Roots

Extending to a Field with Four Elements

Use Kronecker's Theorem to construct a field with four elements by adjoining a suitable root of

x^{4} - x

ℤ_{2}^{%}

First we note that $0, 1 \in ℤ_{2}^{%}$ are roots of the given polynomial and that we have the following factorization. $x^{4} - x = x (x^{3} - 1) = x (x - 1) (x^{2} + x + 1)$

We observe that in $ℤ_{2}^{%}$ that $p (x) = x^{2} + x + 1$ is irreducible, by explicitly plugging in the two possible values for $x$ , therefore $E : = ℤ_{2}^{%} / {(p (x))}_{⋄}$ is a field containing $ℤ_{2}^{%}$ and a root of $p (x)$ , specifically $θ : = X + {(p (x))}_{⋄}$ , where we note that $θ^{2} + θ + 1 = 0$ , we also then note that ${(θ + 1)}^{2} + (θ + 1) + 1 = θ^{2} + 2 θ + 1 + θ + 1 + 1 = θ^{2} + θ + 1 = 0$ so that $θ + 1$ is also a root.

So far we know that $E$ contains the elements $0, 1, θ, θ + 1$ , but we actually note is has no more elements through the addition and multiplication tables:

Addition:

\begin{matrix} + & 0 & 1 & θ & θ + 1 \\ 0 & 0 & 1 & θ & θ + 1 \\ 1 & 1 & 0 & θ + 1 & θ \\ θ & θ & θ + 1 & 0 & 1 \\ θ + 1 & θ + 1 & θ & 1 & 0 \end{matrix}

Multiplication:

\begin{matrix} \cdot & 0 & 1 & θ & θ + 1 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & θ & θ + 1 \\ θ & 0 & θ & θ + 1 & 1 \\ θ + 1 & 0 & θ + 1 & 1 & θ \end{matrix}

A Field with Four elements is not Isomorphic to a subfield of a field with 8

A field with four elements is not ismorphic to a subfield of a field with 8 elements

Suppose that $E, F$ are fields with $4$ and $8$ elements respectively, then let $E_{U}$ denote the set of elements in $E$ which are units, since it's a field, then every non-zero element has an inverse, so then $E_{U}$ has $3$ elements, symetrically $F_{U}$ has 7 elements. Moreover note that $E_{U}, F_{U}$ are multiplicative groups.

Suppose for the sake of contradiction that $E$ was isomorphic to some subfield of $F$ . In this case we note that $E_{U}$ must be a subgroup of $F_{U}$ , this is because $E$ is a subfield, of $F$ , but then by Lagranges theorem, we should have that the order of $E_{U}$ divides $F_{U}$ , but clearly $3$ does not divide $7$ , therefore our assumption was incorrect, so we know that $E$ cannot be isomorphic to any subfield of $F$ .

Let

F = ℤ_{2}^{%}

and let

f (x) = x^{3} + x^{2} + 1 \in F [x]

. Suppose a is a root of

f (x)

in some extension

E

F

. Let

F (a)

denote smallest subfield of

E

containing

F

and

a

How many elements does $F (a)$ have?
Express each element of $F (a)$ in terms of $a$
Write out the multiplication table for $F (a)$

Hint: note that

a, a^{2}, a^{3}, \dots

are all in

F (a)

but that

a^{3}

can be expressed in terms of

a

and

a^{2}

since

a^{3} + a^{2} + 1 = 0

We'll first note that $F (a) ≅ F [x] / {(f (x))}_{⋄}$ , this can be seen by the isomorphism which can be constructed by using the evaluation map at $a$ (which can be seen to be well defined for cosets). Therefore since we know there are $8$ elements in the latter, then we know there are 8 elements in $F (a)$

For each coset from $F [x] / {(f (x))}_{⋄}$ is of the form $c_{2} x_{2} + c_{1} x^{1} + c_{0}$ where $c_{i} \in ℤ_{2}^{%}$ , and so for each evaluation, we obtain an element of $F (a)$ , namely we get the following $0, 1, a, a + 1, a^{2}, a^{2} + a, a^{2} + a + 1$

Note that the multiplication table can be obtained, but will take a while, we can cut down on half of our work by realizing that it is a field and so the table will be symmetric down the diagonal.

Let

F = ℤ_{2}^{%}

and let

f (x) = x^{4} + x + 1 \in F [x]

. Suppose

a

is a root of

f (x)

in some extension

E

F

. Write

f (x)

as a product of linear factors in

E [x]

. Hint: check that

a + 1

is also a root of

f (x)

In order for this to be true, we would need to be able to have all roots existing in the extension $E$ , we now go along this path.

We know that $a$ is a root, therefore we deduce that $a^{4} = a + 1$ , this is beneficial as it will allow us to reduce any power that is greater or equal to $4$ . We'll first note that if $p$ is prime, then for any $k \in [1 \dots p - 1]$ we have that $p | (\binom{p}{k})$ , therefore in the context of $ℤ_{p}^{%}$ when we expand ${(x + y)}^{n}$ , we can ignore all terms execept for the first and last, therefore we obtain ${(a + 1)}^{4} + (a + 1) + 1 = a^{4} + 1 + (a + 1) + 1 = a^{4} + a + 1 = 0$ Similarly we note that $a^{2}$ is a root because ${(a^{2})}^{4} + a^{2} + 1 = {(a^{4})}^{2} + a^{2} + 1 = {(a + 1)}^{2} + a^{2} + 1 = (a^{2} + 1) + (a^{2} + 1) = 0$ We don't have to check $a^{4}$ because we already did as it equals $a + 1$ . Finally note that $a^{8}$ is a root because ${(a^{8})}^{4} + a^{8} + 1 = {(a + 1)}^{8} + a^{8} + 1 = a^{8} + 1 + a^{8} + 1 = 0$ We would continue like this but we see that $a^{16} = {(a^{4})}^{4} = {(a + 1)}^{4} = a^{4} + 1 = a$ , so it cycles back after this point. Anyways we've found all four roots, so we have $f (x) = (x - a) (x - a^{2}) (x - a^{4}) (x - a^{8})$

Double Kronecker

Use Kronecker’s theorem to construct a field with eight elements by adjoining suitable roots of

x^{8} - x

ℤ_{2}^{%}

. Hint:

x^{8} - x = x (x - 1) (x^{3} + x^{2} + 1) (x^{3} + x + 1)

over

ℤ_{2}^{%}

We note that indeed

x^{8} - x = x (x - 1) (x^{3} + x^{2} + 1) (x^{3} + x + 1)

And that since the polynomial

x^{3} + x^{2} + 1

was previously seen to be irreducible in

ℤ_{2}^{%}

, and we can easily verify that

x^{3} + x + 1

is irreducible by checking roots.

Therefore we know that $ℤ_{2}^{%} [x] / {(x^{3} + x + 1)}_{⋄}$ is a field containing $ℤ_{2}^{%}$ and a root $a$ of $x^{3} + x + 1$ .

Notice that this field has order $8$ , meaning that it's multiplicative group has order $7$ . Which means that given any element $x$ in the multiplicative group we know $x^{7} = 1$ , and since $x^{8} - x = x (x^{7} - 1)$ we can conclude that it is also a root to the polynomial.

Therefore we can say that no matter which irredicible polynomial we choose to use Kronecker's on, the extension will contain all $8$ roots.