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In characteristic three, $t^{3} - 1 = (t - 1)^{3}$ and $μ_{6}$ coincides with $μ_{2}$

Example

Let $K$ be a field of characteristic $3$ . Then

$t^{3} - 1 = (t - 1)^{3}, μ_{3} (K) = {1}, μ_{6} (K) = μ_{2} (K) = {1, - 1},$

so $t^{6} - 1$ has only the two distinct roots $1$ and $- 1$ rather than six.

Facts & Assumptions

Given: A field $K$ of characteristic $3$ .

[L1]

For a fixed integer $k \geq 1$ , in characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ (In characteristic $p$ the only $p^{k}$ -th root of unity is $1$ , and $t^{p^{k}} - 1 = (t - 1)^{p^{k}}$ ).

[L2]

$μ_{n} (K) = {x \in K : x^{n} = 1}$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

Verification

technique · direct

1.1

Applying [L1] at $p = 3$ and $k = 1$ gives $t^{3} - 1 = (t - 1)^{3}$ and $μ_{3} (K) = {1}$ .

1.2

Since $(1)^{2} = 1$ and $(- 1)^{2} = 1$ , one has ${1, - 1} \subseteq μ_{2} (K)$ . Conversely, if $x \in μ_{2} (K)$ then $x^{2} = 1$ , so $x^{2} - 1 = (x - 1) (x + 1) = 0$ and therefore $x = 1$ or $x = - 1$ in the field $K$ ; hence $μ_{2} (K) = {1, - 1}$ .

L2algebra

2.1

If $x \in μ_{6} (K)$ then $(x^{2})^{3} = x^{6} = 1$ , so $x^{2} \in μ_{3} (K)$ by [L2]; step 1.1 gives $x^{2} = 1$ , hence $x \in μ_{2} (K)$ by [L2]. Thus $μ_{6} (K) \subseteq μ_{2} (K)$ .

step 1.1step 1.2L2

2.2

Conversely, if $x \in μ_{2} (K)$ then $x^{6} = (x^{2})^{3} = 1$ , so $x \in μ_{6} (K)$ . Therefore $μ_{6} (K) = μ_{2} (K) = {1, - 1}$ .

step 1.2L2algebra

3.1

Using step 1.1, $t^{6} - 1 = (t^{3} - 1) (t^{3} + 1) = (t - 1)^{3} (t + 1)^{3},$ so its only distinct roots are $1$ and $- 1$ , exactly the two elements of step 2.2.

step 1.1step 2.2algebra∎

Remarks

This is why the characteristic hypothesis is load-bearing. The statement " $∣ μ_{n} ∣ = n$ " fails here for two different reasons at once: the polynomial $t^{3} - 1$ is inseparable, and the extra cube roots never appear even after passing to a splitting field because the splitting field is already the base field.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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), Section 1 (standard reference, not scraped)
P. L. Clark, Field Theory (course notes/monograph), Section 9.1.1 (standard reference, not scraped)

In characteristic three, t3−1=(t−1)3t3−1=(t−1)3 and μ6μ6​ coincides with μ2μ2​