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 Feb21 accepted Set of locations where the Hilbert symbol is not equal to $1$ Feb15 revised counterexample for: integral forms that are equivalent as rational forms are also equivalent as integral forms added 10 characters in body Feb15 asked counterexample for: integral forms that are equivalent as rational forms are also equivalent as integral forms Jan7 comment Linear Algebra- independence and dependence When you speak about a vetorspace you always have to say what your "base field" is. That is the set where the scalars come from. In high school we are normally used to the vectorspace $\mathbb{R}^n$ which we normally consider "over the field $\mathbb{R}$", but in general the base field can be everything (for example $\mathbb{F}_2$ or $\mathbb{Q}$) and $V$ does not need to be a power of the field $\mathbb{F}^n$. Jan7 comment Linear Algebra- independence and dependence Is $V$ a vectorspace over some arbitrary field? Then you cannot divide by $2$ (or at least you have to handle this case separately). Jan6 comment Set of locations where the Hilbert symbol is not equal to $1$ Thank you for all your suggestions! I'll think about the name "set of ramified places". But since $p$-adic number theory is not my main focus, I don't know if this name is too fancy. @KCd and CamMcLeman: Am i getting it right that the Hilbert radical is the product of all the $v\in E_{a,b}$? May I furthermore ask if there is a common notation for the "Hilbert support of $_{\mathbb{Q}}$"? Or is something like $E_{a,b}$ convenient? Jan6 comment Set of locations where the Hilbert symbol is not equal to $1$ @KCd: My question was not very precise, I wanted to ask if there is a common name and a common notation :-) Jan6 comment Set of locations where the Hilbert symbol is not equal to $1$ @CamMcLeman thanks for editing! Jan6 comment Set of locations where the Hilbert symbol is not equal to $1$ @Michalis: yes, you are totally right Jan5 asked Set of locations where the Hilbert symbol is not equal to $1$ Jan5 answered Definite integral involving $\sqrt{\log}$ Dec31 asked Hilbert symbol over a ring Dec4 accepted $p$-adic unit for $p\neq 2$ is a square in $\mathbb{Z}_p$ if and only if its first digit is quadratic residue modulo $p$. Dec3 comment $p$-adic unit for $p\neq 2$ is a square in $\mathbb{Z}_p$ if and only if its first digit is quadratic residue modulo $p$. ah, now everything is clear \o/ thank you very much! Dec3 comment $p$-adic unit for $p\neq 2$ is a square in $\mathbb{Z}_p$ if and only if its first digit is quadratic residue modulo $p$. maybe I'm wrong but $(u-x_0^2)/p=x_1^2p+2x_1x_0$ (so the $x_1$ has a square) and therefore it is not a linear equation and does not need to have a solution. And i still don't see that all this has to do with $u$ being a unit or not :-/ Dec3 comment $p$-adic unit for $p\neq 2$ is a square in $\mathbb{Z}_p$ if and only if its first digit is quadratic residue modulo $p$. why exactly is $x_1$ determined uniquely by $(x_1p+x_0)^2\equiv u \pmod{p^2}$? I see, that $p, x_0$ and $u$ are given but I don't see why such an $x_1$ must exist. Dec3 comment $p$-adic unit for $p\neq 2$ is a square in $\mathbb{Z}_p$ if and only if its first digit is quadratic residue modulo $p$. I hoped, that I could avoid this sort of inductive argumentation. I'll lock over it and accept your answer. Thanks a lot! Dec3 comment $p$-adic unit for $p\neq 2$ is a square in $\mathbb{Z}_p$ if and only if its first digit is quadratic residue modulo $p$. where do I use that $u$ is a unit, when \$0