# A field such that no extension has an 11th primitive root of unity

I am asked to find the field described in the title. However, I can't quite understand the question.

For any field $K$, and $\zeta_{11} = e^{\frac{2\pi i}{11}}$ then surely the extension $K \leq K(\zeta_{11})$ contains the 11th primitive root of unity?

So then how can such a field exist?

• The algebraic closure contains always roots of unity – Tsemo Aristide May 8 '18 at 16:03

Hint: what is the order of $\mathbb{F}_q^{\times}$, the roots of unity of a finite field $\mathbb{F}_q$?
An 11th primitive root of unity has order 11 in the multiplicative group $K^*$. So find a field $GF(p^k)$ where $11\not\mid(p^k-1)$
The field $\mathbb{C}$ of complex numbers is not an extension of every field and thus, there exist fields $k$ such that you can't define $k(\zeta_{11})$.
Indeed, if $\mathbb{C}/k$ is a field extension, then $k$ has necessarily null characteristic.
Now, the field $\mathbb{F}_{11}$ with $11$ elements - which is unique up to isomorphism and isomorphic to $\frac{\mathbb{Z}}{11 \mathbb{Z}}$ - has the desired property.
Indeed, if $k$ is a field extension of $\mathbb{F}_{11}$, then the characteristic of $k$ is $11$ and hence, the map $F: x \mapsto x^{11}$ is a field endomorphism of $k$ (called the Frobenius endomorphism) and in particular, is injective.
Therefore, if $k$ contained a primitive $11^{\text{th}}$ root of unity $x \in k$, then $F(x) = 1 = F(1)$ and hence, $x = 1$ which is not a primitive $11^{\text{th}}$ root of unity.
Or alternatively, if $k$ contained a primitive $11^{\text{th}}$ root of unity $x \in k$, then $x$ would be algebraic over $\mathbb{F}_{11}$ and hence, $\mathbb{F}_{11}(x)$ would be a finite extension of $\mathbb{F}_{11}$. Therefore, $\mathbb{F}_{11}(x)$ would be a finite field of cardinal $11^{n}$ for some $n \geq 1$ and $\mathbb{F}_{11}(x)^{\times}$ would be a group of order $11^{n} -1$ which contains $x$ - which is an element of order $11$. Thus, we would have $11 | 11^{n} -1$ which can't happen.