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Jan
25
comment subgroups of ${Z}_{m}\oplus {Z}_{n}$
This does not require $X$ or $Y$ to be abelian. It goes through for any finite groups $X$ and $Y$ having relatively prime orders. You just need to use multiplicative notation instead of additive notation (e.g., $h \mapsto h^{\nu|Y|}$ instead of $h \mapsto \nu|Y|\cdot h$).
Jan
25
comment A set that is an element of itself
I am surprised Bourbaki is asking a question on this site.
Jan
23
comment Is -3 always a quadratic residue in Z/p*Z for a prime p>3?
You write $(-3|p) = ... = (p|3)$, and that much is fine. Your reasoning goes off the rails when you write $(p|3) \Rightarrow p > 3$ and then $p > 3 \Rightarrow (p|3) = 1$. This makes no sense. Do you mean $p \not= 3$ rather than $p > 3$? Concerning your latest comment to your question, 2 is NOT a quadratic residue mod 3!
Jan
18
comment How do I show the eigenvalues of a positive definite matrix are real and positive?
You left out a mathematical expression near the start of the second sentence, probably $\langle Ax,x\rangle$.
Jan
18
comment Is there an analogue of the 15 theorem for cubic forms?
The 15 theorem is about positive-definite quadratic forms. This doesn't have a direct analogue for cubic forms. In fact, there are lots of features of quadratic forms that don't carry over to cubic forms (e.g., diagonalizability over the base field or interesting structure for its orthogonal group). Is there a reason you think there ought to be an analogue, or is this question asked for the sake of curiosity?
Jan
17
comment Being Galois stable under completion?
Write $L = K(\alpha)$. The main point is to show $L_{\mathfrak P} = K_{\mathfrak p}(\alpha)$. If $L/K$ is Galois then the minimal polynomial of $\alpha$ over $K$ is separable and has a full set of roots in $L$. All those roots are in the larger field $L_{\mathfrak P}$ too, and the minimal polynomial of $\alpha$ over $K_{\mathfrak p}$ is a factor of its minimal polynomial over $K$.
Jan
17
comment how to find fourier transform of $\exp(-x^2/2)$
Why don't you show some work, which would help others identify the point where you are making an error. By the way, the title of your question is not what you're asking. You want to know the Fourier transform of $e^{-x^2/2}$ but the title of the question is asking for the Fourier transform of something else (at the time I write this).
Jan
17
comment Question about $p$-adic exponential
If $y = 1 + a + b + c$ and $|a|$, $|b|$, and $|c|$ are all less than 1 then $|y| = 1$. Prove every term in the exponential series besides the constant term has absolute value less than 1 when it is evaluated at a point in the disc of convergence.
Jan
17
comment Application of Hensel's lemma
Hensel's lemma in its basic form is about lifting simple roots mod $p$ to roots in the $p$-adics. The mod 2 reduction has a double root, so the basic version of Hensel's lemma does not apply and tell you anything about this situation.
Jan
17
comment Question about $p$-adic exponential
The exponential is a power series. Look at each term besides the constant term and check it has absolute value less than 1. (Have you done that?) And $\exp(s)$ is a limit of partial sums, each of which is congruent to 1 modulo the maximal ideal of $\mathcal O_K$, so the limit has this property too.
Jan
17
comment Application of Hensel's lemma
In context, surely $O$ means "integers of" and "E" means "residue field" of, so it means $\mathbf Z_2$ and ${\mathbf F}_2$ respectively, in more standard notation.
Jan
17
comment Question about $p$-adic exponential
Of course: look at the series expansion for the exponential and check every term aside from the constant term will be in the maximal ideal.
Jan
17
comment localization in algebraic geometry
Look up the definition of the local ring at a point on a smooth algebraic curve, say. This is used to define the order of vanishing of a function at the point (or the order of a pole there).
Jan
17
revised A question about the QR decomposition
edited title
Jan
17
comment Definition of tame ramification from the ramification groups
Oops, of course it'd be hard for $G_1$ to be trivial in the tamely ratified case if the later indices were not powers of $p$. Yes, I had my description exactly backwards.
Jan
17
comment Definition of tame ramification from the ramification groups
Your question is a standard result about higher ramification groups. The index $[G_0:G_1]$ is a power of $p$ and the indices $[G_i:G_{i+1}]$ are relatively prime to $p$ (when the residue field has characteristic $p$). See books on local fields that discuss higher ramification groups (e.g., the book by Fesenko and Vostokov).
Jan
17
comment Definition of tame ramification from the ramification groups
You should call $S$ the integral closure of $R$ in $L$, not the algebraic closure of $R$ in $L$. Have you looked in books that discuss higher ramification groups?
Jan
16
comment Why does a singular point create a cusp or a node on the trace?
The last sentence of my previous comment is not completely accurate. For instance, if you limit the kinds of singularities that are allowed (e.g., if your curves are algebraic, defined more or less by rational functions) then there are general methods of studying them, but not at the level of the question you're asking (Google "Blow up").
Jan
16
comment Why does a singular point create a cusp or a node on the trace?
Away from singularities there are general theorems (e.g., inverse function theorem, ability to renormalize unit tangent vectors to have length 1, etc.) There is no over-arching general description of singularities that can be codified in one sentence. Sometimes you have what is called a cusp, sometimes it's called a node, and if you work only over the real numbers you may not see anything unusual at all (try $(t,t^3)$ at $t = 0$ over $\mathbf R$). We can say general things about what happens away from such points, not general things that happen at such points.
Jan
16
comment Why does a singular point create a cusp or a node on the trace?
In differential geometry, if you want to construct a unit velocity vector field along a curve then the parametrization must have $\alpha'(t)$ not equal to the zero vector so you can renormalize this along the curve as $\alpha'(t)/|\alpha'(t)|$, which has length 1. That's probably the most important reason for you right now. Other reasons include the implicit function theorem and inverse function theorem; there are hypotheses in them about nonvanishing of derivatives to make the theorems work in general. When all derivatives vanish you can't generally apply such results.