TITLE: Basic question on projective bundles
QUESTION [4 upvotes]: Let $\mathcal{E}$ be a coherent sheaf on an irreducible scheme $S$ ($S$ can be pretty nice, say noetherian of finite type), and let $\mathbf{P}(\mathcal{E}):=\mathrm{Proj}(\mathrm{Sym}(\mathcal{E}))$ be the projective bundle associated to $\mathcal{E}$. Is it true that if the structural map $\mathbf{P}(\mathcal{E})\to S$ is dominant (or even surjective), then $\mathbf{P}(\mathcal{E})$ is irreducible?

REPLY [1 votes]: Let me mention a positive result. Assume

$S$ is Cohen-Macaulay and irreducible;

$\mathcal{E}$ has projective dimension $1$ and generically has rank $r > 0$;

$\mathrm{codim}(\{s \in S \mid \dim(\mathcal{E}_s) \ge r + i \}) \ge i + 1$ for all $i > 0$.

Then $\mathbf{P}(\mathcal{E})$ is irreducible.
To prove this, first, use the projective dimension assumption to write a locally free resolution
$$
0 \to \mathcal{F}_1 \to \mathcal{F}_0 \to \mathcal{E} \to 0
$$
(maybe locally over $S$). Then observe that the epimorphism induces an embedding
$$
\mathbf{P}(\mathcal{E}) \subset \mathbf{P}(\mathcal{F}_0)
$$
as the zero locus of a global section of the vector bundle $p^*\mathcal{F}_1^\vee \otimes \mathcal{O}(1)$ on $\mathbf{P}(\mathcal{F_0})$. Then use the codimension assumption to check that $\mathbf{P}(\mathcal{E})$ has at most one component of dimension $\dim(S) + r - 1$. Then use the Cohen-Macaulay assumption to check that $\mathbf{P}(\mathcal{E})$ is Cohen-Macaulay, hence doesn't have components of dimension less than $\dim(S) + r - 1$.


TITLE: Prove that $4n+2=x^2+y^2+z^2$ for some odd $x,y$ and even $z$
QUESTION [1 upvotes]: Show that for all $n\in \mathbb{N}$, exists $x,y,z \in \mathbb{N}$, such that $x,y$ are odd and $z$ is even, such that $4n+2=x^2+y^2+z^2$.
I started by using the fact that every natural number has a decomposition to a sum of 4 square, and tried to prove that one of the numbers is zero, basically $4n+2=x^2+y^2+z^2+w^2$.
Firstly, I falsely assumed that all of those numbers weren't 0, and then I proved that at least one of those numbers has to be odd, because if they were all even then $4n+2=x^2+y^2+z^2+w^2 \equiv 0 (mod 4)$, contradicting the fact that $4n+2\equiv 2 (mod 4)$. Then I showed that since $4n+2$ is even then there have to be 2 even numbers exactly and 2 odd numbers exactly among $x,y,z,w$.
From here I assumed that $x,y$ are odd, and that $z,w$ are even and tried to prove that at least one of $z,w$ is 0, but I didn't really have ideas in this direction, so this didn't work out.

REPLY [0 votes]: If x and y are odd while z is even, then you can set $x = 2a+1, y = 2b+1, z = 2c$
$4n+2 = (2a+1)^2+(2b+1)^2+(2c)^2 = 4a^2 + 4a+1+4b^2+4b+1+4c^2 = 4(a^2+a+b^2+b+c^2)+2$
Now you're trying to prove that every possible $n$ can be described as $n = a^2+a+b^2+b+c^2 = a(a+1) + b(b+1) + c^2$
Note that $a(a+1)+b(b+1) \equiv 0 (mod 2)$
Does this help?


TITLE: Substitution in ODE
QUESTION [2 upvotes]: A tutorial (page 17) on ODE scaling takes the equation $m\ddot{u}+c\dot{u}+ku=F_0sin(\omega t)$, with $u$ a function of $t$, and substitutes $\eta=u/a$, $\tau=t/b$ to get:
$$\frac{ma}{b^2}\ddot{\eta}+\frac{ca}{b}\dot{\eta}+ka\eta=F_0sin(\omega\tau)$$
Now, when I differentiate $\eta=u/a$, I get $\dot{\eta}=t\dot{u}/a=\tau b\dot{u}/a$. Substituting, the final result is:
$$\frac{ma}{\tau^2b^2}\ddot{\eta}+\frac{ca}{\tau b}\dot{\eta}+ka\eta=F_0sin(\omega\tau b)$$
That's not the same. What am I missing?

REPLY [2 votes]: Gerry has already explained the differences on the left-hand side. On the right-hand side, this is a typo in the book. The next equation has $\sin\omega'\tau$ instead of $\sin\omega\tau$, with $\omega'=\omega\sqrt{m/k}$ and $b=\sqrt{m/k}$, so this is $\sin\omega\tau b$ as you expected. So it should be either $\sin\omega t$ or $\sin\omega'\tau$ on the right-hand side.


TITLE: Are there infinitely many primes $p_n$ for which $p_n< 2n$?
QUESTION [1 upvotes]: I am working in some calculations and I found this kind of problem. That is,
are there infinitely many primes $p_n$ for which $p_n< 2n$ ?
I think the answer is no, but I dont have a formal explanation.

REPLY [5 votes]: First note that $p_5=11>2\cdot 5=10$.
And if $n\geq 5$ and $p_n>2n$, it follows that
$$ p_{n+1}\geq p_n+2>2n+2=2(n+1) $$
so by induction $p_m>2m$ for all $m\geq 5$.