Lubin
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25,684
99/100 score
 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 $AB$?