I am studying localization right now from the lecture notes given by our instructor. In this notes he describes an example of localization what he told us yesterday in the class. Here's this $:$ Let $\mathcal k$ be a field. Let $A= \mathcal k [X_1,X_2, \cdots, X_n].$ Let $\mathfrak p$ be a prime...

Let $R$ be a Noetherian ring and $n \in \mathbb{Z}$ such that for any f.g. $R$-module $M$ and $k > n$ we know that $\text{Ext}^k_R(M,R) = 0$. Does it follow from this that $\text{Ext}^k_R(M,R) = 0$ for arbitrary $R$-modules $M$ as well? I am trying to show that the $R$-module $R$ has injective di...

$S(n,m)$ is a double sequence. Can anyone give me an example where lim$_{m , n \to \infty} S(n,m)$ exists but lim$_{n \to \infty}$( lim$_{m \to \infty} S(n,m)$) , lim$_{m \to \infty}$( lim$_{n \to \infty} S(n,m)$) do not? My Attempt: I thought of an example. $S(1 ,m) =m $, $S(n , 1) ...

Given two real numbers $0<a<1$ and $0<\delta<1$, I want to find a positive integer $i$ (it is better to a smaller $i$) such that $$\frac{a^i}{i!} \le \delta.$$

how would you call the relation like this? $$\ln(\sqrt{5})<\ln(5^2)$$ Is it inequality or inequation? Motivation for this question: In Czech we have different words for a relation involving equality sign and a problem involving equality sign and a variable and the value of the variable is to be ...

The inequation is as following: $$\frac{\pi x^2-(1+\pi^2)x+\pi}{-2x^2+3\pi x}\gt 0$$ So far I was able to factor the inequation, but I don't know how to proceed from now on: $$\frac{(x-1)(x-\pi^2)}{-2x(x-\frac{3\pi}{2})}\gt 0$$

Supppose we have $a$ a real positive number that's not equal to $1$. Solve the following inequation: $$\log_a(x^2-3x)>\log_a(4x-x^2)$$ If it's known that $x=3.75$ is one solution of it.

I have the following inequation : $\sqrt[n]{10} \geq \frac{10}{9}$ and I would want to know for which interval of n the inequation is right, but I have no idea how to solve it. I hope somebody could help me.

How would one proceed in solving this difficult inequation with multiple absolute values? is there a way one should proceed ? $$\frac{x}{||x|-2|} \le \frac{x-1}{|x-3|}$$ thanks

Solve the inequation: $\sin^4x+\cos^4x \geq 1/2$. I did this: $(1-\cos^2x)^2+\cos^4x \geq 1/2$ $-2\cos^2x+2\cos^4x \geq -1/2$ $-2(\cos^2x-\cos^4x) \geq -1/2$ $\cos^2x(1-\cos^2x) \leq 1/4$ $\cos^2x\sin^2x \leq 1/4$ $|\cos x\sin x| \leq 1/2$ Now what? :S I feel it should be easy from he...