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The residue class of $t$ generates $F_{8}^{\times}$

Example

In $F_{8} = F_{2} [t] / (t^{3} + t + 1)$ , the residue class $a$ of $t$ has order $7$ and generates $F_{8}^{\times}$ .

Facts & Assumptions

Given: The quotient $A = F_{2} [t] / (t^{3} + t + 1)$ and the class $a$ .

[L1]

A polynomial quotient over a field is a field exactly when its modulus is irreducible (For a nonconstant $p$ in $F [x]$ , the ideal $(p)$ is maximal and $F [x] / (p)$ is a field exactly when $p$ is irreducible).

[L2]

An irreducible cubic quotient has the power basis $(1, a, a^{2})$ (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

[L3]

The ring $Z / 2$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

[L4]

The nonzero elements of a finite field form a cyclic group (The multiplicative group $F_{q}^{\times}$ of a finite field is cyclic).

Verification

technique · direct

1.1

The cubic has no root in $F_{2}$ , so [L1] and [L3] make the quotient a field, and [L2] gives its eight elements. The defining relation is $a^{3} = a + 1$ .

givenL1L2L3

2.1

Reduction gives $a^{0} = 1$ , $a$ , $a^{2}$ , $a^{3} = a + 1$ , $a^{4} = a^{2} + a$ , $a^{5} = a^{2} + a + 1$ , $a^{6} = a^{2} + 1$ , and $a^{7} = 1$ .

step 1.1algebra

3.1

The first seven displayed powers are distinct and are all nonzero elements, so $a$ has order $7$ and generates $A^{\times}$ , in agreement with [L4].

step 2.1L4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 90 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

K. Conrad, Finite Fields, Section 1 (standard reference, not scraped)

The residue class of tt generates F8×F8×​