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3h
comment Very basic calculus question - do you think there's a typo?
Textbooks and homework sets are written by human beings, all of us all too fallible.
3h
revised Roots of $\Phi_{31}(x)$ as roots of unity.
corrected an embarrassing typo
3h
comment Show that $K=k(u)$ iff $u=\frac{ax+b}{cx+d}$ where $a,b,c,d \in k$ with $ad-bc \neq 0$
I think you really have it already. Certainly, if $u=(ax+b)/(cx+d)$, then $k(u)\subset k(x)$, right? Now use what you have already shown.
3h
answered Fixed Field of $\sigma, \tau$
1d
answered Galois invariants of the Tate module of an elliptic curve over a number field
1d
comment Galois invariants of the Tate module of an elliptic curve over a number field
Let me give an unsatisfactory answer below, which would easily be annihilated by a short answer from an expert, I’m sure.
1d
comment Ramification in $\mathbb Q(i,\sqrt[4]\pi)/\mathbb Q(i)$
Yes, I tried going from the quadratic extensions to the quartic ones, and did not succeed. Others might, however; and I’m not going to be so rash as to claim that the method I showed you is the best or shortest!
1d
comment Galois invariants of the Tate module of an elliptic curve over a number field
Good reduction at $\ell$?
1d
comment Prove that if $f$ and $g$ are analytic at $w$, then so is $fg$
Exactly, since analyticity is just differentiability in the complex sense.
1d
comment Equilateral triangle
Have you drawn a picture?
1d
comment Ramification in $\mathbb Q(i,\sqrt[4]\pi)/\mathbb Q(i)$
Just an addendum: over $\Bbb Q(i)$, you get the fifth roots of unity by adjoining a fourth root of $-15-20i$, so (modulo $8\mathfrak m$) this is what’s needed for a fourth-root extension in which $(1+i)$ remains prime.
2d
comment Notation: $\varphi$ and $\phi$
In total agreement, especially on your last paragraph. I think I’m a coauthor on a paper that uses $\pi$ and $\varpi$, but few recognize the latter as a pi.
2d
answered Ramification in $\mathbb Q(i,\sqrt[4]\pi)/\mathbb Q(i)$
Aug
26
comment Ramification in $\mathbb Q(i,\sqrt[4]\pi)/\mathbb Q(i)$
I’m thinking about this, and am wondering whether you’ve solved the much easier problem of the ramification above $(1+i)$ of the field gotten by adjoining just the square root of $\pi$.
Aug
26
comment Roots of $\Phi_{31}(x)$ as roots of unity.
More precisely, not $\Bbb F_{31}$ but $\Bbb F_{32}$, as you originally specified.
Aug
25
comment Roots of $\Phi_{31}(x)$ as roots of unity.
It’s actually very easy to write down the six irreducible quintics, when you remember that to be reducible, a quintic must be divisible by a linear or a quadratic, and this means that a reducible must either have an $\Bbb F_2$-root, or be divisible by the one-and-only irreducible quadratic $X^2+X+1$.
Aug
25
answered Roots of $\Phi_{31}(x)$ as roots of unity.
Aug
25
comment Is order of poles of functions determined by Laurent series?
Just think of the series for the sine function, and divide by $z^5$. The answer leaps out at you.
Aug
24
comment Find a $u$ so that $k(u)$ is the fixed field of $φ$, determine the minimal polynomial over $k(u)$
I pointed out a root to you in my answer; you should see at a glance what the others are. Since $u$ is in the fixed field, you certainly have $[k(x)\colon k(u)]\ge p$; but you have a polynomial with $x$ for a root that’s of degree $\le p$ (it’s a factor of the polynomial I showed you). So the degree of $x$ over $k(u)$ is both at most and at least $p$. So equal, and the polynomial is irreducible.
Aug
24
answered Find a $u$ so that $k(u)$ is the fixed field of $φ$, determine the minimal polynomial over $k(u)$