Which element of $\mathbb{Q}C_5$ will be mapped onto $(0,4)\in \mathbb{Q} \oplus \mathbb{Q}(\zeta)$

Question

I know that the rational group algebra of $C_5 = \langle g : g^5=1\rangle$ is $\mathbb{Q}C_5 \cong \mathbb{Q} \oplus \mathbb{Q}(\zeta)$ where $\zeta$ is a root of $x^4+x^3+x^2+x+1$. But I was wondering what is the isomorphism. I took $\phi: \sum a_gg \to (\sum a_g, \sum a_g\zeta)$ which satisfies all the conditions. But upon wondering which element of $\mathbb{Q}C_5$ will be mapped onto $(0,4)\in \mathbb{Q} \oplus \mathbb{Q}(\zeta)$, I could not find an answer. How do I find it?

Answer 1

You probably meant your map to be $$\phi: a_0+a_1g+a_2g^2+a_3g^3+a_4g^4 \mapsto (a_0+a_1+a_2+a_3+a_4, a_0+a_1\zeta+a_2\zeta^2+a_3\zeta^3+a_4\zeta^4)$$

If $a_0+a_1\zeta+a_2\zeta^2+a_3\zeta^3+a_4\zeta^4=4$, use that $\zeta^4=-(\zeta^3+\zeta^2+\zeta+1)$ and compare coefficients by using that $1,\zeta,\zeta^2,\zeta^3$ are linearly independant over $\mathbb Q$ to get $a_0-a_4=4 ,a_1-a_4=0, a_2-a_4=0,a_3-a_4=0 $. So $a_1=a_2=a_3=a_4$.

Finally, use that $a_0+a_1+a_2+a_3+a_4=0$ to get $a_0=-4a_4$. Combine this with $a_0-a_4=4$ to get $a_1=a_2=a_3=a_4=-\frac{4}{5}$ and $a_0=\frac{16}{5}$.