# Confusion on Inert Primes in Ireland and Rosen

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!

• A Classic Introduction to Modern Number Theory by Kenneth Ireland and Michael Rosen? – Robert Soupe Nov 5 '14 at 5:33
• Yes, that's the one, on page 191 for reference. – algebraicnumbertheorist Nov 5 '14 at 5:35
• Confusion on Inert Primes in Ireland - Guess that explains the country's troubled history $\ldots$ – Lucian Nov 5 '14 at 17:59
• Is it the Kronecker symbol? – Ricardo Buring Nov 5 '14 at 18:52
• Hm yeah that seems like it would work, it satisfies whatever I/R are talking about... – algebraicnumbertheorist Nov 5 '14 at 21:30

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 $5.5.2$.