I am currently going through a course in Class Field theory and have read that the Artin symbol generalizes the Legendre and Hilbert symbols. I was wondering what field extensions to consider to see this.

I define the Artin and Hilbert maps below for the convenience of users who might not have seen it.

For a cyclic extension $L$ of a number field $K$, define the Artin symbol for an unramified prime $\mathfrak p$, $\displaystyle\left(\frac{L/K}{\mathfrak p}\right)$ to be that element of the Galois group $G$ of $L/K$ that raises an element of $L$ to its $\text{Norm}(\mathfrak p)$-th power. Extend this definition by multiplicativity (in $G$) to all ideals not containing any ramified prime.

For elements $a$ and $b$ in a local field $K$, define the Hilbert symbol $(a,b)$ to be 1 if the equation $z^2 = a x^2 + b y^2$ has a solution $(x,y,z)\in K^3\backslash(0,0,0)$ and -1 otherwise.

  • $\begingroup$ So $\mathfrak{p}\in \text{Spec}(\mathcal{O}_L)$? $\endgroup$ – 54321user Jun 2 '17 at 19:11

To get the Hilbert symbol from the Artin symbol, I believe you either take the extension $K(\sqrt{a})$ and then look at the Artin symbol at the ideal $(b)$, or take the extension $K(\sqrt{b})$ and then look at the Artin symbol at the ideal $(a)$. They should give the same result (assuming the Artin symbols both exist, i.e., the ideals are unramified in the respective extension).

Of course, a Hilbert symbol value of 1 corresponds to the identity, and -1 to the non-identity, element of the Galois group of the extension field. (If either $a$ or $b$ is already a square in $K$ then there is no non-identity element in the respective Galois group, but in that case the Hilbert symbol is always 1, as expected.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.