# Primitive root in quadratic field

• Apr 28th 2008, 06:44 PM
Let $\displaystyle \omega _{m} = e^{ \frac {2 \pi i} {m}}$, then $\displaystyle \omega _{m}$ is one of the primitive m th roots of unity.

a) Show that $\displaystyle \mathbb {Q} [ \omega _{m} ] \subseteq \mathbb {Q} [ \omega _n ]$ if m|n.

Proof.

Let mt = n for some integer t.

Pick an element $\displaystyle a+b \omega _m \in \mathbb {Q} [ \omega _m ]$, with a and b rational numbers.

Then it equals to $\displaystyle a + b e^ { \frac {2 \pi i } {m} } = a + be^{ \frac {2 \pi i (t) } {n} } = a + be^t + be ^ { \frac {2 \pi i } {n} }$ Now, is this element in $\displaystyle \mathbb {Q} [ \omega _n ]$? I just don't know if $\displaystyle e^t$ is a rational number.

b) Show that $\displaystyle \sqrt {p} \in \mathbb {Q} [ \omega _{4p} ]$.

Proof.

There is a theorem that state: Let $\displaystyle \omega$ be a primitive pth root of unity, p an odd prime. $\displaystyle \sqrt {p} \in \mathbb {Q} [ \omega ]$ if $\displaystyle p \equiv 1 \ (mod \ 4 )$

Now, is $\displaystyle \omega _{4p}$ a primitive pth roof of unity?

Thank you.
• Apr 28th 2008, 07:06 PM
ThePerfectHacker
Quote:

a) Show that $\displaystyle \mathbb {Q} [ \omega _{m} ] \subseteq \mathbb {Q} [ \omega _n ]$ if m|n.

Remember the definition of $\displaystyle \mathbb{Q}(\alpha)$ (here $\displaystyle \alpha$ is algebraic number), it is defined to be the smallest field containing $\displaystyle \mathbb{Q}$ and $\displaystyle \alpha$*.

Thus, $\displaystyle \mathbb{Q}(\omega_m)$ is the smallest field containing $\displaystyle \mathbb{Q}$ and $\displaystyle \omega_m$. Thus, if you can show that $\displaystyle \omega_m \in \mathbb{Q}(\omega_n)$ then by definition $\displaystyle \mathbb{Q}(\omega_m)\subseteq \mathbb{Q}(\omega_n)$. Can you show this last step?

*)Meaning if $\displaystyle F$ is any other field containing $\displaystyle \alpha$ and $\displaystyle \mathbb{Q}$ then $\displaystyle \mathbb{Q}(\alpha)\subseteq F$.
• Apr 28th 2008, 07:30 PM
ThePerfectHacker
Quote:

b) Show that $\displaystyle \sqrt {p} \in \mathbb {Q} [ \omega _{4p} ]$.

I do not know this theorem and I only see one way of deriving it.
If $\displaystyle p$ is an odd prime, and $\displaystyle \omega = e^{2\pi i/p}$ then the quadradic Gauss sum, i.e. $\displaystyle g=\sum_{k=1}^{p-1} (k/p)\omega^k = \left\{ \begin{array}{c}\sqrt{p} \mbox{ if }p\equiv 1(\bmod 4) \\ i\sqrt{p} \mbox{ if }p\equiv 3(\bmod 4) \end{array} \right.$

Note that $\displaystyle (k/p) = \pm 1$ because it is the Legendre symbol. This means if $\displaystyle p\equiv 1(\bmod 4)$ then $\displaystyle \sqrt{p} = g \in \mathbb{Q} (\omega_p)$. Thus, then $\displaystyle \sqrt{p} \in \mathbb{Q}(\omega_{4p})$ by excercise #1.

If, however, $\displaystyle p\equiv 3(\bmod 4)$ then $\displaystyle g = i\sqrt{p} \in \mathbb{Q}( \omega_p ) \subseteq \mathbb{Q} ( \omega_{4p} )$. But $\displaystyle i \in \mathbb{Q}(\omega_{4p})$ because $\displaystyle i = \omega_{4p}^p$. Thus, by closure $\displaystyle g/i = \sqrt{p} \in \mathbb{Q}(\omega_{4p})$.
• Apr 30th 2008, 10:05 AM
Please don't laugh for my understanding in this material is very weak.

for a), I have $\displaystyle \omega _{m} = e^{ \frac {2 \pi i }{m} } = e^{ \frac {2 \pi i (t) }{n} }= \omega _n ^t$, this is in $\displaystyle \mathbb {Q} [ \omega _n]$ since it is cyclic, right?

And there is also one more question in this problem:

c) Let m be square-free, show that $\displaystyle \mathbb {Q} [ \sqrt {m} ] \subseteq \mathbb {Q} [ \omega _{4m} ]$

So should I try to get $\displaystyle \sqrt {m} \ | \ \omega _{4m}$

Thanks.
• Apr 30th 2008, 10:41 AM
ThePerfectHacker
Quote:

Please don't laugh for my understanding in this material is very weak.

for a), I have $\displaystyle \omega _{m} = e^{ \frac {2 \pi i }{m} } = e^{ \frac {2 \pi i (t) }{n} }= \omega _n ^t$, this is in $\displaystyle \mathbb {Q} [ \omega _n]$ since it is cyclic, right?

That works.

Quote:

c) Let m be square-free, show that $\displaystyle \mathbb {Q} [ \sqrt {m} ] \subseteq \mathbb {Q} [ \omega _{4m} ]$

So should I try to get $\displaystyle \sqrt {m} \ | \ \omega _{4m}$
If $\displaystyle m$ is square-free then $\displaystyle m=p_1...p_n$ for distinct primes. By other excerise it means $\displaystyle \sqrt{p_k}\in \mathbb{Q}(\omega_{4p_k})$ for $\displaystyle k=1,...,n$. But $\displaystyle \mathbb{Q}(\omega_{4m})$ because $\displaystyle p_k|m$. Thus, $\displaystyle \sqrt{p_k}\in \mathbb{Q}(\omega_{4m})$ for $\displaystyle k=1,...,n$. Since fields are closed under multiplication it means $\displaystyle \sqrt{p_1}\cdot ... \cdot \sqrt{p_m} \in \mathbb{Q}(\omega_{4m})$ thus $\displaystyle \sqrt{m} = \sqrt{p_1...p_m} \in \mathbb{Q}(\omega_{4m})$. Why does this complete the proof?
• Apr 30th 2008, 01:39 PM
The proof is complete because like in the first exercise, $\displaystyle \mathbb {Q} [ \sqrt {m} ]$ is the smallest field that contains both $\displaystyle \mathbb {Q}$ and $\displaystyle \sqrt {m}$, if $\displaystyle \sqrt {m} \in \mathbb {Q} [ \omega _{4m} ]$, it completes the proof.