaves "no room left over" for any other elements besides the subfield:

$F' = \{a \in F: a^{p^n} = a\}$

consisting of the roots.

This is quite remarkable, because it is not immediately obvious that $x^{p^n} - x$ has no multiple roots.

*******************

Suppose we have $p = 2$ and $n = 3$, for a field of 8 elements. We can look at the polynomial:

$x^8 - x = x(x^7 - 1) = x(x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)$. The "obvious" roots 0 and 1 correspond to the first two factors.

Perhaps you may recall that Euge gave a representation of $\Bbb F_8$ as:

$\{0,1,u,1+u,u^2,1+u^2,u+u^2,1+u+u^2\}$ where $u^3 = u + 1$. In that thread, we saw that $u^7 = 1$.

It follows that since $u \neq 0,1$ and $u^3 + u + 1 = 0$, that $u$ must be a root of:

$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$, and indeed we see that:

$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = (x^3 + x + 1)(x^3 + x^2 + 1)$.

So over $\Bbb Z_2$, we have this factorization:

$x^8 - x = x(x - 1)(x^3 + x + 1)(x^3 + x^2 + 1)$ <--these are irreducible factors.

We know one root of the third factor, $u$, and we can use the division algorithm (in $\Bbb F_8[x]$) to find that:

$x^3 + x + 1 = (x - u)(x^2 + ux + 1 + u^2)$ (the first factor could also be written $x + u$).

By trial-and-error (there are only 5 roots left), we find that:

$(u^2)^2 + u(u^2) + 1 + u^2 = u^4 + u^3 + 1 + u^2 = (1 + u)u^3 + 1 + u^2$

$= (1 + u)^2 + 1 + u^2 = 1 + u^2 + 1 + u^2 = 0$, so $u^2$ is a root of:

$x^2 + ux + 1 + u^2$, and thus of $x^3 + x + 1$. Now we can divide $x^3 + x + 1$ by:

$(x - u)(x - u^2) = x^2 + (u + u^2)x + 1 + u$ to get:

$x^3 + x + 1 = (x - u)(x - u^2)(x - (u+u^2))$.

It should thus be the case that:

$x^3 + x^2 + 1 = (x - (1+u))(x - (1+u^2))(x - (1+u+u^2))$

which can be verified by multiplying out the RHS, or checking:

$(1+u)^3 + (1+u)^2 + 1 = (1+u)(1+u)^2 + (1+u)^2 + 1 = u(1+u)^2 + 1 = u(1+u^2) + 1$

$= u + u^3 + 1 = u + u + 1 + 1 = 0$, and similarly for the other two roots.

In short:

$x^8 - x = x(x - 1)(x - u)(x - u^2)(x - (u+u^2))(x - (1+u))(x - (1+u^2))(x - (1+u+u^2))$

this polynomial does indeed split over $\Bbb F_8$, and 3 is prime, so $\Bbb F_8$ has no other subfields between it and $\Bbb Z_2$. Furthermore, we see, in fact, that all the roots are distinct (we have exactly 8, no more, no less, all different).

Thanks Deveno ... most informative ... still reflecting on this post ... BUT ... just a couple of questions of interest ..

In the above post you used \(\displaystyle \mathbb{Z}_2 / ( x^3 + x + 1 ) \) as a model for \(\displaystyle \mathbb{F}_8 \) ... are you able to specify some other models for \(\displaystyle \mathbb{F}_8 \)?

How many models in this sense does \(\displaystyle \mathbb{F}_8 \) have?

Peter

***EDIT*** Just reflecting on my question regarding other models for \(\displaystyle \mathbb{F}_8 \) - presumably one could use \(\displaystyle \mathbb{Z}_2 / ( x^3 + x^2 + 1 ) \) as an alternative model ... but are there others ... ?

I am inclined to think (guess .. :-) ) that there are no other models ... but I am by no means sure of this ... ...

 

[Deveno]

Well, to define $\Bbb F_8$ as a quotient of $\Bbb F_2[x]$, we need an irreducible polynomial of degree 3.

$\Bbb F_2[x]$ only has 8 polynomials of degree 3 (the coefficient of $x^3$ must be $1$, so we can only choose the coefficients of $x^2,x$ and $1$).

We can eliminate the 4 polynomials that have a 0 constant term (as $x$ is a factor of these, so therefore such a polynomial is reducible). That leaves just 4:

$x^3 + x^2 + x + 1$
$x^3 + x^2 + 1$
$x^3 + x + 1$
$x^3 + 1$.

Note that $1$ is a root of the first and last polynomials, so they have a factor of $x - 1 = x + 1$.

So the two in the middle are the only ones that work.

Something interesting to do:

If we write: $\Bbb F_8 \equiv \Bbb F_2(u)$ where $u^3 = u + 1$, and $\Bbb F_8 \equiv \Bbb F_2(b)$ where $b^3 = b^2 + 1$, what is the isomorphism between them?

It's clearly NOT: $u \mapsto b$.

 

[Math Amateur]

It is possible for a polynomial to have" multiple roots", for example the polynomial:

$x^2 - 1 = (x - 1)^2$ in $\Bbb Z_2[x]$. 1 is a "double root" for this polynomial, so the cardinality of the set of its roots is less than its degree.

If a polynomial does not have such multiple roots, then the cardinality of the set of roots equals the degree of the polynomial.

So if $x^{p^n} - x$ has no multiple roots, we may conclude it has exactly $p^n$ roots. This means that every element of a field of cardinality $p^n$, being a root of said polynomial, leaves "no room left over" for any other elements besides the subfield:

$F' = \{a \in F: a^{p^n} = a\}$

consisting of the roots.

This is quite remarkable, because it is not immediately obvious that $x^{p^n} - x$ has no multiple roots.

*******************

Suppose we have $p = 2$ and $n = 3$, for a field of 8 elements. We can look at the polynomial:

$x^8 - x = x(x^7 - 1) = x(x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)$. The "obvious" roots 0 and 1 correspond to the first two factors.

Perhaps you may recall that Euge gave a representation of $\Bbb F_8$ as:

$\{0,1,u,1+u,u^2,1+u^2,u+u^2,1+u+u^2\}$ where $u^3 = u + 1$. In that thread, we saw that $u^7 = 1$.

It follows that since $u \neq 0,1$ and $u^3 + u + 1 = 0$, that $u$ must be a root of:

$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$, and indeed we see that:

$x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = (x^3 + x + 1)(x^3 + x^2 + 1)$.

So over $\Bbb Z_2$, we have this factorization:

$x^8 - x = x(x - 1)(x^3 + x + 1)(x^3 + x^2 + 1)$ <--these are irreducible factors.

We know one root of the third factor, $u$, and we can use the division algorithm (in $\Bbb F_8[x]$) to find that:

$x^3 + x + 1 = (x - u)(x^2 + ux + 1 + u^2)$ (the first factor could also be written $x + u$).

By trial-and-error (there are only 5 roots left), we find that:

$(u^2)^2 + u(u^2) + 1 + u^2 = u^4 + u^3 + 1 + u^2 = (1 + u)u^3 + 1 + u^2$

$= (1 + u)^2 + 1 + u^2 = 1 + u^2 + 1 + u^2 = 0$, so $u^2$ is a root of:

$x^2 + ux + 1 + u^2$, and thus of $x^3 + x + 1$. Now we can divide $x^3 + x + 1$ by:

$(x - u)(x - u^2) = x^2 + (u + u^2)x + 1 + u$ to get:

$x^3 + x + 1 = (x - u)(x - u^2)(x - (u+u^2))$.

It should thus be the case that:

$x^3 + x^2 + 1 = (x - (1+u))(x - (1+u^2))(x - (1+u+u^2))$

which can be verified by multiplying out the RHS, or checking:

$(1+u)^3 + (1+u)^2 + 1 = (1+u)(1+u)^2 + (1+u)^2 + 1 = u(1+u)^2 + 1 = u(1+u^2) + 1$

$= u + u^3 + 1 = u + u + 1 + 1 = 0$, and similarly for the other two roots.

In short:

$x^8 - x = x(x - 1)(x - u)(x - u^2)(x - (u+u^2))(x - (1+u))(x - (1+u^2))(x - (1+u+u^2))$

this polynomial does indeed split over $\Bbb F_8$, and 3 is prime, so $\Bbb F_8$ has no other subfields between it and $\Bbb Z_2$. Furthermore, we see, in fact, that all the roots are distinct (we have exactly 8, no more, no less, all different).

Deveno ... Thanks for the fully worked example regarding the roots of the field of 8 elements ... It really assisted my understanding of finite fields ... wish texts would do more of this!

Peter

 

[Math Amateur]

Well, to define $\Bbb F_8$ as a quotient of $\Bbb F_2[x]$, we need an irreducible polynomial of degree 3.

$\Bbb F_2[x]$ only has 8 polynomials of degree 3 (the coefficient of $x^3$ must be $1$, so we can only choose the coefficients of $x^2,x$ and $1$).

We can eliminate the 4 polynomials that have a 0 constant term (as $x$ is a factor of these, so therefore such a polynomial is reducible). That leaves just 4:

$x^3 + x^2 + x + 1$
$x^3 + x^2 + 1$
$x^3 + x + 1$
$x^3 + 1$.

Note that $1$ is a root of the first and last polynomials, so they have a factor of $x - 1 = x + 1$.

So the two in the middle are the only ones that work.

Something interesting to do:

If we write: $\Bbb F_8 \equiv \Bbb F_2(u)$ where $u^3 = u + 1$, and $\Bbb F_8 \equiv \Bbb F_2(b)$ where $b^3 = b^2 + 1$, what is the isomorphism between them?

It's clearly NOT: $u \mapsto b$.

Thanks Deveno ...

So to sum up regarding the number of models for $\Bbb F_8$ ... only two exist since $\Bbb F_8$ as a quotient of $ \Bbb F_3[x] \text{ or } \Bbb F_4[x] $ ... ... etc would obviously not work ...

Is that correct?

Peter

 

[Math Amateur]

Well, to define $\Bbb F_8$ as a quotient of $\Bbb F_2[x]$, we need an irreducible polynomial of degree 3.

$\Bbb F_2[x]$ only has 8 polynomials of degree 3 (the coefficient of $x^3$ must be $1$, so we can only choose the coefficients of $x^2,x$ and $1$).

We can eliminate the 4 polynomials that have a 0 constant term (as $x$ is a factor of these, so therefore such a polynomial is reducible). That leaves just 4:

$x^3 + x^2 + x + 1$
$x^3 + x^2 + 1$
$x^3 + x + 1$
$x^3 + 1$.

Note that $1$ is a root of the first and last polynomials, so they have a factor of $x - 1 = x + 1$.

So the two in the middle are the only ones that work.

Something interesting to do:

If we write: $\Bbb F_8 \equiv \Bbb F_2(u)$ where $u^3 = u + 1$, and $\Bbb F_8 \equiv \Bbb F_2(b)$ where $b^3 = b^2 + 1$, what is the isomorphism between them?

It's clearly NOT: $u \mapsto b$.

Hi Deveno,

I have been reflecting on your question:

"If we write: $\Bbb F_8 \equiv \Bbb F_2(u)$ where $u^3 = u + 1$, and $\Bbb F_8 \equiv \Bbb F_2(b)$ where $b^3 = b^2 + 1$, what is the isomorphism between them?"

I have not been able to come up with an answer ... can you help ...?

(Maybe another MHB member can take up your challenge??)

Peter***EDIT***

I would really love to say that I have now answered your question by my own efforts ... but on reading Beachy and Blair, I came upon this worked example:View attachment 2862I am now working through this carefully to make sure I fully understand it.

Peter***EDIT 2***

The above example refers to Lemma 6.4.3. For MHB members who may be interested, I am now including this Lemma and its proof:View attachment 2863
View attachment 2864

 

[Deveno]

Indeed, all that is true.

As you will discover later, it turns out that:

$\{\theta \in \text{Aut}(\Bbb F_8/\Bbb F_2)\}$

the set of field automorphisms of $\Bbb F_8$ that FIX the subfield $\Bbb F_2$, form a group.

How large do you suppose that group is?

Of course such a group contains the identity automorphism $1_{\Bbb F_8}$.

Since $\Bbb F_2$ and $u$ completely determine the elements of $\Bbb F_8$, the real question is: where can such a field automorphism send $u$?

Note that since:

$\theta(u^3 + u + 1) = \theta(0) = 0$ and:

$\theta(u^3 + u + 1) = \theta(u^3) + \theta(u) + \theta(1) = (\theta(u))^3 + \theta(u) + 1$

it follows that $\theta(u)$ is ALSO a root of $x^3 + x + 1$.

What can you deduce from this?

 

[Math Amateur]

Indeed, all that is true.

As you will discover later, it turns out that:

$\{\theta \in \text{Aut}(\Bbb F_8/\Bbb F_2)\}$

the set of field automorphisms of $\Bbb F_8$ that FIX the subfield $\Bbb F_2$, form a group.

How large do you suppose that group is?

Of course such a group contains the identity automorphism $1_{\Bbb F_8}$.

Since $\Bbb F_2$ and $u$ completely determine the elements of $\Bbb F_8$, the real question is: where can such a field automorphism send $u$?

Note that since:

$\theta(u^3 + u + 1) = \theta(0) = 0$ and:

$\theta(u^3 + u + 1) = \theta(u^3) + \theta(u) + \theta(1) = (\theta(u))^3 + \theta(u) + 1$

it follows that $\theta(u)$ is ALSO a root of $x^3 + x + 1$.

What can you deduce from this?

Thanks Deveno ... some intriguing questions ... I will get onto these as soon as I can ... BUT ...

... ... unfortunately I am having real trouble following Beachy and Blair's solution to your problem ... in particular I am having trouble with the mechanics regarding exactly how Beachy and Blair are using Lemma 6.4.3 (which I posted above) ...

In particular, I am confused when Beachy and Blair finish their first construction of an isomorphism between \(\displaystyle F_1 \)and \(\displaystyle F_2\) with the statement:

" ... ... Thus if we choose v + 1 + u , we then map an element \(\displaystyle a w^2 + bw + c \in F_2 \text{ to } av^2 + bv + c \in F_1 \).

How exactly does this demonstrator an isomorphism between F_1 and F_2?

Is this just establishing the conditions of Lemma 6.4.3 - I must match these conditions to the example more clearly if this is the case.Do you have a simpler/clearer demonstration of the isomorphism and how to determine/specify it than Beachy and Blair's 'explanation' that I posted ...

Peter

 

[Math Amateur]

Thanks Deveno ... some intriguing questions ... I will get onto these as soon as I can ... BUT ...

... ... unfortunately I am having real trouble following Beachy and Blair's solution to your problem ... in particular I am having trouble with the mechanics regarding exactly how Beachy and Blair are using Lemma 6.4.3 (which I posted above) ...

In particular, I am confused when Beachy and Blair finish their first construction of an isomorphism between \(\displaystyle F_1 \)and \(\displaystyle F_2\) with the statement:

" ... ... Thus if we choose v + 1 + u , we then map an element \(\displaystyle a w^2 + bw + c \in F_2 \text{ to } av^2 + bv + c \in F_1 \).

How exactly does this demonstrator an isomorphism between F_1 and F_2?

Is this just establishing the conditions of Lemma 6.4.3 - I must match these conditions to the example more clearly if this is the case.Do you have a simpler/clearer demonstration of the isomorphism and how to determine/specify it than Beachy and Blair's 'explanation' that I posted ...

Peter

As I mentioned in the post above I am having difficulty in understanding Beachy and Blair Example 6.5.4. (see the Example and the associated Lemma 6.4.3 given in full in a post in this thread.)

Since Beachy and Blair state:

"... ... ...It follows from Lemma 6.4.3 that to construct an isomorphism from \(\displaystyle F_2\) to \(\displaystyle F_1\), we only need to find a root \(\displaystyle v\) of \(\displaystyle x^3 + x^2 + 1\) in \(\displaystyle F_1\), and then map \(\displaystyle w\) to \(\displaystyle v\). ... ..."

This statement of Beachy and Blair seems to indicate that this problem/example is 'solved/proved' through an application of Lemma 6.4.3. However, I am having trouble matching the conditions of the Lemma to the example ... I hope someone can help by showing exactly how Lemma 6.4.3 applies to the Example ...

To indicate my thinking I will indicate how I have so far matched up the conditions of the Lemma with the situation in the Example ... I know what follows is rather laborious but I cannot think of another way to indicate my thinking so far ...We are trying to demonstrate an isomorphism from \(\displaystyle F_2\) to \(\displaystyle F_1\), that is a bijective mapping \(\displaystyle \phi : \ F_2 \longrightarrow F_1 \) ... so, in terms of Lemma 6.4.3 we are looking to construct an isomorphism \(\displaystyle \phi : \ F \longrightarrow E\) where \(\displaystyle F = F_2\) and \(\displaystyle E = F_1\).Now in terms of the Example we have that\(\displaystyle F_2 = \mathbb{Z}_2 [x] / (x^3 + x^2 + 1)\)

and we let \(\displaystyle w = x + (x^3 + x^2 + 1)\)

so that w is a root of \(\displaystyle (x^3 + x^2 + 1)\) in \(\displaystyle F_2 \)

In terms of the Lemma, \(\displaystyle K = \mathbb{Z}_2\).and\(\displaystyle F_1 = \mathbb{Z}_2 [x] / (x^3 + x + 1) \)

and we let \(\displaystyle u = x + (x^3 + x + 1)\)

so that u is a root of \(\displaystyle (x^3 + x + 1)\)

In terms of the Lemma, \(\displaystyle L = \mathbb{Z}_2\).
Now I match statements in the Lemma with statements in the situation of the Example in order to carefully and exactly identify that the conditions of the Lemma are met in the Example.
The first statement of the Lemma:

"Let \(\displaystyle \theta : \ K \longrightarrow L \) be an isomorphism of fields"

translates, in the situation of the Example to:

"Let \(\displaystyle \theta : \ \mathbb{Z}_2 \longrightarrow \mathbb{Z}_2 \) be an isomorphism of fields"
The second statement of the Lemma:

"\(\displaystyle F\) is an extension field of \(\displaystyle K\)"

translates, in the situation of the Example to:

"\(\displaystyle F_2\) is an extension field of \(\displaystyle \mathbb{Z}_2\)"

and we have that (see above)

\(\displaystyle F_2 = \mathbb{Z}_2 [x] / (x^3 + x^2 + 1) \cong \mathbb{Z}_2 (w) \)
The third statement of the Lemma:

"Let \(\displaystyle p(x)\) be the minimal polynomial of \(\displaystyle u\) over \(\displaystyle K\)"

translates, in the situation of the Example to:

"Let \(\displaystyle p(x) = (x^3 + x^2 + 1) \) be the minimal polynomial of \(\displaystyle w\) over \(\displaystyle \mathbb{Z}_2\)"
The fourth statement of the Lemma:

"Let \(\displaystyle q(x)\) be the image of \(\displaystyle p(x)\) under \(\displaystyle \theta\)"

translates, essentially, in the situation of the Example, to:

\(\displaystyle q(x) = \theta (p(x)) = \theta (x^3 + x^2 + 1) \)

[? but what is the exact nature of \(\displaystyle \theta \) and how do I actually evaluate q(x) ?]
The fifth statement of the Lemma:

"If v is any root of q(x) in the extension field of L and E = L(v) ... ... "

translates, essentially, in the situation of the Example, to:

"If v is any root of q(x) in the extension field of \(\displaystyle \mathbb{Z}_2\) and \(\displaystyle F_1 = \mathbb{Z}_2(v)\) ... ... "

But at this point I am getting confused ... particularly about the exact nature of \(\displaystyle \theta \) ...

Can anyone help clarify exactly how Lemma 6.4.3 applies to Example 6.5.4.

Peter

 

[Deveno]

Ok, we have two different extensions of $\Bbb F_2$, namely:

$F = \Bbb F_2(u)$, where $u$ is a root of $x^3 + x + 1$ and:

$E = \Bbb F_2(v)$, where $v$ is a root of $x^3 + x^2 + 1$.

In this case, both our "base fields" are the same field, $K = L = \Bbb F_2$, and we may use the IDENTITY isomorphism for what B&B call $\theta$.

This means that isomorphism $\Bbb F_2[x] \to \Bbb F_2[x]$ is also the identity map.

Suppose we map $u \mapsto v+1$. We need to show first that $v+1$ is a root of $x^3 + x + 1$.

So...

$(v+1)^3 + (v + 1) + 1 = (v+1)(v+1)^2 + (v+1) + 1 = (v+1)(v^2 + 1) + (v + 1) + 1$

$= v^3 + v + v^2 + 1 + v + 1 + 1$ (so far, we are still just expanding)

$= (v^3 + v^2 + 1) + v + v + 1 + 1$ (the stuff in the parentheses = 0, since $v$ is a root of $x^3 + x^2 + 1$)

$ = 0 + v + v + 1 + 1 = 0 + 0 + 0 = 0$ (since $v + v = (1 + 1)v$ and $1 + 1 = 0$, since our field has characteristic 2).

This shows that indeed, $v+1$ is a root of $\hat{\theta}(x^3 + x + 1) = x^3 + x + 1$.

What we should do now is show $\Bbb F_2(v+1) = \Bbb F_2(v)$. Hopefully this is trivial.

 

[Math Amateur]

Ok, we have two different extensions of $\Bbb F_2$, namely:

$F = \Bbb F_2(u)$, where $u$ is a root of $x^3 + x + 1$ and:

$E = \Bbb F_2(v)$, where $v$ is a root of $x^3 + x^2 + 1$.

In this case, both our "base fields" are the same field, $K = L = \Bbb F_2$, and we may use the IDENTITY isomorphism for what B&B call $\theta$.

This means that isomorphism $\Bbb F_2[x] \to \Bbb F_2[x]$ is also the identity map.

Suppose we map $u \mapsto v+1$. We need to show first that $v+1$ is a root of $x^3 + x + 1$.

So...

$(v+1)^3 + (v + 1) + 1 = (v+1)(v+1)^2 + (v+1) + 1 = (v+1)(v^2 + 1) + (v + 1) + 1$

$= v^3 + v + v^2 + 1 + v + 1 + 1$ (so far, we are still just expanding)

$= (v^3 + v^2 + 1) + v + v + 1 + 1$ (the stuff in the parentheses = 0, since $v$ is a root of $x^3 + x^2 + 1$)

$ = 0 + v + v + 1 + 1 = 0 + 0 + 0 = 0$ (since $v + v = (1 + 1)v$ and $1 + 1 = 0$, since our field has characteristic 2).

This shows that indeed, $v+1$ is a root of $\hat{\theta}(x^3 + x + 1) = x^3 + x + 1$.

What we should do now is show $\Bbb F_2(v+1) = \Bbb F_2(v)$. Hopefully this is trivial.

Thanks Deveno ... Just working through this carefully now,

Peter

 

[Deveno]

A word about polynomial rings, and their related quotient rings:

Polynomial rings have a "universal" property, which goes something like this:

If $\phi:R \to S$ is a ring homomorphism, and $a \in S$, there is a UNIQUE homomorphism:

$\hat{\phi}:R[x] \to S$ such that:

$\hat{\phi}(r) = \phi(r)$ and $\hat{\phi}(x) = a$.

This homomorphism is easy to state, explicitly:

$\hat{\phi}(c_0 + c_1x +\cdots + c_nx^n) = \phi(c_0) + \phi(c_1)a +\cdots + \phi(c_n)a^n$.

The case that is usually of most interest, is when $R \subseteq S$, so that $\phi = 1_R$ as an inclusion map.

We then get a mapping $\hat{\phi}: R[x] \to R(a)$, (some texts call this $R[a]$) given by:

$\hat{\phi}(f(x)) = f(a)$

This map is typically called the "evaluation at $a$" mapping.

Here is an example:

Suppose that $R = \Bbb Z$, and that $S = \Bbb R$. We might take $a = \sqrt{2}$.

Then $\hat{\phi}(x + 1) = \sqrt{2} + 1$, while $\hat{\phi}(x^2 - 2) = 0$.

It is convenient to think of elements of $\hat{\phi}(R[x])$ as "polynomial expressions in $a$", although they might possesses additional properties that "pure" polynomials do not. In my example above, the polynomials:

$x^2 - 2$ and $0$ are definitely NOT the same, but the "evaluations":

$(\sqrt{2})^2 - 2$ and $0$ are.

So there's definitely some "collapsing" going on, two unequal polynomials may evaluate to the same ring element.

Now, it is possible to view the quotient:

$R[x]/\text{ker }\hat{\phi}$ as the main object of interest, but the objects in the isomorphic ring:

$R(a)$ are more tractable to deal with.

For example, elements of the form $a + b\sqrt{2}$ for $a,b \in \Bbb Z$ are far more easier to deal with than performing "polynomial long-division" on an integral polynomial to find out which coset of the ideal generated by $x^2 - 2$ it's in. The important thing to realize is that all the algebraic rules (inherited from our original polynomial ring) still apply, but we can simplify $(\sqrt{2})^2$ (and thus higher powers) by replacing it with the integer 2:

$(a + b\sqrt{2})(c + d\sqrt{2}) = ac + ad\sqrt{2} + bc\sqrt{2} + ad(\sqrt{2})^2$

$= ac + 2bd + (ad + bc)\sqrt{2}$ <--we collect "powers of $\sqrt{2}$" the same way we collect "powers of $x$".

*************************
Of course, when the coefficients of our polynomials are in a FIELD, we have a bit more flexibility, we can divide by non-zero coefficients (and so make "any non-zero term" monic). The interesting part of this, is that now the RING $F(a)$ is often a field itself.

For example, if we allow RATIONAL coefficients in $a + b\sqrt{2}$, then since $\sqrt{2}$ is a root of $x^2 - 2$, we have:

$2 = (\sqrt{2})^2$

$1 = \dfrac{\sqrt{2}}{2}\cdot \sqrt{2}$

so that:

$\dfrac{1}{\sqrt{2}}$ is also of the form $a + b\sqrt{2}$, for $a = 0$ and $b = \frac{1}{2}$.

In fact, we can realize $\dfrac{1}{a + b\sqrt{2}}$ as:

$c + d\sqrt{2}$ where:

$c = \dfrac{a}{a^2 - 2b^2}$ and $d = -\dfrac{b}{a^2 - 2b^2}$.

These happy occurrences are because $\sqrt{2}$ is algebraic, and because Euclidean integral domains (such as $F[x]$, for any field $F$) are "almost fields". As with $\Bbb Z$, all we need to do is "mod out" a maximal ideal.

****************************

The notation in B&B is not always entirely helpful (not to put too much blame, a lot of mathematical notation could use some work). When you see a field, such as $\Bbb F_p(\alpha)$, you shouldn't be daunted, these are just expressions like:

$[k_0]_p + [k_1]_p\alpha + \cdots + [k_n]_p\alpha^n$ (Often we write just the integers $k_j$, the residues mod $p$ being understood)

which behave "pretty much like polynomials" (except sometimes they simplify).

 

[Math Amateur]

Ok, we have two different extensions of $\Bbb F_2$, namely:

$F = \Bbb F_2(u)$, where $u$ is a root of $x^3 + x + 1$ and:

$E = \Bbb F_2(v)$, where $v$ is a root of $x^3 + x^2 + 1$.

In this case, both our "base fields" are the same field, $K = L = \Bbb F_2$, and we may use the IDENTITY isomorphism for what B&B call $\theta$.

This means that isomorphism $\Bbb F_2[x] \to \Bbb F_2[x]$ is also the identity map.

Suppose we map $u \mapsto v+1$. We need to show first that $v+1$ is a root of $x^3 + x + 1$.

So...

$(v+1)^3 + (v + 1) + 1 = (v+1)(v+1)^2 + (v+1) + 1 = (v+1)(v^2 + 1) + (v + 1) + 1$

$= v^3 + v + v^2 + 1 + v + 1 + 1$ (so far, we are still just expanding)

$= (v^3 + v^2 + 1) + v + v + 1 + 1$ (the stuff in the parentheses = 0, since $v$ is a root of $x^3 + x^2 + 1$)

$ = 0 + v + v + 1 + 1 = 0 + 0 + 0 = 0$ (since $v + v = (1 + 1)v$ and $1 + 1 = 0$, since our field has characteristic 2).

This shows that indeed, $v+1$ is a root of $\hat{\theta}(x^3 + x + 1) = x^3 + x + 1$.

What we should do now is show $\Bbb F_2(v+1) = \Bbb F_2(v)$. Hopefully this is trivial.

Thanks again for the help Deveno ...

Sorry to be slow but I still have some difficulties ...

I will refer to your post and (as I have done before) stick religiously to the terminology of Lemma 6.4.3 since the proof or demonstration of the isomorphism between F and E rests on satisfying the conditions of the Lemma.

Just repeating the Lemma for the convenience of those reading this post: (for the proof as well see one of the posts above in this thread)https://www.physicsforums.com/attachments/2873

Now given F and E (and the associated roots u and v) as you have defined in your post we are trying (again, in the notation of the Lemma) to demonstrate that there is an isomorphism \(\displaystyle \phi : \ F \longrightarrow E \) where \(\displaystyle F = \mathbb{F}_2 (u)\) and \(\displaystyle E = \mathbb{F}_2 (v) \).

We let \(\displaystyle \theta : \ \mathbb{F}_2 \longrightarrow \mathbb{F}_2 \) be the isomorphism of fields required as a condition of the Lemma.

As you point out \(\displaystyle \theta (a) = a \) for all \(\displaystyle a \in \mathbb{F}_2 \)

Further, we have that \(\displaystyle p(x) = x^3 + x + 1 \) is the minimal polynomial of \(\displaystyle u\) over \(\displaystyle K = \mathbb{F}_2 \)

and so

\(\displaystyle q(x) = \theta ( p(x) ) = \theta ( x^3 + x + 1 ) = x^3 + x + 1\)

So, now, following the Lemma, if \(\displaystyle v\) is any root of \(\displaystyle q(x) = x^3 + x + 1\) in an extension field of \(\displaystyle L = \mathbb{F}_2 \) and \(\displaystyle E = L(v) = \mathbb{F}_2(v) \) ...

But now I am in difficulties (which may be only notational) because we want a mapping \(\displaystyle \phi : \ F \longrightarrow E \) such that \(\displaystyle \phi (u) = v\) and \(\displaystyle \phi (a) = \theta (a) = a\) ...

but your mapping is \(\displaystyle \phi (u) = v + 1 \) ... ?
Just thinking ... ...

... ... I could, in trying to resolve this by restating my sentence above that:

"So, now, following the Lemma, if \(\displaystyle v\) is any root of \(\displaystyle q(x) = x^3 + x + 1\) in an extension field of \(\displaystyle L = \mathbb{F}_2\) and E = \mathbb{F}_2(v)"

as

"So, now, following the Lemma, if \(\displaystyle v'\) is any root of \(\displaystyle q(x) = x^3 + x + 1\) in an extension field of \(\displaystyle L = \mathbb{F}_2\) and \(\displaystyle E = \mathbb{F}_2(v')\) where \(\displaystyle v' = v+1\) but this redefines \(\displaystyle E\)???

Can you help me to clarify this matter?

Peter

 

[Math Amateur]

A word about polynomial rings, and their related quotient rings:

Polynomial rings have a "universal" property, which goes something like this:

If $\phi:R \to S$ is a ring homomorphism, and $a \in S$, there is a UNIQUE homomorphism:

$\hat{\phi}:R[x] \to S$ such that:

$\hat{\phi}(r) = \phi(r)$ and $\hat{\phi}(x) = a$.

This homomorphism is easy to state, explicitly:

$\hat{\phi}(c_0 + c_1x +\cdots + c_nx^n) = \phi(c_0) + \phi(c_1)a +\cdots + \phi(c_n)a^n$.

The case that is usually of most interest, is when $R \subseteq S$, so that $\phi = 1_R$ as an inclusion map.

We then get a mapping $\hat{\phi}: R[x] \to R(a)$, (some texts call this $R[a]$) given by:

$\hat{\phi}(f(x)) = f(a)$

This map is typically called the "evaluation at $a$" mapping.

Here is an example:

Suppose that $R = \Bbb Z$, and that $S = \Bbb R$. We might take $a = \sqrt{2}$.

Then $\hat{\phi}(x + 1) = \sqrt{2} + 1$, while $\hat{\phi}(x^2 - 2) = 0$.

It is convenient to think of elements of $\hat{\phi}(R[x])$ as "polynomial expressions in $a$", although they might possesses additional properties that "pure" polynomials do not. In my example above, the polynomials:

$x^2 - 2$ and $0$ are definitely NOT the same, but the "evaluations":

$(\sqrt{2})^2 - 2$ and $0$ are.

So there's definitely some "collapsing" going on, two unequal polynomials may evaluate to the same ring element.

Now, it is possible to view the quotient:

$R[x]/\text{ker }\hat{\phi}$ as the main object of interest, but the objects in the isomorphic ring:

$R(a)$ are more tractable to deal with.

For example, elements of the form $a + b\sqrt{2}$ for $a,b \in \Bbb Z$ are far more easier to deal with than performing "polynomial long-division" on an integral polynomial to find out which coset of the ideal generated by $x^2 - 2$ it's in. The important thing to realize is that all the algebraic rules (inherited from our original polynomial ring) still apply, but we can simplify $(\sqrt{2})^2$ (and thus higher powers) by replacing it with the integer 2:

$(a + b\sqrt{2})(c + d\sqrt{2}) = ac + ad\sqrt{2} + bc\sqrt{2} + ad(\sqrt{2})^2$

$= ac + 2bd + (ad + bc)\sqrt{2}$ <--we collect "powers of $\sqrt{2}$" the same way we collect "powers of $x$".

*************************
Of course, when the coefficients of our polynomials are in a FIELD, we have a bit more flexibility, we can divide by non-zero coefficients (and so make "any non-zero term" monic). The interesting part of this, is that now the RING $F(a)$ is often a field itself.

For example, if we allow RATIONAL coefficients in $a + b\sqrt{2}$, then since $\sqrt{2}$ is a root of $x^2 - 2$, we have:

$2 = (\sqrt{2})^2$

$1 = \dfrac{\sqrt{2}}{2}\cdot \sqrt{2}$

so that:

$\dfrac{1}{\sqrt{2}}$ is also of the form $a + b\sqrt{2}$, for $a = 0$ and $b = \frac{1}{2}$.

In fact, we can realize $\dfrac{1}{a + b\sqrt{2}}$ as:

$c + d\sqrt{2}$ where:

$c = \dfrac{a}{a^2 - 2b^2}$ and $d = -\dfrac{b}{a^2 - 2b^2}$.

These happy occurrences are because $\sqrt{2}$ is algebraic, and because Euclidean integral domains (such as $F[x]$, for any field $F$) are "almost fields". As with $\Bbb Z$, all we need to do is "mod out" a maximal ideal.

****************************

The notation in B&B is not always entirely helpful (not to put too much blame, a lot of mathematical notation could use some work). When you see a field, such as $\Bbb F_p(\alpha)$, you shouldn't be daunted, these are just expressions like:

$[k_0]_p + [k_1]_p\alpha + \cdots + [k_n]_p\alpha^n$ (Often we write just the integers $k_j$, the residues mod $p$ being understood)

which behave "pretty much like polynomials" (except sometimes they simplify).

In your post that begins:

"A word about polynomial rings, and their related quotient rings: ... ... ... "

you write:

" ... ... The case that is usually of most interest, is when $R \subseteq S$, so that $\phi = 1_R$ as an inclusion map.

We then get a mapping $\hat{\phi}: R[x] \to R(a)$, (some texts call this $R[a]$) given by:

$\hat{\phi}(f(x)) = f(a)$

This map is typically called the "evaluation at $a$" mapping. ... ..."
Are you implying that when $R \subseteq S$ that it follows that $\phi = 1_R$ as an inclusion map - if so I cannot see why this must follow?

Or do you mean that the case of most interest is when $R \subseteq S$ AND $\phi = 1_R$ is the inclusion map?

Can you please clarify this issue for me?

Would appreciate the help.

Peter

 

[Deveno]

If we are to have a field isomorphism, the fields should "act just alike".

Now our $u$ and our $v$ DON'T, for $u$ we have:

$u^3 = u + 1$, while for $v$ we have:

$v^3 = v^2 + 1$.

In the lemma, THEIR $v$ has to be a root of $\hat{\theta}(p(x)) = q(x)$. In the matter at hand, since the base fields are "the same", the polynomials are also the same.

In other words, whatever we map $u$ to, HAS to be a root of $x^3 + x + 1$.

So, given only that we know (about "our" $v$) that $v^3 = v^2 + 1$, we have to find some expression:

$a + bv + cv^2$ (where $a,b,c \in \Bbb F_2$) which is a root of $x^3 + x + 1$.

Since we need some element "outside" of $\Bbb F_2$, this precludes the possibilities $b = c = 0$. Since we know $v$ itself doesn't work, this precludes $a = 0, b= 1, c = 0$. So here's what's left:

$a = 0, b = 1, c = 1 \iff v+v^2$
$a = 0, b = 0, c = 1 \iff v^2$
$a = b = 1, c = 0 \iff 1+v$ <--this one works. Two others should, as well.
$a = b = c = 1 \iff 1+v+v^2$
$a = 1, b = 0, c = 1 \iff 1+v^2$.

We could use trial and error, since we only have 5 possible candidates.

It is unfortunate that the $v$ WE have been using is different than the $v$ in the lemma.

But this is the important thing: EVERY root of $x^8 - x$ is contained in $\Bbb F_8$ (since that polynomial SPLITS in $\Bbb F_8$), and every element of $\Bbb F_8$ is a root of $x^8 - x$ (using group properties of a finite set, and facts about how many roots a polynomial over a field can have).

It should be easy to see that ANY of the six elements of $\Bbb F_8\setminus \Bbb F_2$ generate $\Bbb F_8$ as a field.

3 of these roots have minimal polynomial $x^3 + x + 1$, the other 3 have minimal polynomial $x^3 + x^2 + 1$.

If we have two different "models" of $\Bbb F_8$ (using different ideals to quotient $\Bbb F_2[x]$ by), to ensure we have a ring-isomorphism, all the multiplication rules need to be the same. In particular, we need to find generators in each model that satisfy the SAME polynomial.

THAT is why I say we should show $\Bbb F_2(v+1) = \Bbb F_2(v)$. Call $v+1$ by "some other letter" like $y$, so that we have, from, $y = v + 1$, that:

$v = v + 1 + (-1) = v + 1 + 1 = y + 1$.

Since $v$ satisfies the polynomial $x^3 + x^2 + 1$, we have:

$(y + 1)^3 + (y + 1)^2 + 1 = 0$

$y^3 + y^2 + y + 1 + y^2 + 1 + 1 = 0$

$y^3 + y + 1 = 0$, as desired.

Then our isomorphism can be written as:

$\Bbb F_2(u) \to \Bbb F_2(y)$ (here, we are using $y$ instead of $v$ like B&B do), given by:

$a + bu + u^2 \mapsto a + by + cy^2 = a + b(v+1) + c(v+1)^2 = (a+b+c) + bv + cv^2$

- - - Updated - - -

In your post that begins:

"A word about polynomial rings, and their related quotient rings: ... ... ... "

you write:

" ... ... The case that is usually of most interest, is when $R \subseteq S$, so that $\phi = 1_R$ as an inclusion map.

We then get a mapping $\hat{\phi}: R[x] \to R(a)$, (some texts call this $R[a]$) given by:

$\hat{\phi}(f(x)) = f(a)$

This map is typically called the "evaluation at $a$" mapping. ... ..."
Are you implying that when $R \subseteq S$ that it follows that $\phi = 1_R$ as an inclusion map - if so I cannot see why this must follow?

Or do you mean that the case of most interest is when $R \subseteq S$ AND $\phi = 1_R$ is the inclusion map?

Can you please clarify this issue for me?

Would appreciate the help.

Peter

Option number 2.

Sometimes, it is desirable to "reduce the coefficients" mod $I$, for some ideal of $R$, which is when the general form comes in handy.

There may be, even when $R \subseteq S$, MANY possible homomorphisms. But in any case, there is ALWAYS the "inclusion homomorphism" to fall back on. I meant the "AND" statement.

 

[Math Amateur]

If we are to have a field isomorphism, the fields should "act just alike".

Now our $u$ and our $v$ DON'T, for $u$ we have:

$u^3 = u + 1$, while for $v$ we have:

$v^3 = v^2 + 1$.

In the lemma, THEIR $v$ has to be a root of $\hat{\theta}(p(x)) = q(x)$. In the matter at hand, since the base fields are "the same", the polynomials are also the same.

In other words, whatever we map $u$ to, HAS to be a root of $x^3 + x + 1$.

So, given only that we know (about "our" $v$) that $v^3 = v^2 + 1$, we have to find some expression:

$a + bv + cv^2$ (where $a,b,c \in \Bbb F_2$) which is a root of $x^3 + x + 1$.

Since we need some element "outside" of $\Bbb F_2$, this precludes the possibilities $b = c = 0$. Since we know $v$ itself doesn't work, this precludes $a = 0, b= 1, c = 0$. So here's what's left:

$a = 0, b = 1, c = 1 \iff v+v^2$
$a = 0, b = 0, c = 1 \iff v^2$
$a = b = 1, c = 0 \iff 1+v$ <--this one works. Two others should, as well.
$a = b = c = 1 \iff 1+v+v^2$
$a = 1, b = 0, c = 1 \iff 1+v^2$.

We could use trial and error, since we only have 5 possible candidates.

It is unfortunate that the $v$ WE have been using is different than the $v$ in the lemma.

But this is the important thing: EVERY root of $x^8 - x$ is contained in $\Bbb F_8$ (since that polynomial SPLITS in $\Bbb F_8$), and every element of $\Bbb F_8$ is a root of $x^8 - x$ (using group properties of a finite set, and facts about how many roots a polynomial over a field can have).

It should be easy to see that ANY of the six elements of $\Bbb F_8\setminus \Bbb F_2$ generate $\Bbb F_8$ as a field.

3 of these roots have minimal polynomial $x^3 + x + 1$, the other 3 have minimal polynomial $x^3 + x^2 + 1$.

If we have two different "models" of $\Bbb F_8$ (using different ideals to quotient $\Bbb F_2[x]$ by), to ensure we have a ring-isomorphism, all the multiplication rules need to be the same. In particular, we need to find generators in each model that satisfy the SAME polynomial.

THAT is why I say we should show $\Bbb F_2(v+1) = \Bbb F_2(v)$. Call $v+1$ by "some other letter" like $y$, so that we have, from, $y = v + 1$, that:

$v = v + 1 + (-1) = v + 1 + 1 = y + 1$.

Since $v$ satisfies the polynomial $x^3 + x^2 + 1$, we have:

$(y + 1)^3 + (y + 1)^2 + 1 = 0$

$y^3 + y^2 + y + 1 + y^2 + 1 + 1 = 0$

$y^3 + y + 1 = 0$, as desired.

Then our isomorphism can be written as:

$\Bbb F_2(u) \to \Bbb F_2(y)$ (here, we are using $y$ instead of $v$ like B&B do), given by:

$a + bu + u^2 \mapsto a + by + cy^2 = a + b(v+1) + c(v+1)^2 = (a+b+c) + bv + cv^2$

- - - Updated - - -
Option number 2.

Sometimes, it is desirable to "reduce the coefficients" mod $I$, for some ideal of $R$, which is when the general form comes in handy.

There may be, even when $R \subseteq S$, MANY possible homomorphisms. But in any case, there is ALWAYS the "inclusion homomorphism" to fall back on. I meant the "AND" statement.

Thanks Deveno .. I am still working on and reflecting on the first of your posts above, but even on a first quick read what you have said is EXTREMELY helpful to me ... you definitely should be teaching at a University or authoring algebra books ...

Thanks again for your generous help ...

Peter