since I do not know algebra very well, I have some troubles with the following concepts:
I want to ask something about quadratic fields and quadratic forms. Actually my main question is the connection between the primitive Dirichlet characters and the theory of quadratic forms.
I could not understand the relation between the discriminant of quadratic fields and quadratic forms, actually I do not understand basically what the discriminant of the quadratic fields is.
After all, It has been stated that in theory of quadratic fields $(\frac{d}{p})$ determines the way in which a prime $p$ factorizes in the quadratic filed of discriminant $d$.
Also, we know the all the primitive Dirichlet characters are $(\frac{-4}{n}), (\frac{8}{n}), (\frac{-8}{n})$ and $(\frac{p(-1)^\frac{p-1}{2}}{n}) $ when $p\neq 2$.
Assuming $\chi$ is a real primitive Dirichlet character, what the value of $\chi(-1)$ gives us, I could not understand, to connect the primitive real character concept with the statement above about the factorization of prime numbers.
$\endgroup$1 Answer
$\begingroup$I do not really know what you want; this is a start.
Take a discriminant of binary quadratic forms, that is $\Delta \equiv 0,1 \pmod 4,$ also $\Delta,$ if non-negative, is not a square.
Take a prime $p$ such that $\Delta \neq 0 \pmod p$ and Legendre symbol $(\Delta| p ) = 1.$ It is a consequence of reduction methods that there is a primitive binary form of discriminant $\Delta$ that represents $p.$ That is, there are integers $A,B,C$ with $\gcd(A,B,C) = 1,$ and $B^2 - 4 AC = \Delta,$ with integers $u,v$ and $$ A u^2 + B uv + C v^2 = p. $$
For example, if $(-4|p) = 1,$ we can solve $u^2 + v^2 = p.$
For $(-23|p) = 1,$ we can solve one of the three $x^2 + xy + 6 y^2 = p,$ or $2x^2 \pm xy + 3 y^2 = p,$ not both. It is possible to say which: if $$ z^3 - z + 1 \equiv 0 \pmod p $$ has a root ( since $(-23|p) = 1$ it will actually have three roots), then we have $x^2 + xy + 6 y^2 = p.$
Suggest the book by Cox as well as
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