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The divisor-sum identity at $q = 2$ , $n = 3$ finds exactly two monic irreducible cubics

Example

Over $F_{2}$ the divisor-sum identity ( $\sum_{d ∣ n} d N_{q} (d) = q^{n}$ for the counts $N_{q} (d)$ of monic irreducibles of degree $d$ over $F_{q}$ ) at $n = 3$ reads

$N_{2} (1) + 3 N_{2} (3) = 2^{3} = 8,$

and $N_{2} (1) = 2$ , so $N_{2} (3) = 2$ . The two monic irreducible cubics in $F_{2} [t]$ are

$t^{3} + t + 1 and t^{3} + t^{2} + 1.$

Facts & Assumptions

Given: The field $F_{2}$ with two elements and the counts $N_{2} (d)$ of monic irreducible polynomials of degree $d$ in $F_{2} [t]$ (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L1]

$\sum_{d ∣ n} d N_{q} (d) = q^{n}$ for every $n \geq 1$ , the sum over the positive divisors of $n$ ( $\sum_{d ∣ n} d N_{q} (d) = q^{n}$ for the counts $N_{q} (d)$ of monic irreducibles of degree $d$ over $F_{q}$ , Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L2]

A polynomial of degree $2$ or $3$ over a field is irreducible if and only if it has no root in that field (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field).

[L3]

For a commutative ring $R$ , $a \in R$ and $f \in R [x]$ : $f (a) = 0$ if and only if $x - a$ divides $f$ (Factor theorem over a commutative ring, Evaluation and roots of a polynomial in a commutative target ring).

Verification

technique · direct

1.1

The monic polynomials of degree one in $F_{2} [t]$ are $t$ and $t + 1$ , and each is irreducible, having degree one; so $N_{2} (1) = 2$ .

givenalgebra

1.2

A monic cubic over $F_{2}$ is $f = t^{3} + a t^{2} + b t + c$ with $a, b, c \in F_{2}$ , so there are eight of them. Such an $f$ has no root in $F_{2}$ exactly when $f (0) = c \neq 0$ and $f (1) = 1 + a + b + c \neq 0$ , that is exactly when $c = 1$ and $a + b = 1$ .

givenalgebra

2.1

The positive divisors of $3$ are $1$ and $3$ , so [L1] at $q = 2$ and $n = 3$ reads $N_{2} (1) + 3 N_{2} (3) = 8$ ; with step 1.1 this gives $3 N_{2} (3) = 6$ and $N_{2} (3) = 2$ .

step 1.1L1algebra

3.1

The pairs $(a, b)$ with $a + b = 1$ in $F_{2}$ are $(0, 1)$ and $(1, 0)$ , so exactly two monic cubics have no root in $F_{2}$ , namely $t^{3} + t + 1$ and $t^{3} + t^{2} + 1$ ; by [L2] these two are irreducible and by [L2] and [L3] the other six are not, each having a root and hence a linear factor. This agrees with the count $N_{2} (3) = 2$ of step 2.1.

step 2.1step 1.2L2L3algebra∎

Remarks

The identity is a recursion, not a formula. It determines $N_{2} (3)$ only because $N_{2} (1)$ is already known; at $n = 4$ it would read $N_{2} (1) + 2 N_{2} (2) + 4 N_{2} (4) = 16$ and would need $N_{2} (2)$ first.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

K. Conrad, Roots and Irreducibles (expository blurb), Example 6.2 (standard reference, not scraped)
K. Conrad, Finite Fields (expository blurb), Section 6 (standard reference, not scraped)

The divisor-sum identity at q=2q=2, n=3n=3 finds exactly two monic irreducible cubics