Reputation
25,684
Next tag badge:
99/100 score
32/20 answers
Badges
2 25 56
Impact
~335k people reached

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.
Apr
20
comment The irreducibility of $ X^q - 3 $ in the field extension $ \mathbb{Q}(\zeta_q, 2^{1/q}) $
Certainly if $q$ is relatively prime to $3$. Then, the field $\Bbb Q(\zeta_q,2^{1/q})$ is unramified above $3$, but you have $q$ of ramification at $3$ when you adjoin $3^{1/q}$. To get insight, you might try $q=9$ and $q=27$.
Apr
19
comment Proving that $-1$ has no square root in $\mathbb Z_3$ (3-adic integers)
Not merely fine, but economical.
Apr
19
comment An operation on a whole is equal to an operation on each part?
Excellent response. I like to say that IN MATHEMATICS, NOTHING IS TRUE. Except when we have a proof that it’s true. OP’s teacher should have explained why $\sqrt a/\sqrt b=\sqrt{a/b}$. If(s)he didn’t, (s)he abdicated all responsibility for teaching.
Apr
19
comment How to find inverse of a relation if the inverse isn't a function?
As a function defined on $\Bbb Z^{>0}$, this looks perfectly one-to-one and onto to me. There should not be any problem at all. Have you written out the values for inputs from (say) $1$ through $12$?
Apr
19
comment How to find inverse of a relation if the inverse isn't a function?
There’s always an inverse relation, as subset of, in this case, $\Bbb Z\times\Bbb Z^+$.
Apr
19
answered Splitting field of an irreducible polynomial of degree four
Apr
18
comment $4$th root of unity: 5-adic
The next step is to calculate $7^5\pmod{125}$. These computations do get tedious, more so as you go on, unless you have access to some kind of symbolic calculation package.
Apr
18
comment Stronger form of Hensel's lemma?
Just as a comment, I hope that you recognize that this is the Newton-Raphson method applied to the problem of finding a root in $\Bbb Z_p$ of $f$. And as to your Idea, I find it’s much easier, once you’re used to it, to use the additive valuation $v_p$ than the absolute value.
Apr
17
comment Proving $f(x)$ is not a square in $k[x]$
The polynomial rings are isomorphic, and the corresponding fraction fields likewise.
Apr
17
answered If $a \in \mathbb{Z}_5$ and $a \equiv \pm1 \text{ }(\text{mod }5)$, does there exist $x \in \mathbb{Z}_5$ where $x^2 = a$?
Apr
17
comment Proving $f(x)$ is not a square in $k[x]$
I’m not sure I understand. Aren’t the fields $k(X)$ and $k(x)$ isomorphic?
Apr
17
comment How to show that $\{x,y,,z\}$ are linearly independent $\Rightarrow$ $\{x+y,x+z,y+z\}$ is independent does not hold for arbitrary field $F$?
Alternatively, the third vector $y+z$ is the sum of the other two, if the field is $\Bbb F_2$.
Apr
16
comment If $L=K(a_1,a_2,..,a_n)$ can we find an irreducible polynomial in $K[x]$ s.t $p(a_i)=0$?
You mean one irreducible polynomial good for all the $a_i$’s? If that’s what you mean, then as @quid says, Of Course Not.
Apr
16
comment Find the degree measure of an arc of a circle
Do you know the theorem that the if $A$, $B$, and $C$ are three points on the circumference of a circle, then the measure of angle $ACB$ is half the measure of the arc $AB$?