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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.
Aug
19
comment What exactly is a convex set in two dimensional space?
This is good, because OP clearly was led astray by the use of the word “convex” also for functions, which the convexity of a set has little to do with.
Aug
17
comment Is frequency a scalar quantity?
What are the units for describing frequency?
Aug
15
comment If $E/F$ is finite divisible, then it is separable.
Then “divisibility” is not a property of extensions, but of the individual fields. Did you mean for both $E$ and $F$ to be “divisible” or just $E$?
Aug
14
answered height of formal group of an elliptic curves
Aug
13
comment Does $\cos (\pi/5)$ belong to $\mathbb{Q} (\sin(\pi/5))$?
Easiest to write twice the cosine as $\zeta+\zeta^{-1}$, where $\zeta$ is a primitive tenth root of unity, satisfying a fourth-degree polynomial.
Aug
13
answered Does $\cos (\pi/5)$ belong to $\mathbb{Q} (\sin(\pi/5))$?
Aug
13
comment Finding Galois group of $K=\Bbb{Q}(\omega,\sqrt2)$, showing that $K=\Bbb{Q}(\omega\sqrt2)$, and finding $\operatorname{min}(\omega\sqrt2,\Bbb{Q})$
To do that is much easier once you realize that $\Bbb Q(\omega)=\Bbb Q(\sqrt{-3}\,)$.
Aug
12
comment Adjoining all roots of unity to an arbitrary field $F$, is an abelian extension?
This is, I think, called the Theorem on Natural Irrationalities. It says that if $N\supset K$ is Galois and $F\supset K$ is an extension of any kind, then $NF\supset K$ is Galois, and the Galois group is isomorphic to the group of $N\supset(N\cap K)$. (Hope I’ve quoted it correctly.) It’s rather easy to prove (if true).
Aug
11
comment When is a Morphism between Curves a Galois Extension of Function Fields
This is very satisfying.
Aug
11
comment Finding Galois group of $K=\Bbb{Q}(\omega,\sqrt2)$, showing that $K=\Bbb{Q}(\omega\sqrt2)$, and finding $\operatorname{min}(\omega\sqrt2,\Bbb{Q})$
No, I told you what the $a$ and the $b$ had to be.
Aug
11
answered Show that $\mathbb{Q}(\sqrt{2},\sqrt{3},\dots,\sqrt{p},\dots)$ is an algebraic extension of $\mathbb{Q}$, for $p$ prime.