@MattN. That would be all of $\Bbb R$. Perhaps you want something like, letting $\{q_i\}$ enumerate the rationals, the open dense set $\displaystyle \bigcup_{i\ge1}B(q_i,2^{-i})$.
math.stackexchange.com/questions/257271/… is a question of mine, which I will soon edit (because it's mostly answered). But I have another question which uses the construction in the first paragraph there
@Mechanicalsnail Oh, didn't even think about that :D Yeah, I could do this.
well, on to my "real" question ;) later, the author considers the k-algebra $k[y_1,y_2,y_3]$ as homogeneous coordinate ring of the projective plane.
and then takes the points $(a_{1i}:a_{2i}:a_{3i})$; he says they are "independent generic points over the prime field K"
Do you maybe know what this notion means? I haven't really come across generic points, but these should be points whose closures are the whole projective plane. What really confuses me is the "over the prime field K" part
@anon yes, I know that :) sorry for the bad formulation there
the $a_{ij}$ are algebraically independent elements over K
"independent generic points over the prime field" sounds to me like there is some "change of base field" going on
maybe the points viewed as points of a certain "variety" over K this time; although I guess the independence over K just comes from algebraic indepence over K. But I have not found a rigorous definition of this notion
maybe it's just all because of the age of the paper :)
but I didn't want to seemingly stop all the rest of you from chatting, feel free to ignore me and go on ;D
here are some algorithms. they are meant for large numbers. I am not familiar with any of them, or which methods are meant for accessibility for smaller examples
Let $m$ be the smallest positive integer such that $g^m\equiv1$ mod $a$. (This is $\mathrm{ord}_a(g)$ by definition.) Then every solution $n$ to $g^n\equiv b$ mod $a$ differs by an integer multiple of $m$.
This is a good exercise if you are becoming familiar with modular arithmetic.
Also, in general, I don't think it is easy to compute the order of an element mod a. For example, we still have a very incomplete picture of what primitive roots look like.
Your best bet will be just taking powers until you get g^m=1, probably.
I believe there are some reasonable algorithms for solving the discrete logarithm problem $\pmod{a}$ so long as all the prime factors of $a-1$ are small
@Ethan Because then the absolute difference between them would be a positive number $k$ for which $g^k\equiv1$, and $k$ would be smaller than the smallest positive number $m$ such that $g^m\equiv1$, a contradiction.
now could u give me a fast proof of why, there can be no solutions between [n,n+e], where e is the order of g modulo a, in the equations $g^n\equiv b$ mod a, and $g^{n+e}\equiv b$ mod a, where $g^e\equiv 1$ mod a, 'e' is the primitive root for g modulo a
Let $m$ be the smallest positive integer such that $g^m\equiv 1$. Let $n$ be a number such that $g^n\equiv1$. Write $n=rm+s$ via the Euclidean algorithm, where $0\le s<m$. Thus $1\equiv g^n\equiv g^{rm+s}\equiv (g^m)^rg^s\equiv g^s$ so $g^s\equiv 1$. If $s>0$ then $s$ is smaller than $m$ yet $m$ is the smallest positive integer such that $g^k\equiv1$; you cannot go smaller than smallest, so we have a contradiction, which proves that $s=0$.
Now suppose $g^{n_1}\equiv g^{n_2}$. Then $g^{n_1-n_2}\equiv1$ and by the previous lemma $n_1-n_2=rm$ for some $r$ is a multiple of $m$, the order of $g$.
