# How does the Artin symbol generalize Legendre and Hilbert symbols?

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.

• So $\mathfrak{p}\in \text{Spec}(\mathcal{O}_L)$? – 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.)