@BalarkaSen consider $S : (\sqrt{x^2+y^2}-1)^2 + z^2 \le 1$ with parametrization $\begin{cases} x &=& (1+\sin\varphi\cos\omega)\cos\theta \\ y &=& (1+\sin\varphi\cos\omega)\sin\theta \\ z &=& \sin\varphi\sin\omega \end{cases}$. Then, map $(\varphi,\theta,\omega)$ to $(\sin\varphi\exp(i\omega),
\cos\varphi\exp(i\theta)\left(\dfrac{\cos\varphi\sin\omega-i\sin\varphi-i\cos\omega}{1+\sin\varphi\cos\omega}\right)) \in \Bbb S^3$. Then, all the preimages of the Hopf fibration are unit circles.
\cos\varphi\exp(i\theta)\left(\dfrac{\cos\varphi\sin\omega-i\sin\varphi-i\cos\omega}{1+\sin\varphi\cos\omega}\right)) \in \Bbb S^3$. Then, all the preimages of the Hopf fibration are unit circles.
Let $R$ be a PID. An ideal $P$ in $R$ is said to be primary if $ab \in P$ and $a \notin P$ implies $b^n \in P$ for some $n \in \Bbb{N}$. Show that $P$ is primary if and only if $P = (p^n)$ for some $n \in \Bbb{N}$ and some prime element $p \in P$.
Here is my attempt:
Assume that $P = (p...
According to wikipedia, we have that
$$
x^{\log_b(y)} = y^{\log_b(x)}
$$
because
$$
x^{\log_b(y)} = b^{\log_b(x) \log_b(y)} = b^{\log_b(y) \log_b(x)} = y^{\log_b(x)}
$$
But what justifies that first leap?
$$
x^{\log_b(y)} = b^{\log_b(x) \log_b(y)}
$$
