$\text{Nil}(R)=\text{rad}(0), I$ - prime ideal $\iff R/I$ integral domain, I - max ideal $\iff R/I$ field, and other questions on Ideals and Rings

Question

So I've analysed my last lecture in Commutative Algebra and those are the following questions I'd have:$\DeclareMathOperator{\Max}{Max}$

1. Like e.g. $I+J=\left\{a+b\::\:a\:\in \:I,\:b\:\in \:J\right\}$, is there a way to precisely define the elements in $\bigcap _{\alpha \in A}\:I_{\alpha }$?

2. As in the title, why is $\operatorname{Nil}(R) = \operatorname{rad}(0)$?

3. I - prime ideal $\iff R/I$ integral domain and $I$ - max ideal $\iff R/I$ is a field - Why? What's the proof for that?

4. We defined something as a multiplicative set S, which is if $1\in S, 0\notin S, a,b \in S$ then $ab \in S$. Why do we even bother to define something like that, what's the practical use and interpretation here?

5. Following that: If $I$ - prime ideal $\iff R/I$ multiplicative set, why?

6. If $I_n\subset I_{n+1}$ are all ideals, why is $\bigcup _{n\in N}\:I_n$ also an ideal?

7. We had such a theorem:

Let $S\subset R$ be a multiplicative set and let $J$ be a family of ideals $I$ from ring $R$ so that $I\cap S=\varnothing $. Then $\forall _{I_0\:\in \:J}\:\:\exists \,{P\:\in \:J-\text{ prime and maximal ideal}},\:\:I_o \subset P$.

I kind of understand why P is a maximal Ideal (J fulfills Zorn's lemma thus if we take a chain of ideals so that $I_a\subset I_b$ and $a\le b$ we can say that $P = \bigcup _{\alpha \in A}\:I_{\alpha }$). But why is it also a prime ideal?

8. We had those statements and like to know why they are true:

• (a) If $f\::\:R\:\rightarrow \:T$ is an epimorphism, and $I\subset R$ is an ideal, then $f\left(I\right)$ is an ideal in T

• (b) If I take an ideal $J\supset I$, and $\pi :\:R\:\rightarrow \:R/I$ is an epimorphism, then $\pi \left(J\right)$ is also an ideal

• (c) The function $\pi$ is a bijection which retains the inclusion relation

• (d) $\left\{J\supset I\::\:J\:\in \Max\:R\right\}\ni L\:\rightarrow \:\pi \left(L\right)\in \:\left\{M\:\subset R/I\::\:M\:\in \:Max\:R/I\right\}$ is also a bijection which retains the inclusion relation (the same goes for Spec R and Spec R/I)

• (e) $I \in \Max\:R\:\Leftrightarrow \:\pi \left(I\right)=\left(0\right)\in \Max R/I \Leftrightarrow R/I\text{ is a field}$

Sorry that it's so much but I would really really appreciate any effort here, I'm just curious and want to understand every detail and I'm not satisfied accepting something without a reason.

Answer 1

Some answers (or questions):

1. What are the $I_\alpha$?
2. This is the definition of the nilradical. Remember the radical of an ideal $I$ of the ring $R$ is $\{x\in R\mid\exists\, n\in\mathbf N, \,x^n\in I\}$.
3. The definition of a prime ideal $I$ is that, for all $x,y\in R$, $xy\in I\implies x\in I\text{ or }y\in R$. Translated in terms of congruence classes modulo $I$, it becomes $\overline{xy}=\bar x\bar y=\bar x=\bar 0\implies \bar x=\bar0$ or $\bar y=\bar 0$, which corresponds to the definition of an integral domain. As to the maximality of the ideal $I$, a field can be characterised by the property that it has only two ideals – $\{0\}$ and the field itself.
4. Multiplicative sets are the basis to defines rings of fractions od a given ring. Among these are the localisations of a ring at a prime ideal, which is an essential tool in algebraic geometry.
5. If $I$ is a prime ideal, it is not $R/I$ which is a multiplicative set, but $R\setminus I$
6. $\bigcup\limits_{n\in\mathbf N}I_n$ satisfies the properties defining an ideal – the sum of two elements of the union is in the union essentially because by hypothesis, the set of ideals is totally ordered by inclusion.
7. A maximal ideal is also a prime ideal (because, considering the quotient $R/I$, a field is an integral domain).