Lubin
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 Apr 24 answered Extensions of the quadratic closure of $\mathbb{Q}$ Apr 24 comment Extensions of the quadratic closure of $\mathbb{Q}$ Because now every finite extension of $\Bbb Q$ in $\Bbb Q^q$ is of degree a power of two. Apr 24 comment Extensions of the quadratic closure of $\mathbb{Q}$ Do you see that every element of $\Bbb Q^q$ is of degree a power of two over $\Bbb Q$? Apr 24 answered describe explicitly all the ideals of $R/(f(x))$ Apr 23 comment Find all points of finite order on the elliptic curve $y^2+7xy=x^3+16x$. And 3-division points are the points of inflection. Apr 23 comment Find all points of finite order on the elliptic curve $y^2+7xy=x^3+16x$. Oh, 2-division is easiest: that's where the tangent is vertical. Apr 23 comment Find all points of finite order on the elliptic curve $y^2+7xy=x^3+16x$. Has good reduction at $5$, $7$, and (I think) at $11$, with $8$, $8$, and $16$ points, respectively, but the $2$ division over $\Bbb Q$ is cyclic. I’m willing to wager that it has eight rational torsion points, cyclic. Apr 22 comment Finding side of a triangle, given two sides and angle bisector Oh yes, now I remember seeing this relatively late in life; but Mrs. Stewart didn’t teach us that one. Thanks. Apr 22 comment Finding side of a triangle, given two sides and angle bisector By a tedious computation, I got $\tan\alpha=\sqrt5/2$, and you got $\cos\alpha=2/3$, so I have to agree. But I don’t see the reason for your first ratio. Could you remind me? Apr 22 comment Finding a möbius transformation satisfying the following mappings yes. Now solve! (Answer is not unique.) Apr 22 comment Rational Expression equivalent form I think that only this answer is unmysterious. Plus one. Apr 22 comment Radius of convergence of product of two power series The sum you have written is not the product of two power series. Apr 21 comment Is the absolute Galois Group of $\Bbb Q$ countable? There are no countably infinite Galois groups. Apr 21 comment Determining automorphisms of this extension As @skmehta says, $\omega$ must go to a root of $X^2+X+1$, and your chosen cube root of $2$ (you’ve called it $2^{1/3}$) must go a root of $X^3-2$. There’s your six members of the Galois group. Apr 21 comment The irreducibility of $X^q - 3$ in the field extension $\mathbb{Q}(\zeta_q, 2^{1/q})$ I contest your claim of having misled, @knsam. We never know which route will take us to the end; reaching it, we think the successful route was obvious. It wasn’t. Specifically, I thought the conjecture was true, and was about to spend a lot of time trying to prove it. Apr 20 comment The irreducibility of $X^q - 3$ in the field extension $\mathbb{Q}(\zeta_q, 2^{1/q})$ It seems to me that when the base is not $\Bbb Q$ or some other field with unique factorization, Kummer is much harder to use. We know so little in general about the arithmetic in OP’s field $L$ that I for one wouldn’t know where to start when kummerizing. Apr 20 answered The irreducibility of $X^q - 3$ in the field extension $\mathbb{Q}(\zeta_q, 2^{1/q})$ Apr 20 comment Find the Subextensions of $\Bbb Q(\sqrt2, \sqrt3, \sqrt5)/\Bbb Q$ Since there are seven subgroups of order $2$, there ought to be seven subgroups of index $2$ as well. Apr 20 comment Limit of p-adic numbers I’m not sure I see your argument clear. I see that $a^{(p-1)!}=1+pz$ for some $p$-adic integer $z$, but have you shown that the higher powers get closer to $1$ than that? Apr 20 comment The irreducibility of $X^q - 3$ in the field extension $\mathbb{Q}(\zeta_q, 2^{1/q})$ For using Kummer Theory, you’d need to show that $3$ remains square-free in $L$, or some such thing that looks to me very difficult.