Nature and character of Finite Fields of small order

[Math Amateur]

i am studying finite fields and trying to get an idea of the nature of finite fields.

In order to achieve this understanding I am bring to determine the elements and the addition and multiplication tables of some finite fields of small order.

For a start I am trying to determine the elements andthe addition and multiplication tables of the finite field of order 8.

Can someone show me how to get started on this problem?

Peter

***EDIT*** Just reflecting that maybe I should have started with \(\displaystyle \mathbb{F}_2 \text{ and } \mathbb{F}_3 \) but \(\displaystyle \mathbb{F}_8 \) seemed a bit less "special" and more general!

 

[Deveno]

$\Bbb F_8$ will have a "medium-large" multiplication table (64 entries) but it's do-able with some patience. Ditto for addition.

The key to creating such a table is finding an irreducible polynomial of degree 4 in $\Bbb Z_2[x]$, and expressing every element of $\Bbb F_8$ as $\Bbb Z_2$-linear combinations of $1,u,u^2,u^3$, where $u$ is a root of the irreducible polynomial you found. This will make the addition table easy.

Now, if you happen to find an element $b \in \Bbb F_8$ of order 7, the multiplication table will be easy. Good luck!

(Hint: try $x^4 + x^3 + x^2 + x + 1$. It clearly has no linear factors...does it have quadratic factors?).

 

[Euge]

I think Deveno's example has 16 elements. A model for $\mathbb{F}_8$ is $F := Z_2[x]/(x^3+x+1)$. Letting $u = x + (x^3+x+1)$, we have $u^3 = u + 1$ and we can write

$F = \{0,1,u, u+1,u^2,u^2+1, u^2+u, u^2+u+1\}$.

With this, you can make a table.

 

[Math Amateur]

$\Bbb F_8$ will have a "medium-large" multiplication table (64 entries) but it's do-able with some patience. Ditto for addition.

The key to creating such a table is finding an irreducible polynomial of degree 4 in $\Bbb Z_2[x]$, and expressing every element of $\Bbb F_8$ as $\Bbb Z_2$-linear combinations of $1,u,u^2,u^3$, where $u$ is a root of the irreducible polynomial you found. This will make the addition table easy.

Now, if you happen to find an element $b \in \Bbb F_8$ of order 7, the multiplication table will be easy. Good luck!

(Hint: try $x^4 + x^3 + x^2 + x + 1$. It clearly has no linear factors...does it have quadratic factors?).

Thanks Deveno … think I get this … BUT, just to be sure:

You write:

"The key to creating such a table is finding an irreducible polynomial of degree 4 in $\Bbb Z_2[x]$, and expressing every element of $\Bbb F_8$ as $\Bbb Z_2$-linear combinations of $1,u,u^2,u^3$, where $u$ is a root of the irreducible polynomial you found."

What is the theorems/logic behind this reasoning?

Peter

- - - Updated - - -

I think Deveno's example has 16 elements. A model for $\mathbb{F}_8$ is $F := Z_2[x]/(x^3+x+1)$. Letting $u = x + (x^3+x+1)$, we have $u^3 = u + 1$ and we can write

$F = \{0,1,u, u+1,u^2,u^2+1, u^2+u, u^2+u+1\}$.

With this, you can make a table.

Thanks Euge … appreciate your help!

I guess the logic here is that \(\displaystyle \mathbb{F}_8 \) is isomorphic to $F := Z_2[x]/(x^3+x+1)$ so that you can use this 'model' to determine/calculate the elements of \(\displaystyle \mathbb{F}_8 \). Is that correct?

What logic and results/theorems did you use to establish that $F := Z_2[x]/(x^3+x+1)$ was a 'model' for (isomorphic to) \(\displaystyle \mathbb{F}_8 \)?Thanks again for your help.

Peter

 

[Euge]

Yes, Peter, I meant $F$ is isomorphic to $\mathbb{F}_8$. I used the fact that the quotient of a commutative ring by a maximal ideal is a field to claim that $F$ is a field. Since $x^3+x+1$ is irreducible over $\mathbb{Z}_2$ (since it doesn't have any roots in $\mathbb{Z}_2$), the ideal $(x^3+x+1) $ is maximal in $\mathbb{Z}_2$. So I can use fact I mentioned to deduce that $F$ is a field.

 

[Math Amateur]

Yes, Peter, I meant $F$ is isomorphic to $\mathbb{F}_8$. I used the fact that the quotient of a commutative ring by a maximal ideal is a field to claim that $F$ is a field. Since $x^3+x+1$ is irreducible over $\mathbb{Z}_2$ (since it doesn't have any roots in $\mathbb{Z}_2$), the ideal $(x^3+x+1) $ is maximal in $\mathbb{Z}_2$. So I can use fact I mentioned to deduce that $F$ is a field.

Thanks Euge … OK, yes … but how did you figure that \(\displaystyle \mathbb{Z}_2 {x}/ (x^3 + x + 1} \) had 8 elements … and further, how did you come up with this field in the first place … what mental process did you work though ...

Peter

 

[Euge]

To answer both your questions, Peter, let's consider the polynomial ring $\mathbb{Z}_p[x]$, where $p$ is a prime. Let $f(x)$ be any polynomial of degree $n$ in $\mathbb{Z}_p[x]$. Since $pg = 0$ for all $g \in Z_p$, it follows that $pv = 0$ for all $v$ in $\mathbb{Z}_p[x]/(f(x))$. So the factor ring $\mathbb{Z}_p[x]/(f(x))$ has the structure of a vector space over $\mathbb{Z}_p$. You can also see that $\mathbb{Z}_p[x]/(f(x))$ is a $\mathbb{Z}_p$-vector space by observing that it is a module over $\mathbb{Z}_p$ and $\mathbb{Z}_p$ is a field.

Let

$f(x) = a_0 + a_1 x + \cdots + a_n x^n$

where $a_n \neq 0$. If $u = x + (f(x))$, then

$u^n = -(a_0/a_n) - (a_1/a_n)u - \cdots - (a_{n-1}/a_n)u^{n-1}$.

By the division algorithm for polynomials, it follows that every element of $\mathbb{Z}_p[x]/(f(x))$ can be represented uniquely as a polynomial in $\mathbb{Z}_p$ of degree less than $n$. This shows that the set $\{1,u, u^2,\ldots,u^{n-1}\}$ is a $\mathbb{Z}_p$-basis for $\mathbb{Z}_p[x]/(f(x))$. Therefore $\mathbb{Z}_p[x]/(f(x))$ has $|\mathbb{Z}_p|^n = p^n$ elements. Choosing $f(x)$ to be irreducible will make $\mathbb{Z}_p[x]/(f(x))$ a field with $p^n$ elements.

In the special case that $p = 2$ and $f(x) = x^3 + x + 1$, we get the field $F$ with 8 elements.

 

[Deveno]

I think Deveno's example has 16 elements. A model for $\mathbb{F}_8$ is $F := Z_2[x]/(x^3+x+1)$. Letting $u = x + (x^3+x+1)$, we have $u^3 = u + 1$ and we can write

$F = \{0,1,u, u+1,u^2,u^2+1, u^2+u, u^2+u+1\}$.

With this, you can make a table.

My bad, for some reason I got $4\cdot 2( = 2^3)$ confused with $2^4$

Thanks Euge … OK, yes … but how did you figure that \(\displaystyle \mathbb{Z}_2[x]/ (x^3 + x + 1} \) had 8 elements … and further, how did you come up with this field in the first place … what mental process did you work though ...

Peter

You only need an irreducible polynomial of the proper degree. For consider, we can write every element in:

$\Bbb Z_2[x]/(x^3 + x + 1)$ as a coset of the form:

$a + bx + cx^2 + (x^3 + x + 1)$

This can be represented as the triple: $(a,b,c)$.

We have two choices for $a$, two choices for $b$, two choices for $c$, giving $2 \cdot 2\cdot 2 = 8$ distinct cosets.

(this is because we can for any polynomial, write:

$p(x) = q(x)(x^3 + x + 1) + r(x)$, where $r(x) = a + bx + cx^2$ (possibly any of $a,b,c$ are 0)).

So we can list the 8 elements of $\Bbb F_8$ as either:

$(0,0,0) \iff 0$
$(1,0,0) \iff 1$
$(0,1,0) \iff u = x + (x^3 + x + 1)$
$(1,1,0) \iff 1+u$
$(0,0,1) \iff u^2$
$(1,0,1) \iff 1+u^2$
$(0,1,1) \iff u+u^2$
$(1,1,1) \iff 1+u+u^2$

**********

Let's try to find an element of order 7.

$u^2 \neq 1$ so $u$ is not order 2.
$u^3 = 1+u$ (since $u^3 + u + 1 = 0$, so $u^3 + (u + u) + (1+1) = u + 1$, and the last two terms on the left are 0).

So $u$ is not of order 3.

$u^4 = u(u^3) = u(1+u) = u+u^2 \neq 1$
$u^5 = u(u^4) = u(u+u^2) = u^2 + u^3 = 1+u+u^2$
$u^6 = u(u^5) = u(1+u+u^2) = u+u^2+u^3 = 1+u+u+u^2 = 1+u^2$
$u^7 = u(u^6) = u(1+u^2) = u+u^3 = 1+u+u = 1$

so $u$ is an element of order 7.

So, for example, to calculate:

$(1+u)(1+u^2) = u^3u^6 = u^9 = u^2$.

We could use $u^3 = 1+u$ to do this, too:

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

$ = 0 + 0 + u^2 = u^2$.

 

[Deveno]

Some other things to consider with finite fields:

Every field has a multiplicative identity, $1$.

It follows that the subgroup of $(F,+)$ for a finite field $F$, generated by $1$ is a finite cyclic group. Thus, $1$ has finite additive order in $F$.

Since this is an abelian group, we have a natural $\Bbb Z$-action:

$\displaystyle k \cdot 1 = \sum_{i = 1}^k 1$, often written as:

$k \cdot 1 = 1 + 1 + \cdots + 1$ ($k$ summands).

Since $\langle 1\rangle$ is cyclic, every element can be written in this form.

This action gives $\langle 1\rangle$ a natural ring multiplication, as:

$(k\cdot 1)(m\cdot 1) = (km)\cdot 1$.

It follows that $\langle 1\rangle \cong \Bbb Z_n$, for some integer $n$ (called the characteristic of $F$).

Now we know $F$ contains no zero-divisors (since all non-zero elements are units, and units are NEVER zero-divisors, and vice versa). Hence our copy of $\Bbb Z_n$ inside $F$ likewise contains no zero-divisors.

Hence, $n$ is prime. For emphasis: EVERY finite field has prime characteristic.

We can thus regard $F$ as an extension of the ring $\Bbb Z_p$, for some prime $p$, and this defines a $\Bbb Z_p$-action on $F$:

$(k \cdot 1)\cdot x = (k\cdot 1)x$ (the multiplication on the right, is that of $F$).

As Euge notes, this means we have a vector space over the field $\Bbb Z_p$.

As this vector space is finite, it must have a finite basis, say: $\{x_1,x_2,\dots,x_n\}$.

This means every element of $F$ is UNIQUELY expressible as:

$a_1x_1 + a_2x_2 +\cdots + a_nx_n$, for $a_1,a_2,\dots,a_n \in \Bbb Z_p$.

Well, we can simply COUNT these elements: there are $p$ choices for each "coefficient" of the $x_j$, which gives us:

$|F| = p^n$.

So every finite field has $p^n$ elements for some prime $p$, and some natural number $n > 0$.

So, for example, there is NO field that has 6 elements, or 12, or 15.

One might wonder...well, how many fields of order $p^n$ are there? The answer is rather startling: exactly one (up to isomorphism).

Not only is every finite field of order $p^n$, but for every $p$ and every $n$, there IS a field of that order, and every other field of that order is isomorphic to it.

There is another surprising fact about finite fields: the multiplicative group of a finite field is CYCLIC.

Finite fields have a simple subfield structure: for example, a finite field of order 16, has a very simple lattice of subfields:

$\Bbb F_{16} - \Bbb F_8 - \Bbb F_2$.

At this point you might want to hazard a guess about what the group $\text{Gal}(\Bbb F_{16}/\Bbb F_2)$ looks like, and try to conjecture what polynomial in $\Bbb F_2[x]$ that $\Bbb F_{16}$ might be the splitting field of.

 

[Math Amateur]

To answer both your questions, Peter, let's consider the polynomial ring $\mathbb{Z}_p[x]$, where $p$ is a prime. Let $f(x)$ be any polynomial of degree $n$ in $\mathbb{Z}_p[x]$. Since $pg = 0$ for all $g \in Z_p$, it follows that $pv = 0$ for all $v$ in $\mathbb{Z}_p[x]/(f(x))$. So the factor ring $\mathbb{Z}_p[x]/(f(x))$ has the structure of a vector space over $\mathbb{Z}_p$. You can also see that $\mathbb{Z}_p[x]/(f(x))$ is a $\mathbb{Z}_p$-vector space by observing that it is a module over $\mathbb{Z}_p$ and $\mathbb{Z}_p$ is a field.

Let

$f(x) = a_0 + a_1 x + \cdots + a_n x^n$

where $a_n \neq 0$. If $u = x + (f(x))$, then

$u^n = -(a_0/a_n) - (a_1/a_n)u - \cdots - (a_{n-1}/a_n)u^{n-1}$.

By the division algorithm for polynomials, it follows that every element of $\mathbb{Z}_p[x]/(f(x))$ can be represented uniquely as a polynomial in $\mathbb{Z}_p$ of degree less than $n$. This shows that the set $\{1,u, u^2,\ldots,u^{n-1}\}$ is a $\mathbb{Z}_p$-basis for $\mathbb{Z}_p[x]/(f(x))$. Therefore $\mathbb{Z}_p[x]/(f(x))$ has $|\mathbb{Z}_p|^n = p^n$ elements. Choosing $f(x)$ to be irreducible will make $\mathbb{Z}_p[x]/(f(x))$ a field with $p^n$ elements.

In the special case that $p = 2$ and $f(x) = x^3 + x + 1$, we get the field $F$ with 8 elements.

Thanks Euge … most helpful … a most important piece of logic for me was when you write:

"This shows that the set $\{1,u, u^2,\ldots,u^{n-1}\}$ is a $\mathbb{Z}_p$-basis for $\mathbb{Z}_p[x]/(f(x))$. Therefore $\mathbb{Z}_p[x]/(f(x))$ has $|\mathbb{Z}_p|^n = p^n$ elements. Choosing $f(x)$ to be irreducible will make $\mathbb{Z}_p[x]/(f(x))$ a field with $p^n$ elements. "

This links the degree of the minimal polynomial and the number of elements in \(\displaystyle \mathbb{Z}_p \) to the number of elements in the field of interest … previously I was missing that link! … ...

Thanks again,

Peter

 

[Math Amateur]

Some other things to consider with finite fields:

Every field has a multiplicative identity, $1$.

It follows that the subgroup of $(F,+)$ for a finite field $F$, generated by $1$ is a finite cyclic group. Thus, $1$ has finite additive order in $F$.

Since this is an abelian group, we have a natural $\Bbb Z$-action:

$\displaystyle k \cdot 1 = \sum_{i = 1}^k 1$, often written as:

$k \cdot 1 = 1 + 1 + \cdots + 1$ ($k$ summands).

Since $\langle 1\rangle$ is cyclic, every element can be written in this form.

This action gives $\langle 1\rangle$ a natural ring multiplication, as:

$(k\cdot 1)(m\cdot 1) = (km)\cdot 1$.

It follows that $\langle 1\rangle \cong \Bbb Z_n$, for some integer $n$ (called the characteristic of $F$).

Now we know $F$ contains no zero-divisors (since all non-zero elements are units, and units are NEVER zero-divisors, and vice versa). Hence our copy of $\Bbb Z_n$ inside $F$ likewise contains no zero-divisors.

Hence, $n$ is prime. For emphasis: EVERY finite field has prime characteristic.

We can thus regard $F$ as an extension of the ring $\Bbb Z_p$, for some prime $p$, and this defines a $\Bbb Z_p$-action on $F$:

$(k \cdot 1)\cdot x = (k\cdot 1)x$ (the multiplication on the right, is that of $F$).

As Euge notes, this means we have a vector space over the field $\Bbb Z_p$.

As this vector space is finite, it must have a finite basis, say: $\{x_1,x_2,\dots,x_n\}$.

This means every element of $F$ is UNIQUELY expressible as:

$a_1x_1 + a_2x_2 +\cdots + a_nx_n$, for $a_1,a_2,\dots,a_n \in \Bbb Z_p$.

Well, we can simply COUNT these elements: there are $p$ choices for each "coefficient" of the $x_j$, which gives us:

$|F| = p^n$.

So every finite field has $p^n$ elements for some prime $p$, and some natural number $n > 0$.

So, for example, there is NO field that has 6 elements, or 12, or 15.

One might wonder...well, how many fields of order $p^n$ are there? The answer is rather startling: exactly one (up to isomorphism).

Not only is every finite field of order $p^n$, but for every $p$ and every $n$, there IS a field of that order, and every other field of that order is isomorphic to it.

There is another surprising fact about finite fields: the multiplicative group of a finite field is CYCLIC.

Finite fields have a simple subfield structure: for example, a finite field of order 16, has a very simple lattice of subfields:

$\Bbb F_{16} - \Bbb F_8 - \Bbb F_4 - \Bbb F_2$.

At this point you might want to hazard a guess about what the group $\text{Gal}(\Bbb F_{16}/\Bbb F_2)$ looks like, and try to conjecture what polynomial in $\Bbb F_2[x]$ that $\Bbb F_{16}$ might be the splitting field of.

Thanks so much Deveno … that was an extremely informative post … rather an excellent basic tutorial on finite fields … thanks!

I will now be able to move on in field theory feeling more confident my understanding of the basic nature of finite fields ...

Peter

 

[Math Amateur]

To answer both your questions, Peter, let's consider the polynomial ring $\mathbb{Z}_p[x]$, where $p$ is a prime. Let $f(x)$ be any polynomial of degree $n$ in $\mathbb{Z}_p[x]$. Since $pg = 0$ for all $g \in Z_p$, it follows that $pv = 0$ for all $v$ in $\mathbb{Z}_p[x]/(f(x))$. So the factor ring $\mathbb{Z}_p[x]/(f(x))$ has the structure of a vector space over $\mathbb{Z}_p$. You can also see that $\mathbb{Z}_p[x]/(f(x))$ is a $\mathbb{Z}_p$-vector space by observing that it is a module over $\mathbb{Z}_p$ and $\mathbb{Z}_p$ is a field.

Let

$f(x) = a_0 + a_1 x + \cdots + a_n x^n$

where $a_n \neq 0$. If $u = x + (f(x))$, then

$u^n = -(a_0/a_n) - (a_1/a_n)u - \cdots - (a_{n-1}/a_n)u^{n-1}$.

By the division algorithm for polynomials, it follows that every element of $\mathbb{Z}_p[x]/(f(x))$ can be represented uniquely as a polynomial in $\mathbb{Z}_p$ of degree less than $n$. This shows that the set $\{1,u, u^2,\ldots,u^{n-1}\}$ is a $\mathbb{Z}_p$-basis for $\mathbb{Z}_p[x]/(f(x))$. Therefore $\mathbb{Z}_p[x]/(f(x))$ has $|\mathbb{Z}_p|^n = p^n$ elements. Choosing $f(x)$ to be irreducible will make $\mathbb{Z}_p[x]/(f(x))$ a field with $p^n$ elements.

In the special case that $p = 2$ and $f(x) = x^3 + x + 1$, we get the field $F$ with 8 elements.

In the above post Euge writes:

"Since $pg = 0$ for all $g \in Z_p$, it follows that $pv = 0$ for all $v$ in $\mathbb{Z}_p[x]/(f(x))$. So the factor ring $\mathbb{Z}_p[x]/(f(x))$ has the structure of a vector space over $\mathbb{Z}_p$."

Can someone please explain the reason why $pv = 0$ for all $v$ in $\mathbb{Z}_p[x]/(f(x))$ implies that the factor ring $\mathbb{Z}_p[x]/(f(x))$ has the structure of a vector space over $\mathbb{Z}_p$?

Peter

 

[Deveno]

Every extension field $E$ of another field $F$ has the structure of a vector space over $F$.

I believe that in this case, Euge is saying $\Bbb Z_p[x]/(f(x))$ has characteristic $p$, and thus its prime field is $\Bbb Z_p$

(every field is a vector space over its prime field).

 

[Math Amateur]

Every extension field $E$ of another field $F$ has the structure of a vector space over $F$.

I believe that in this case, Euge is saying $\Bbb Z_p[x]/(f(x))$ has characteristic $p$, and thus its prime field is $\Bbb Z_p$

(every field is a vector space over its prime field).

Thanks Deveno ...

Yes, indeed … thanks …

I was mainly puzzling over how "$pv = 0$ for all $v$ in $\mathbb{Z}_p[x]/(f(x))$" led to such a conclusion, but you have given an explanation of that too … thanks ...

Peter

 

[Euge]

In the above post Euge writes:

"Since $pg = 0$ for all $g \in Z_p$, it follows that $pv = 0$ for all $v$ in $\mathbb{Z}_p[x]/(f(x))$. So the factor ring $\mathbb{Z}_p[x]/(f(x))$ has the structure of a vector space over $\mathbb{Z}_p$."

Can someone please explain the reason why $pv = 0$ for all $v$ in $\mathbb{Z}_p[x]/(f(x))$ implies that the factor ring $\mathbb{Z}_p[x]/(f(x))$ has the structure of a vector space over $\mathbb{Z}_p$?

Peter

I used the fact that if $G$ is an abelian group and $p$ is a prime such that $px$ = 0 for all $x \in G$, then $G$ is a vector space over $\mathbb{Z}_p$. To see this, define a scalar multiplication in $G$ by setting $[k]_p \cdot x = kx$, for all $k\in \mathbb{Z}$ and $x\in G$. The multiplication is well-defined: if $k = m \pmod p$ then $k = m + np$ for some integer $n$, and thus

$[k]_p \cdot x = kx = (m + np)x = mx + n(px) = mx + n(0) = mx = [m]_p \cdot x$

for all $x \in G$. Note $[1]_p \cdot x = 1x = x$ and

$[m]_p \cdot ([n]_p \cdot x) = [m]_p \cdot nx = m(nx) = (mn)x = [mn]_p \cdot x = ([m]_p \cdot [n]_p) \cdot x$

for all $x\in G$. The distributive laws follow from the distributive properties of the natural $\mathbb{Z}$-action on $G$. So $G$ satisfies the axioms for a vector space over $\mathbb{Z}_p$.

 

[Deveno]

I used the fact that if $G$ is an abelian group and $p$ is a prime such that $px$ = 0 for all $x \in G$, then $G$ is a vector space over $\mathbb{Z}_p$. To see this, define a scalar multiplication in $G$ by setting $[k]_p \cdot x = kx$, for all $k\in \mathbb{Z}$ and $x\in G$. The multiplication is well-defined: if $k = m \pmod p$ then $k = m + np$ for some integer $n$, and thus

$[k]_p \cdot x = kx = (m + np)x = mx + n(px) = mx + n(0) = mx = [m]_p \cdot x$

for all $x \in G$. Note $[1]_p \cdot x = 1x = x$ and

$[m]_p \cdot ([n]_p \cdot x) = [m]_p \cdot nx = m(nx) = (mn)x = [mn]_p \cdot x = ([m]_p \cdot [n]_p) \cdot x$

for all $x\in G$. The distributive laws follow from the distributive properties of the natural $\mathbb{Z}$-action on $G$. So $G$ satisfies the axioms for a vector space over $\mathbb{Z}_p$.

That is a fascinating construction, which implies any finite abelian group of exponent $p$, is in fact, isomorphic to $\Bbb (Z_p)^n$ (the structure theorem also gives this result, but still...). I would like to know which text you saw it in, or if you devised it yourself.

The thought naturally occurred to me that any finite abelian group $A$ with exponent dividing $n$ could be turned into a $\Bbb Z_n$-module in the same way.

Of course, it is natural to, given a ring-homomorphsim $R \to R/I$, turn an $R/I$-module into an $R$-module, but this gives a sufficient condition to turn an $R$-module $M$ into an $R/I$-module:

if $x\cdot m = 0$, for all $x \in I$. In other words, $I \subseteq \text{Ann}_R(M)$.

Is this necessary, as well?

What bothers me (just a bit) about this, is that it regards the elements $[k]_p$ as something "external" we hit the elements of $\Bbb Z_p[x]/(f(x))$ with, but, in fact, we have the elements of $\Bbb Z_p$ "already in" this quotient ring (as the cosets of constant polynomials), and for scalar multiplication, we can just use the "internal" multiplication of the quotient ring. While this is fine for finite fields, it seems to me to be cumbersome for fields of characteristic 0.

 

[Euge]

That is a fascinating construction, which implies any finite abelian group of exponent $p$, is in fact, isomorphic to $\Bbb (Z_p)^n$ (the structure theorem also gives this result, but still...). I would like to know which text you saw it in, or if you devised it yourself.

The thought naturally occurred to me that any finite abelian group $A$ with exponent dividing $n$ could be turned into a $\Bbb Z_n$-module in the same way.

Of course, it is natural to, given a ring-homomorphsim $R \to R/I$, turn an $R/I$-module into an $R$-module, but this gives a sufficient condition to turn an $R$-module $M$ into an $R/I$-module:

if $x\cdot m = 0$, for all $x \in I$. In other words, $I \subseteq \text{Ann}_R(M)$.

Is this necessary, as well?

What bothers me (just a bit) about this, is that it regards the elements $[k]_p$ as something "external" we hit the elements of $\Bbb Z_p[x]/(f(x))$ with, but, in fact, we have the elements of $\Bbb Z_p$ "already in" this quotient ring (as the cosets of constant polynomials), and for scalar multiplication, we can just use the "internal" multiplication of the quotient ring. While this is fine for finite fields, it seems to me to be cumbersome for fields of characteristic 0.

I came up with the construction myself, and it can be generalized to show that a left $R$-module $M$ is a faithful $R/\text{Ann}(M)$-module. The result I mentioned earlier follows from this since the given conditions imply Ann($G$) = $p\Bbb Z$.

The answer to your second question is no. For example, $\Bbb Z_4$ is a $\Bbb Z$-module with a $\Bbb Z_6$-action given by $[k]_6 [m]_4 = [0]_4$ for all $k, m \in \Bbb Z$. However, $6\Bbb Z$ is not contained in Ann($\Bbb Z_4$): $6 \cdot [3]_4 = [2]_4 \neq [0]_4$. In general, given a left $R$-module $M$ and an ideal $I$ of $R$ which is not contained in Ann($M$), we can make $M$ into a left $R/I$-module by choosing the trivial representation $T : R/I \to \text{End}(M)$.