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Jun
16
comment Irreducibility of $x^n-x-1$ over $\mathbb Q$
@Corvus — To echo TedShifrin, the polynomial ring $\mathbb{F}_{p}[X]$ is infinite, the set $\mathrm{Map}(\mathbb{F}_{p}, \mathbb{F}_{p})$ is finite. Hence the natural map $\mathbb{F}_{p}[X] \to \mathrm{Map}(\mathbb{F}_{p}, \mathbb{F}_{p})$ is not injective. To stress the difference, let me ‘prove’ that $\mathbb{F}_{2}$ is algebraically closed: Take a non-constant polynomial $f$, then $f$ is not constant $1$, and as $\mathbb{F}_{2}$ has only two elements, $f(x) = 0$ for some $x \in \mathbb{F}_{2}$. Thus $f$ has a zero, hence $\mathbb{F}_{2}$ is algebraically closed. $\square$ Spot the error.
Jun
3
comment When do we have that Absolute hodge classes= Hodge classes for complex projective manifolds?
I think, if the Lefschetz operators are algebraic, then André's motivated cycles are algebraic.
May
27
comment A simpler expression that is always smaller or larger
Well, write your function as $an^2 + bn + c$, where $a,b,c$ are fractions. Then it is already pretty simple. Next, you can consider if you want your estimate on a particular domain, or on all of $\mathbb{R}$. Depending on the answer, you might be able to round $a$ to an integer, and get rid of $b$ and $c$.
May
27
answered A simpler expression that is always smaller or larger
May
27
answered Why do the addition of linear equations all pass through the same point
May
27
suggested suggested edit on can someone give me the solution with the proof?
May
27
comment Dimensions of eigenspaces.
Figure out what it means to be in either of the eigenspaces: How are matrices $A$ called that satisfy $A = A^{\perp}$, and what if $A = -A^{\perp}$? It shouldn't be too hard to figure out a basis/the dimension for the eigenspaces, once you have done that.
May
26
comment What topological restrictions are there for a topological space to be a group?
I don't think your statement is correct. For example $\mathbb{C}$ is non-compact, a Riemann-surface, and carries a group structure. Probably the statement you mean is: If $X$ (your Riemann surface) is compact, then it admits a group structure if and only if the genus is $1$.
May
12
answered Equivalence between exact sequence of module and its induced one.
May
12
comment Equivalence between exact sequence of module and its induced one.
I think you need to add “for all $Y$” after the exact sequence of $\mathrm{Hom}$-sets. Otherwise, take $Y = 0$; then the sequence of $\mathrm{Hom}$-sets is exact, independent of what $X$, $X'$, and $X''$ are.
May
12
answered Singular points query
May
10
comment Positive and negative complex numbers?
In other words: squares are positive, and everything is a square.
May
10
comment Hanging a picture on the wall using two nails in such a way that removing any nail makes the picture fall down
@Awesome, how about soldering the nails together?
May
9
comment Proving that Z(p) is a ring
@GeorgesElencwajg – Right again. After thinking twice, I realise it wasn't that I misread the question (because my answer speaks about a subring of $\mathbb{Q}$). So indeed, something in my brain shortwired, due to, as you say "absent-mindedness". Thanks for correcting me. [By the way, on the notations that you mention: I fully agree that 3) is not advisable. The coincidence of 1) and 2) is unfortunate. I think $\mathbb{Z}_{p}$ should always mean 1)...]
May
9
revised Proving that Z(p) is a ring
Corrected stupid error
May
9
comment Proving that Z(p) is a ring
@GeorgesElencwajg – Wait; me stupid... You're absolutely right. I didn't even read the question properly.
May
8
comment Proving that Z(p) is a ring
@Lynnie — You should use curly braces {} instead of parentheses () around the $p$. So \mathbb{Z}_{p}.
May
8
answered Proving that Z(p) is a ring
May
8
revised Proving that Z(p) is a ring
Fixed grammar, mathematics, formatting
May
8
comment Proving that Z(p) is a ring
This ring is usually denoted $\mathbb{Z}_{p}$.