# Set of locations where the Hilbert symbol is not equal to $1$

Let $V$ be the set of prime together with the symbol $\infty$. For a prime $v=p$, denote the $p$-adic numbers by $\mathbb{Q}_p$ and the real numbers by $\mathbb{Q}_\infty$. For $v\in V$ the Hilbert symbol is defined for $a,b\in\mathbb{Q}^*_v$ as

\begin{align*} (a,b)_v=\begin{cases}+1,&\text{ if }ax^2+by^2=z^2\text{ has a non-zero solution }(x,y,z)\in \mathbb{Q}_v^3;\\-1,&\text{ else.}\end{cases} \end{align*}

Furthermore for $v\in V, a,b\in\mathbb{Q}^*$ we denote by $(a,b)_v$ the Hilbert symbol of $(\bar a,\bar b)_v$ where $\bar a,\bar b$ are the images of $a,b$ in $\mathbb{Q}_v$.

Now a theorem by Hilbert says that $(a,b)_v=1$ for almost all $v\in V$ (and that furthermore $\prod_{v\in V}(a,b)_v=1$, but I'm not interested in this at the moment). The theorem can be found in "A course in Arithmetic" by Jean-Pierre Serre for example.

It basically says that there is a finite set $E\subseteq V$ such that \begin{align*} (a,b)_v=\begin{cases}+1,&\text{ if }v\notin E\\-1,&\text{ if }v\in E\end{cases} \end{align*}

My question is if this set $E$ has a common name in the literature. Something like $E_{a,b}$ would make sense to me (since it depends on $a$ and $b$). If there is no widely used name, what are your suggestions?

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For purely aesthetic reasons, I've grown quite fond of putting $+1$, instead of a naked 1, in the first case of the definition of piecewise-defined $\pm 1$-functions like you have above. Feel free to overrule. – Cam McLeman Jan 5 '13 at 15:06
I think you mean $(a,b)_v=+1,\quad \text{ if }v\notin E$. – Michalis Jan 5 '13 at 15:29
Are you asking for the common name or the common notation? Writing "Something like $E_{a,b}$ would make sense" is referring to notation for the set rather than a word for it. – KCd Jan 5 '13 at 19:15
@Michalis: yes, you are totally right – born Jan 6 '13 at 3:19
@CamMcLeman thanks for editing! – born Jan 6 '13 at 3:19

Note that $(a,b)_v=-1$ if and only if the quaternion algebra $\langle a,b\rangle_\mathbb{Q}$ (among hundreds of other notations) is ramified at $v$, i.e., if $v$ divides its discriminant. So if you were in a setting where, say, algebraic geometry language was convenient, you might call this set the "Hilbert support of $\langle a,b\rangle$", or maybe reference the Hilbert radical if it were more convenient to refer to the product of such primes (which is occasionally useful).
I agree that the term "ramified" will sound unusual if someone never heard of ramification in any context, but it's not exactly like the usage of "ramified" in algebraic number theory would make intuitive sense either to someone who doesn't know about its similarity to a situation in complex analysis. As von Neumann might say, you just get used to it. Hopefully later you'd learn where it comes from. If we don't want to create new terms for this set but use what people would actually call it, then "the set of ramified places" is the common name for $E$ in the literature. – KCd Jan 5 '13 at 19:10
@born: The standard name I'd suggest for the product of the places where the symbol is $-1$ is the discriminant. See the last section of en.wikipedia.org/wiki/Quaternion_algebra. – KCd Jan 6 '13 at 20:43