In Ireland and Rosen, the following law for inert rational primes in a quadratic field is stated as: if $p\nmid \delta_K$, where $\delta_K$ is the discriminant of the quadratic field, and $d$ is a quadratic non-residue $\mod p$, then the ideal $(p)$ is prime. They then go on to derive an easier law for $d\equiv 1\pmod{4}$,applying quadratic reciprocity to get $$\left(\frac{\delta_K}{p}\right)=\left(\frac{p}{\delta_K}\right)(-1)^{(p-1)(\delta_K-1)/4}=\left(\frac{p}{\delta_K}\right).$$

While at a quick glance this appears to work, what does it mean to have the Legendre symbol $\left(\frac{p}{\delta_K}\right)$? Here, $\delta_K$ can be negative, so what does it even mean to have a negative number in the bottom of a Legendre symbol? Are they missing something or am I being silly? Any help is appreciated!

  • $\begingroup$ A Classic Introduction to Modern Number Theory by Kenneth Ireland and Michael Rosen? $\endgroup$ Nov 5, 2014 at 5:33
  • 1
    $\begingroup$ Yes, that's the one, on page 191 for reference. $\endgroup$ Nov 5, 2014 at 5:35
  • $\begingroup$ Confusion on Inert Primes in Ireland - Guess that explains the country's troubled history $\ldots$ $\endgroup$
    – Lucian
    Nov 5, 2014 at 17:59
  • $\begingroup$ Is it the Kronecker symbol? $\endgroup$ Nov 5, 2014 at 18:52
  • $\begingroup$ Hm yeah that seems like it would work, it satisfies whatever I/R are talking about... $\endgroup$ Nov 5, 2014 at 21:30

1 Answer 1


Ireland and Rosen really mean the Legendre symbol $(\delta_F/p)$, for the characterization of inert, ramified and split primes for the quadratic number field $F=\mathbb{Q}(\sqrt{d})$. But later, when they argue in the case $d\equiv 1(4)$, they want to invert the symbol using quadratic reciprocity, and that is no longer the Legendre symbol, if $d=\delta_F$ is negative. It should be the Kronecker symbol. On the other hand, I think that Ireland and Rosen never define the Kronecker symbol in the book, but only the Jacobi symbol, see Proposition $5.5.1$ and $5.5.2$.


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