Lubin
Reputation
99/100 score
 2h comment consider the table of indices of 13 to the base 2 are @AndréNicolas, thanks. Just shows how parochial and provincial my upbringing has been. 1d comment How can I “groupify” an elliptic curve over a non-field? Even if $R=\Bbb Z[1/2,1/3]$, ordinarily when you add two points, you need denominators to express the sum. Maybe he’s just referring to the projective plane over $R$, for putting $E$ into. 1d comment Center of finite dimensional division $\mathbb{R}$-algebra? Center will be a field, finite over $\Bbb R$. There are only two such. 1d comment All possible degree 3 field extensions To extend @Gro-Tsen’s comment, I’ll say that any normal cubic extension of $\Bbb Q$ must be totally real (and Gro-Tsen exhibits just one of the infinitely many of these), while a normal extension involving cube root will automatically have the cube roots of unity, quadratic over $\Bbb Q$, so such a field will be of even total degree over $\Bbb Q$. Apr 29 comment Splitting field of an irreducible polynomial of degree four I'm away from home right now, it may take a while. Apr 29 comment Does rationality of $\cosh(nx)$ and $\cosh((n+1)x)$ imply rationality of $\cosh(x)$? I think your argument about the extensions $\Bbb Q(\cosh x)\supset\Bbb Q(\cosh(nx))$ is the most convincing, but I don’t hink it’s quite complete. It seems clear to me that the field-extension degree in this case is $n$, and the fact that $\Bbb Q(\cosh(nx),\cosh((n+1)x))$ is sandwiched between fields of degree $n$ and $n+1$ beneath $\Bbb Q(\cosh x)$ shows immediately that $\Bbb Q(\cosh(nx),\cosh((n+1)x))=\Bbb Q(\cosh x)$. This argument works as well for the ordinary trigonometric functions, too. Apr 28 comment Does rationality of $\cosh(nx)$ and $\cosh((n+1)x)$ imply rationality of $\cosh(x)$? Is the corresponding proposition true for the ordinary trigonometric functions? Apr 28 comment P-adic expansion of rational Alternatively, you can identify the repeating part as $150/(1-100)$ in $7$-ary, that is, $84/(1-49)=-7/4$ in decimal, which you add to the nonrepeating part, namely $2$, to get the right value. Apr 28 comment Are we allowed to compare infinities? I’m sorry, but I think this is a bad quote, because it says “infinite numbers” when it should say “infinitely many numbers”. It is not pedantic to insist on the distinction. Notice that OP did not make this error, her youth notwithstanding. Apr 28 comment Why do we study real numbers? Some may consider your question naive, but I don’t. There are all sorts of reasons, many coming from physics. The alternative, I guess, would be to try to do everything in rational numbers. But we don’t have an Intermediate Value Theorem there, and I think that that’s one of the most useful properties of the reals. (Others will have different opinions.) Apr 28 comment simple proof for principle of pigeons True, @lulu, but I think it’s using a pile driver to crack a peanut. Apr 27 comment Find $\theta \neq \sqrt{3}+\sqrt{5}$ such that $\mathbb{Q}(\theta) = \mathbb{Q}(\sqrt{3},\sqrt{5})$. Need a hint to get started. The union of the three two-dimensional subspaces in the four-dimensional space is not $K$. You’re confusing the field-theoretic join (“compositum”) with set-theoretic join, which is the union. The rest of your argument is correct. Apr 27 comment Proving that $\mathbb{Q}_p$ is not formally real. For $p>3$, it’s easily seen, without Hensel nor Binomial Series, that $\Bbb Q_p$ contains the $(p-1)$-th roots of unity. Start with any $p$-adic unit, say $n$, and take $p$-th powers, successively. You get a sequence $p$-adically convergent to the root of unity that’s $\equiv n\pmod p$. (The method works for $p=2,3$ as well, it’s just that the roots of unity there are real.) Apr 26 comment Give an example of a polynomial whose Galois group is isomorphic to $D_{10}$ Please learn enough TeX for your question to be readable. Apr 26 comment Describing a Galois group for a field with the given roots adjoined Not correct in that you think $i$ is in the splitting field, and it isn’t. Nor is $\sqrt3$. What is in the splitting field is $\sqrt{-3}$. Apr 26 comment II. p-adic equations [2.1. Solutions] (J.-P. Serre) Well, in parts 1 and 2, it looked like you were thinking abstractly only. The good example is $\cdots\to\Bbb Z/(p^n)\to\Bbb Z/(p^{n-1})\to\cdots$. To show nonemptiness, I’d think you’d have to use Axiom of Choice. Apr 25 comment II. p-adic equations [2.1. Solutions] (J.-P. Serre) I hope you have explicit example in mind. What is it? Apr 25 comment How do I compute the first few digits of √11 in $\Bbb Q_5$ You can do it by direct computation, and learn a lot along the way. Just take $1+5n$ and square it, and see what $n$ must be to make the result congruent to $11$ modulo $25$. Keep $n$ in the range $0,\cdots,4$, of course. Apr 24 comment Extensions of the quadratic closure of $\mathbb{Q}$ Because now every finite extension of $\Bbb Q$ in $\Bbb Q^q$ is of degree a power of two. Apr 24 comment Extensions of the quadratic closure of $\mathbb{Q}$ Do you see that every element of $\Bbb Q^q$ is of degree a power of two over $\Bbb Q$?