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$Φ_{5}$ has four roots in $F_{11}$

Example

Over $F_{11}$ the fifth cyclotomic polynomial

$Φ_{5} (t) = t^{4} + t^{3} + t^{2} + t + 1$

splits into four distinct linear factors:

$Φ_{5} (t) = (t - 3) (t - 4) (t - 5) (t - 9) .$

The four roots $3, 4, 5, 9$ are exactly the primitive fifth roots of unity in $F_{11}$ .

Facts & Assumptions

Given: The field $F_{11}$ and the polynomial $Φ_{5} (t) = t^{4} + t^{3} + t^{2} + t + 1$ .

[L1]

If $\gcd (n, q) = 1$ , the reduction of $Φ_{n}$ in $F_{q} [t]$ is a product of distinct monic irreducibles, each of degree the order of $[q]$ modulo $n$ (For $\gcd (n, q) = 1$ the reduction of $Φ_{n}$ in $F_{q} [t]$ is a product of distinct monic irreducibles, each of degree the order of $[q]$ modulo $n$ ).

[L2]

For $\gcd (n, q) = 1$ , the image of $Gal (F_{q} (μ_{n}) / F_{q})$ in $(Z / n)^{\times}$ is generated by $[q]$ , so the extension degree is the order of $[q]$ modulo $n$ (For $\gcd (n, q) = 1$ the image of $Gal (F_{q} (μ_{n}) / F_{q})$ in $(Z / n)^{\times}$ is generated by $[q]$ ).

Verification

technique · direct

1.1

In $(Z / 5)^{\times}$ one has $[11] = [1]$ , so the order of $[11]$ modulo $5$ is $1$ . Hence [L1] and [L2] say every irreducible factor of $Φ_{5}$ over $F_{11}$ is linear, and the factors are distinct.

L1L2algebra

1.2

The powers of $3$ in $F_{11}^{\times}$ are $3^{2} = 9$ , $3^{3} = 5$ , $3^{4} = 4$ and $3^{5} = 1$ , so the four nontrivial fifth roots of unity in $F_{11}$ are $3, 9, 5, 4$ .

givenalgebra

2.1

Each of $3, 4, 5, 9$ is therefore a root of $t^{5} - 1$ and is not $1$ , so each is a root of $Φ_{5} (t) = (t^{5} - 1) / (t - 1)$ . Since $Φ_{5}$ is monic of degree $4$ , it follows that $Φ_{5} (t) = (t - 3) (t - 4) (t - 5) (t - 9) .$

step 1.2algebra

3.1

The roots $3, 4, 5, 9$ all have multiplicative order $5$ by step 1.2, so they are exactly the primitive fifth roots of unity in $F_{11}$ .

step 1.2algebra∎

Remarks

This is the order-one case of the finite-field factorisation theorem. When $[q]$ has order one modulo $n$ , every irreducible factor has degree one and the whole cyclotomic polynomial splits over the base field.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

28 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, Cyclotomic Extensions (expository blurb), Example 5.5 (standard reference, not scraped)
P. L. Clark, Field Theory (course notes/monograph), Section 9.1.2 (standard reference, not scraped)

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