1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ The algebraic closure contains always roots of unity $\endgroup$ – Tsemo Aristide May 8 '18 at 16:03
1
$\begingroup$

Hint: what is the order of $\mathbb{F}_q^{\times}$, the roots of unity of a finite field $\mathbb{F}_q$?

$\endgroup$
1
$\begingroup$

An 11th primitive root of unity has order 11 in the multiplicative group $K^*$. So find a field $GF(p^k)$ where $11\not\mid(p^k-1)$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.