happy new year

I have this statement: "By quadratic reciprocity there are the integers $a$ and $b$ such that $(a,b)=1$, $(a-1,b)=2$, and all prime $p$ with $p\equiv a$ (mod $b$) splits in $K$ (where $K$ is a real quadratic field)".

I have tried with many properties of quadratic reciprocity but couldn't even get to the first conclusion.

Thank you very much in advance, for any idea or advice for approach the problem

  • $\begingroup$ I'm confused. For the first conclusion, just take $b=2$ and $a$ to be odd. (Also, I think there needs to be some more quantifiers here. Are you talking about a specific quadratic field $K$? Or some quadratic field?) $\endgroup$ Commented Jan 5, 2012 at 14:18
  • $\begingroup$ Sorry, the first condition is trivial but the i need fulfilled the three conditions at the same time and for all real quadratic field Thanks $\endgroup$
    – Farnsworth
    Commented Jan 5, 2012 at 14:40

2 Answers 2


The point is that (because of quadratic reciprocity), if $D$ is the discriminant of $K$, then splitting in $K$ is determined by the congruence class of $p$ mod $D$ (more precisely, by the Jacobi symbol mod $D$). If $D$ is even, take $b = D$, if $D$ is odd, take $b = 2D$.

The problem is now to find a residue class in $(\mathbb Z/b)^{\times}$ such that the Jacobi symbol of $a$ mod $D$ is trivial (this is easy; half of the elements in $(\mathbb Z/b)^{\times}$ has this property), and such that $(a-1,b) = 2$.

For simplicitly, suppose that $D = q$, an odd primes. Then you can take $a$ to be any residue class with trivial Legendre symbol such that $a-1$ is not zero mod $q$. Since $q \equiv 1 \bmod 4$ (it is a fundamental discriminant), it is at least $5$, so there is a quadratic residue mod $q$ beside $1$. Let $a$ be any odd representative of this quadratic residue. Then $(a,2q) = 1,$ $(a-1,2q) = 2$, and if $p \equiv a \bmod b$, then $p$ splits in $\mathbb Q(\sqrt{q}).$

Presumably the general case can be handled in a similar way, using the Chinese Remainder theorem (applied to the prime factorization of $D$).


Edited to address some bizarrely horrible errors in the first version.

Here's a simple case from which it should not be too hard to generalize. Suppose that $K$ is a quadratic field of prime discriminant $q$. Since $q\equiv 1\pmod{4}$, note that a prime $p$ splits in $K$ if and only if $\left(\frac{p}{q}\right)=1$.

Let $b=2q$ and choose an integer $a\not\equiv 1\pmod{q}$ which is an odd quadratic residue mod $q$. Such a thing exists since there are $\frac{q-1}{2}\geq 2$ (since $q\geq 5$) residues mod $q$, and if $a'$ is any not-conguent-to-1 residue mod $q$, then one of $a'=a$ and $a'=q+a$ gives you an odd residue. For example, when $q=5$, take $a'=4$ and then $a=9$. (Actually, $p=5$ is the only example where you have to add $q$: for all other $q$ there is an odd quadratic residue $a$ in the range $2\leq a\leq p-1$).

Now $a$ is odd and not divisible by $q$, so $(a,b)=(a,2q)=1$, and since $a\not\equiv 1\pmod{q}$, we also have $(a-1,b)=2$. Finally, if $p\equiv a\pmod{b}$, then $p\equiv a\pmod{q}$, so $\left(\frac{p}{q}\right)=\left(\frac{1}{q}\right)=1$, and $p$ splits in $K$.

Just some minor commentary on where this came from: Your $\gcd$ conditions force $b$ to be even, and for a congruence-mod-b condition to determine splitting in $K$, your $b$ needs to be a multiple of the discriminant (in this case, $q$), and preferrably as small a multiple as possible to prevent extra congruence classes from slipping in. The value of $b=2q$ satisfies all of these requirements, and since clearly one must choose $a$ to be an odd quadratic residue mod $q$, you're left with essentially the above construction.

  • $\begingroup$ I am quite confused here: why is $(a-1,b)=2$ instead of $2q$? Maybe to let $a=2q+3, b=2q$ suffices? $\endgroup$
    – awllower
    Commented Feb 7, 2012 at 12:55
  • $\begingroup$ @awllower Yeah, thanks, this is complete nonsense -- I was trying to get things to work out for the residue 1, but I think Matt's answer correctly implies that you have to pick a different residue. I don't think your fix works since then $p\equiv 3$ mod $q$, which is not necessarily a square. I'll either fix or delete this soon. $\endgroup$ Commented Feb 7, 2012 at 13:59
  • $\begingroup$ Sorry, I forgot the other conditions... $\endgroup$
    – awllower
    Commented Feb 7, 2012 at 14:23
  • $\begingroup$ So any other such odd residue class than the trivial one will do? $\endgroup$
    – awllower
    Commented Feb 7, 2012 at 14:33
  • $\begingroup$ Right. The answer is updated to address this. It occurs to me that it's now rather identical to Matt E's answer. Oh well. $\endgroup$ Commented Feb 7, 2012 at 14:38

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