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Aug
27
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 $k(u)$
Aug
24
comment Find a $u$ so that $k(u)$ is the fixed field of $φ$, determine the minimal polynomial over $k(u)$
Not $\Bbb Q(u)$, but $k(u)$.
Aug
24
answered Find the cube roots of $ -8 i $ and plot them on a plane.
Aug
24
comment Elliptic curve is self dual.
That should have something to do with the $e_{p^n}$ pairing on the points of order $p^n$, I guess?
Aug
24
comment Degree of $\mathbb{Q}_p$-irreducible representations of a cyclic group of order $p^n$ for a prime p
Doesn’t it have something to do with the degrees of the irreducible factors of $X^{p^n}-1\,$?
Aug
23
comment What's so special about quadratic extensions?
I was thinking of algebraic number theory, where it’s very easy to determine which numbers in the extension field are algebraic integers; and beyond that very easy to find the units in the ring of algebraic integers, and to see how the primes of $\mathbb Z$ split in the ring of integers upstairs. More!
Aug
23
answered Two definitions of ramification groups, why are they equivalent?
Aug
23
comment What's so special about quadratic extensions?
NONE of those statements holds for extensions of degree $>2$. The upshot is that quadratic extensions are easy to run examples on, the other extensions not so much.
Aug
21
comment Complete field and field extension.
I don’t understand. If $\pi\colon K_u\to L$ is a morphism of fields, doesn’t $\pi$ already embed $K_u$ into $L$?
Aug
20
comment Proving that the set of separable elements over a field is a field itself.
We’re on the same page, @KCd. It’s nowhere as easy or direct as showing that the sum, product, etc. of algebraic elements is algebraic. Ayush, if you’d like me to sketch out my strategy, I can do it in an answer.
Aug
19
comment Proving that the set of separable elements over a field is a field itself.
I would disagree with @KCd on the hardness of the result. Seems to me that it can be proved by the correct strategy of going back and forth between separability of an element over the field $k$ and separability of an extension $K\supset k$. Large number of steps, no one of them particularly hard. (Now I do my homework and see if I can write out all the details.)
Aug
19
answered Understanding $Gal(\bar k /k)$
Aug
19
comment Does linearly independent imply all elements are orthogonal?
Indeed, any two vectors in $\Bbb R^2$ that are not in the same (or opposite) direction, no matter how small the angle between them.