Question from a Qualifying Exam:
- Show that $\sqrt 5$ can be expressed as a polynomial in $e^{(\frac{2\pi i}{5})}$ over $\Bbb Z$
- If in a field the equation $x^2-5$ has no solution then $x^5-1$ also has no non-trivial solution.
I am unable to show how to find the polynomial
Please givesome hints
To your first question: Here is a high-faluting answer. If $p$ is any odd prime number (i.e., any prime number $>2$), then the Gauss sum is defined to be the number \begin{equation} g\left( 1;p\right) :=\sum_{n=0}^{p-1}e^{2\pi in^{2}/p}. \end{equation} Gauss proved that \begin{equation} g\left( 1;p\right) = \begin{cases} \sqrt{p}, & \text{if }p\equiv1\operatorname{mod}4;\\ i\sqrt{p}, & \text{if }p\equiv3\operatorname{mod}4 \end{cases} \end{equation} (and this has been re-proven many times since Gauss; see a post by David Speyer on SBSeminar for my favorite proof, although he denotes $g\left( 1;p\right) $ by $g\left( \zeta\right) $ and defines it somewhat differently).
Applying this to $p=5$, we obtain $g\left( 1;5\right) =\sqrt{5}$ (since $5\equiv1\operatorname{mod}4$). Hence, \begin{align*} \sqrt{5} & =g\left( 1;5\right) =\sum_{n=0}^{4}e^{2\pi in^{2}/5}=e^{2\pi i\cdot0^{2}/5}+e^{2\pi i\cdot1^{2}/5}+e^{2\pi i\cdot2^{2}/5}+e^{2\pi i\cdot3^{2}/5}+e^{2\pi i\cdot4^{2}/5}\\ & =z^{0^{2}}+z^{1^{2}}+z^{2^{2}}+z^{3^{2}}+z^{4^{2}},\qquad\text{where }z=e^{2\pi i/5}. \end{align*} This is, of course, a polynomial in $e^{2\pi i/5}$ over $\mathbb{Z}$. Hence, your first question is answered.
To your second question: Let $K$ be a field. We shall show that if $x^{2}-5$ has no solution in $K$, then $x^{5}-1$ has no non-trivial solution in $K$.
Indeed, let us prove the contrapositive: Let us prove that if $x^{5}-1$ has a non-trivial solution in $K$, then $x^{2}-5$ has a solution in $K$.
So we assume that $x^{5}-1$ has a non-trivial solution in $K$. Fix such a solution, and denote it by $z$. Thus, $z^{5}-1=0$ but $z\neq1$.
Inspired by the above answer to the first question, we set $w=z^{0^{2} }+z^{1^{2}}+z^{2^{2}}+z^{3^{2}}+z^{4^{2}}$. We shall now prove that $w^{2}-5=0$.
Indeed, $z-1\neq0$ (since $z\neq1$). Hence, we can cancel $z-1$ from the equality $\left( z-1\right) \left( z^{4}+z^{3}+z^{2}+z+1\right) =z^{5}-1=0$. We thus obtain $z^{4}+z^{3}+z^{2}+z+1=0$, so that $z^{4} =-z^{3}-z^{2}-z-1$. Also, from $z^{5}-1=0$, we obtain $z^{5}=1$, thus $z^{8}=z^{3}$ and $z^{9}=z^{4}$ and $z^{16}=z^{11}=z^{6}=z$. Hence, \begin{align*} w & =\underbrace{z^{0^{2}}}_{=z^{0}=1}+\underbrace{z^{1^{2}}}_{=z^{1} =z}+\underbrace{z^{2^{2}}}_{=z^{4}}+\underbrace{z^{3^{2}}}_{=z^{9}=z^{4} }+\underbrace{z^{4^{2}}}_{=z^{16}=z}\\ & =1+z+z^{4}+z^{4}+z=1+2z+2z^{4}. \end{align*} Squaring this equality, we find \begin{align*} w^{2} & =\left( 1+2z+2z^{4}\right) ^{2}=1+4z+4z^{2}+4z^{4} +8\underbrace{z^{5}}_{=1}+4\underbrace{z^{8}}_{=z^{3}}\\ & =1+4z+4z^{2}+4z^{4}+8+4z^{3}=5+4\underbrace{\left( z^{4}+z^{3} +z^{2}+z+1\right) }_{=0}=5. \end{align*} In other words, $w^{2}-5=0$. Hence, $x^{2}-5$ has a solution in $K$ (namely, $w$). This answers the second question.