I am having a bit of an issue with an example I was looking at. The question states: "What is the smallest field of characteristic 2 that contains a primitive 11th root of unity?". I am not familiar with how we would go about finding such a field, but my thought processes is going a bit like this:

1) By 'smallest field', I am thinking they mean a field of smallest order? Therefore I want to find the smallest ordered field, with characteristic 2, that contains a primitive 11th root of unity.

2) Having a primitive 11th root of unity means that there $\exists x \in \mathbb{F}_q$ such that $x^{11}=1$ and that $x^s\ne 1$ for $0\lt s \lt 11$. I am not quite sure how I would go about finding the $\mathbb{F}_1$ that contains this primitive root of unity, other than manually checking each $\mathbb{F}_q$, $q=2,3,...$.

3) The fact is given that the characteristic of $\mathbb{F}_q$ is $p=2$. I know that this means that in the field, $1+1=0$, I am not really sure how I would use this fact to find the field. Even so, other than $\mathbb{F}_2$, I am not sure what other fields would have this characteristic.

So yeah, I am a bit lost overall how I would go about attacking this problem (or similar ones - different characteristic, or different n-th roots).

Thanks for any guidance.

  • $\begingroup$ If you take an irreducible polynomial $f$ in $\mathbb F_2[x]$, then the quotient $\mathbb F_2[x]/\langle f\rangle $ is a field extension of $\mathbb F_2$ of size $2^{\deg f}$. $\endgroup$ – RKD Mar 2 '16 at 1:24
  • $\begingroup$ Try and guess how you should choose $f$ so that you have an element $x$ in the extension satisfying the requirement that its $11$-th power is 1 and all smaller powers are not. $\endgroup$ – RKD Mar 2 '16 at 1:26
  • $\begingroup$ The nonzero elements of a field form a group under multiplication. The order of an element (of a finite group) divides the order of the group. That should get you started. $\endgroup$ – Gerry Myerson Mar 2 '16 at 1:53

For any prime $p$ and any $n \geq 1$, there is a (unique) field $\mathbb{F}_{p^n}$ of order $p^n$. You can construct such a field by taking the splitting field of $x^{p^n} - x$ over $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$, but the construction doesn't really matter for the question, just the fact that there is such a thing. (See here for more information on finite fields.)

Suppose $\zeta \in \mathbb{F}_{2^n}$ for some $n$, where $\zeta$ is a primitive $11^\text{th}$ root of $1$. As you point out, then $\zeta$ has order $11$ in the unit group ${\mathbb{F}_{2^n}}^\times$. Since $\mathbb{F}_{2^n}$ is a field, every element except for $0$ is a unit, so $|{\mathbb{F}_{2^n}}^\times| = 2^n - 1$. By Lagrange's Theorem, then $11 \mid 2^n - 1$, which provides a necessary condition. Since ${\mathbb{F}_{2^n}}^\times$ is a cyclic group then for each divisor $d$ of $2^n - 1$ there is an element of order $d$, so this condition is also sufficient.

Given $11 \mid 2^n - 1$, then $2^n - 1 = 11k$ for some $k \in \mathbb{Z}$, so $2^n \equiv 1 \pmod{11}$. Thus to determine the smallest such $n$, we simply have to compute the order of $2$ in $(\mathbb{Z}/11\mathbb{Z})^\times$. Since $|(\mathbb{Z}/11\mathbb{Z})^\times| = 10$, then $2^{10} \equiv 1 \pmod{11}$. To ensure that this is the smallest such $n$, we simply check that $2^{10/5} = 2^2 = 4 \not \equiv 1 \pmod{11}$ and $2^{10/2} = 2^5 = 32 \equiv -1 \not \equiv 1 \pmod{11}$.

  • $\begingroup$ So basically, as long as I follow the process above, I can always find the smallest such field, assuming $p$ is prime, correct? $\endgroup$ – user3784030 Mar 2 '16 at 2:48
  • $\begingroup$ Yes, this should work for any prime $p$ except for $11$, since $x^{11} - 1 = (x-1)^{11}$ in $\mathbb{Z}/11\mathbb{Z}$, so there is no such thing as a primitive $11^\text{th}$ root of unity mod $11$. $\endgroup$ – Viktor Vaughn Mar 2 '16 at 2:49

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