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2d
comment Complex extension isomorphic to $\mathbb{R}$?
… for example because the algebraic closure of $\mathbb C(X)$ is (non explicitly) isomorphic to $\mathbb C$. You get this way other subfields $F \subset \mathbb C$ isomorphic to $\mathbb R$, but such that $\mathbb C/F$ is an infinite extension.
2d
comment Complex extension isomorphic to $\mathbb{R}$?
Actually, an automorphism of $\mathbb C$ sending $\mathbb R$ to itself can only be the identity or the complex conjugation (that's not hard to prove), so in your sentence, "wild enough" simply means $\neq \{\mathrm{id},\bar\cdot\}$. With this construction, you get a lot of fields $F \subset \mathbb C$ isomorphic to $\mathbb R$ (actually you get all of those fields such that $\mathbb C/F$ is a finite extension). But there is another source of examples, still not explicit: not only $\mathbb C$ has nontrivial automorphisms, it also has nontrivial selfembeddings $\mathbb C \to \mathbb C$…
2d
comment Complex extension isomorphic to $\mathbb{R}$?
Such fields exist, but none of them is explicit. They come from the wild automorphisms of $\mathbb C$, see e.g. math.stackexchange.com/questions/412010/… and maa.org/sites/default/files/pdf/upload_library/22/Ford/…
Mar
22
comment If $M\subseteq\Bbb R^n$ has $\dim(M)<n$, is $\Bbb R^n-M$ dense in $\Bbb R^n$?
Yes. You only have to prove that points of $M$ are limit points of sequences living outside of $M$. If $M = \mathbb R^d \varsubsetneq \mathbb R^n$, that's easy: $(x,0) = \lim (x,1/n)$. The general case reduces to this one through a coordinate chart.
Mar
16
comment $x^TMx \geq \lambda_{min}x^Tx$ is true?
It's a bit weird to use your order on the matrix, is it not? On a diagonalisation basis, $\frac{x^T M x}{x^T x} = \sum \lambda_i \frac{x_i^2}{\sum x_i^2}$ is a convex combination of all the $\lambda_i$, so it's larger than the smallest one. And, by the way, the positivity of the $\lambda_i$'s is irrelevant.
Mar
8
comment Why a sphere cannot have a Lorentzian Metric?
S³ does have a Lorentzian metric, given the answers of mathoverflow.net/questions/44861/…
Mar
8
comment Why a sphere cannot have a Lorentzian Metric?
Did he mean a two-dimensional sphere? If so, it's because the light cone could be used to define a non-vanishing vector field, contradicting the Hairy Ball Theorem.
Feb
14
awarded  Yearling
Feb
11
comment $-1$ as the only negative prime.
Small remark: even if it's extremely unusual, the great John Conway makes some interesting points on the interest of considering $-1$ as prime in the preface of The Sensual (Quadratic) Form. They may be a bit off-topic for the OP, but are worth reading nevertheless.
Feb
11
comment Group of exponent $2$.
It is correct. In an Abelian group, "conjugate" is another name for "equal" so the property easily follows from the fact that exponent-2 groups are Abelian.
Feb
2
comment Finding char polynomial in $Z_3$
Your formula for the determinant looks suspicious... Sarrus's rule is only valid for $3 \times 3$ matrices → en.wikipedia.org/wiki/Rule_of_Sarrus
Jan
29
comment Theorem 2.27 (a) in Baby Rudin: Is his proof complete enough?
Always relevant: abstrusegoose.com/12
Jan
29
answered Are 3#4 and 3*#4composite knots isotopic?
Jan
29
asked What does it take to have a precise definition of volume?
Jan
6
comment Term for an infinite sequence with both a beginning and an end
I would call that a $(\omega + \omega^*)$-sequence, see e.g. mathworld.wolfram.com/OrderType.html. But I don't think a canonical name exists.
Jan
5
awarded  Enlightened
Jan
5
awarded  Nice Answer
Dec
18
comment Set consisting of all unoriented cobordism classes of smooth closed $n$-manifolds can be made into additive group?
Addition is disjoint union, inverse is the manifold itself.
Nov
20
awarded  Nice Answer
Oct
29
revised Alien Mathematics
edited body