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One field for each prime power and none of any other size: the quotient construction that builds them, the cyclic multiplicative group, and the subfields that match the divisors of the exponent.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to say why a finite field has prime power size, build one as a quotient of a polynomial ring by an irreducible polynomial, complete its multiplication table and read off whether the quotient is a field, list its subfields from the divisors of the exponent, describe the Frobenius map and the cyclic multiplicative group, and prove that one finite field sits inside another exactly when the exponents divide.
The first course built $\mathbb{Z}/p\mathbb{Z}$ and found it a field exactly when $p$ is prime, and built $F[x]/(f)$ and found it a field exactly when $f$ is irreducible. This lesson puts the second on top of the first and finds that the combination produces every finite field there is — and that the list is astonishingly short.
The characteristic of a field is the least $p > 0$ with $p \cdot 1 = 0$, or zero if there is none; for a finite field it is prime. The prime subfield is the copy of $\mathbb{F}_p$ generated by $1$. The Frobenius map is $x \mapsto x^{p}$, which is a field automorphism in characteristic $p$. A primitive element generates the multiplicative group.
Size. A finite field $F$ has prime characteristic $p$ — if the characteristic were composite, $p = ab$ would give zero divisors. So $F$ contains $\mathbb{F}_p$, and as a vector space over it $F$ has some finite dimension $n$. Choosing coordinates,
$$|F| = p^{n}.$$
There is no field with $6$, $10$ or $12$ elements.
Existence and uniqueness. For every prime power $q = p^{n}$ there is exactly one field with $q$ elements, up to isomorphism. It is the splitting field of
$$x^{q} - x$$
over $\mathbb{F}_p$: the roots of that polynomial form a field (the Frobenius map makes the set closed under the operations) and there are exactly $q$ of them because the derivative is $-1$, so there are no repeated roots. Uniqueness of splitting fields does the rest.
Concretely, $\mathbb{F}_{p^{n}} \cong \mathbb{F}_p[x]/(f)$ for any irreducible $f$ of degree $n$ over $\mathbb{F}_p$; different choices of $f$ give isomorphic fields.
The multiplicative group is cyclic. $F^{\times}$ has $q - 1$ elements and is cyclic: a finite subgroup of the multiplicative group of any field is. A generator is a primitive element, and its powers run through every non-zero element — which is how finite field arithmetic is tabulated in practice.
Subfields are divisors. $\mathbb{F}_{p^{m}} \subseteq \mathbb{F}_{p^{n}}$ exactly when $m \mid n$, and then there is exactly one such subfield. The proof is the tower law one way and a divisibility of polynomials the other. So the subfield lattice of $\mathbb{F}_{64}$ is the divisor lattice of $6$: $\mathbb{F}_2, \mathbb{F}_4, \mathbb{F}_8, \mathbb{F}_{64}$, with $\mathbb{F}_4$ and $\mathbb{F}_8$ incomparable.
Frobenius. $\varphi(x) = x^{p}$ satisfies $\varphi(x + y) = \varphi(x) + \varphi(y)$, because the binomial coefficients $\binom{p}{k}$ for $0 < k < p$ are divisible by $p$. It is an automorphism fixing $\mathbb{F}_p$ pointwise, it has order $n$ on $\mathbb{F}_{p^{n}}$, and it generates the whole automorphism group — cyclic of order $n$. The subfields are its fixed fields, which is the Galois correspondence in its simplest instance.
A warning. $\mathbb{F}_{p^{n}}$ is not $\mathbb{Z}/p^{n}\mathbb{Z}$ for $n > 1$. In $\mathbb{F}_4$ every element added to itself gives zero; in $\mathbb{Z}/4\mathbb{Z}$ the element $2$ is a zero divisor and there is no field at all.
Another way: picture
Take the clock arithmetic of $\mathbb{Z}/p\mathbb{Z}$ as the ground floor. To build a bigger field, do not take a bigger clock — that introduces zero divisors. Instead take polynomials over the small field and impose one irreducible relation, exactly as $\mathbb{C}$ is built from $\mathbb{R}$ by imposing $i^{2} = -1$. The result has $p$ choices for each of $n$ coordinates, so $p^{n}$ elements, and it is the only field of that size.
Another way: steps
To work with $\mathbb{F}_{p^{n}}$: 1. Find an irreducible polynomial of degree $n$ over $\mathbb{F}_p$. 2. Represent elements as polynomials of degree below $n$, with $p^{n}$ of them. 3. Add coordinatewise modulo $p$; multiply as polynomials and reduce modulo the irreducible one. 4. Find a primitive element by trying small ones and checking its order is $p^{n} - 1$. 5. Read the subfields off the divisors of $n$.
Take $p = 2$ and $n = 3$, so a field of eight elements is wanted.
Choose an irreducible cubic over $\mathbb{F}_2$. A cubic is irreducible exactly when it has no root, since a factorisation would force a linear factor. There are eight cubics with leading coefficient $1$; those with constant term $0$ have the root $0$, and those with an even number of non-zero coefficients have the root $1$. What survives is $x^{3} + x + 1$ and $x^{3} + x^{2} + 1$. Either will do; they give isomorphic fields.
The elements. Take $f = x^{3} + x + 1$ and write $\alpha$ for the class of $x$, so $\alpha^{3} = \alpha + 1$. The eight elements are
$$0, \ 1, \ \alpha, \ \alpha + 1, \ \alpha^{2}, \ \alpha^{2} + 1, \ \alpha^{2} + \alpha, \ \alpha^{2} + \alpha + 1.$$
The multiplicative group. It has seven elements, and $7$ is prime, so every non-identity element is a generator. Powers of $\alpha$: $\alpha, \alpha^{2}, \alpha + 1, \alpha^{2} + \alpha, \alpha^{2} + \alpha + 1, \alpha^{2} + 1, 1$. All seven, as promised.
Subfields. $3$ has divisors $1$ and $3$, so the only subfields are $\mathbb{F}_2$ and the whole field. $\mathbb{F}_8$ does not contain $\mathbb{F}_4$, even though $4 < 8$ and both are powers of $2$ — because $2 \nmid 3$. This is the fact that surprises people, and it is why the divisor condition is on the exponents rather than on the sizes.
Frobenius. $\varphi(x) = x^{2}$ permutes the roots of $f$: $\alpha \mapsto \alpha^{2} \mapsto \alpha^{4} = \alpha^{2} + \alpha \mapsto \alpha^{8} = \alpha$. So $\varphi$ has order $3$, the automorphism group is cyclic of order $3$, and its only subgroups are the trivial one and everything — matching the two subfields exactly.
Taking $\mathbb{F}_{p^{n}}$ to be $\mathbb{Z}/p^{n}\mathbb{Z}$. The second is not a field for $n > 1$. The right construction is a quotient of a polynomial ring.
Expecting a field of any size. Only prime powers. There is no field with $6$ elements, and no amount of cleverness produces one.
Reading the subfield condition off the sizes. $\mathbb{F}_4$ is not a subfield of $\mathbb{F}_8$; the divisibility is on the exponents, $2 \nmid 3$.
Thinking different irreducible polynomials give different fields. They give isomorphic ones. The field depends only on its size.
Forgetting that Frobenius is additive. $(x + y)^{p} = x^{p} + y^{p}$ in characteristic $p$ — sometimes called the freshman's dream, and in this one setting it is correct.
Assuming the additive group is cyclic. It is elementary abelian: $(\mathbb{Z}/p\mathbb{Z})^{n}$, in which every non-zero element has order $p$. It is the multiplicative group that is cyclic.
$x^{2} + x + 1$ is the only irreducible quadratic over $\mathbb{F}_2$: it has no root, since $0$ and $1$ both give $1$.
One candidate.
$\mathbb{F}_4 = \mathbb{F}_2[x]/(x^{2} + x + 1) = \{0, 1, \alpha, \alpha + 1\}$ with $\alpha^{2} = \alpha + 1$.
Four elements.
The three non-zero elements form a cyclic group of order $3$: $\alpha^{2} = \alpha + 1$ and $\alpha^{3} = \alpha^{2} + \alpha = 1$. And $\alpha + \alpha = 0$, so this is nothing like $\mathbb{Z}/4\mathbb{Z}$.
Cyclic multiplicative group.
A finite field has prime characteristic $p$, and contains $\mathbb{F}_p$ as its prime subfield.
The characteristic is prime.
It is then a vector space over $\mathbb{F}_p$ of some dimension $n$, so its size is $p^{n}$.
Count coordinate vectors.
$6 = 2 \times 3$ is not a prime power, so no such field exists. The same rules out $10, 12, 14, 15$ and every other size with two distinct prime factors.
No field of size six.
Subfields correspond to divisors of the exponent $12$.
Count divisors, not sizes.
The divisors of $12$ are $1, 2, 3, 4, 6, 12$.
Six of them.
So there are six subfields: $\mathbb{F}_3, \mathbb{F}_9, \mathbb{F}_{27}, \mathbb{F}_{81}, \mathbb{F}_{729}$ and the field itself. The lattice is the divisor lattice of $12$, which is also the subgroup lattice of a cyclic group of order $12$ — as the Galois correspondence will explain.
$\mathbb{F}_{64}$ has $64 = 2^{6}$ elements. Each row gives a divisor of $6$. Fill in the size of the corresponding subfield and the degree of $\mathbb{F}_{64}$ over it.
| The divisor $d$ | Elements in the subfield | Degree of the big field over it | |
|---|---|---|---|
| The smallest divisor | 1 | ||
| The next divisor | 2 | ||
| The next divisor | 3 | ||
| The largest divisor | 6 |
How many subfields does $\mathbb{F}_8$ have, counting itself and the prime subfield?
Answer:
Here is $\mathbb{F}_2[x]/(x^{2} + 1)$, writing $2$ for the class of $x$ and $3$ for the class of $x + 1$. Its four elements are labelled $0, 1, 2, 3$. Complete the multiplication table; the row and column of $1$ are given.
| 1 | 2 | 3 | |
|---|---|---|---|
| 1 | 1 | 2 | 3 |
| 2 | 2 | ||
| 3 | 3 |
Is this ring a field: $\mathbb{F}_2[x]/(x^{2} + x)$, writing $2$ for the class of $x$ and $3$ for the class of $x + 1$?
Select every statement that is true of finite fields.
This task has no paper form; do it on a device.
Build the proof that $\mathbb{F}_{p^{m}}$ sits inside $\mathbb{F}_{p^{n}}$ exactly when $m$ divides $n$.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Here is $\mathbb{Z}[x]/(x, 4)$, which is $\mathbb{Z}/4\mathbb{Z}$ with $2$ and $3$ as its other two elements. Its four elements are labelled $0, 1, 2, 3$. Complete the multiplication table; the row and column of $1$ are given.
| 1 | 2 | 3 | |
|---|---|---|---|
| 1 | 1 | 2 | 3 |
| 2 | 2 | ||
| 3 | 3 |
You can build a small finite field, multiply in it, and list its subfields. Say in your own words why there is no field with six elements, and why $\mathbb{F}_4$ is not inside $\mathbb{F}_8$. Next: turning degrees into a statement about what ruler and compasses can reach.
9. Your turn: how many subfields does $\mathbb{F}_{3^{12}}$ have?, step 3