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In Ireland and Rosen's Number theory book, ch.8, p94, it explains how to use Jacobi sums to find the number of solutions to the equation $x^{3}+y^{3}=1$.

From the book: $N(x^{3}+y^{3}=1)=p-\chi(-1)-\chi^{2}(-1)+2ReJ(\chi,\chi)=p-2+2ReJ(\chi,\chi)$.

My question is, how do we get the part $2ReJ(\chi,\chi)$? I would like a detailed explanation, although I think it must be easy since it was omitted. thanks

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up vote 3 down vote accepted

The formula you ask about is on top of page 95 of my copy of the book. It's derived from a formula on the bottom of page 94 of my copy by using the observation that $\chi^2$ is the complex conjugate of $\chi$, from which it follows that $J(\chi,\chi)+J(\chi^2,\chi^2)$ is twice the real part of $J(\chi,\chi)$.

You start with the formula in the middle of page 93, which expresses $N(x^3+y^3=1)$ as a sum of 9 Jacobi sums. You use Theorem 1 on page 93 to evaluate 7 of those 9 sums, and that gets you to the formula with $J(\chi,\chi)+J(\chi^2,\chi^2)$.

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It is not clear to me why $J(\chi,\chi)+J(\chi^{2},\chi^{2})=2ReJ(\chi,\chi)$. Can you explain this equality? Thanks. – Edison Mar 5 '12 at 5:26
As I wrote, $\chi^2$ is the conjugate of $\chi$. Then from the definition of the Jacobi sum, $J(\chi^2,\chi^2)$ is the conjugate of $J(\chi,\chi)$. Then when you add a number to its complex conjugate, you get twice the real part of the number. – Gerry Myerson Mar 5 '12 at 5:54
Thank you for your reply. – Edison Mar 5 '12 at 6:04

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