Conditions on ideal b for fields or integral domains

Question

Let $A$ be a ring and $b$ be an ideal of $A$. Prove that
1. $A/b$ is a field $\iff b$ is maximal
2. $A/b$ is an integral domain $\iff b$ is prime

I figure that the first is derived from the fact there is only the trivial ideal in $A/b \iff b$ is the only ideal in A containing $b$.

It is the second proof I am having trouble with. Intuitively I realise that from the definition of integral domain, $\forall x,y \in A/b, xy=0 \implies x=0$ or $y=0$, means something about the cosets of A that make up $A/b$, but I can't quite see it. I want to avoid use of equivalence classes and use the coset notation instead.

I have begun by stating the definition of an integral domain and then do I need to claim that $x=0 \implies x=x+b$ where x is an element of b in A? Or is x just an element of A? If $x \in b$ then the rest would be clear since ideals are closed under addition so $x+b \in b $ which, once the same argument is applied to y also, forms the definition of prime ideal.

If this is correct, I do not understand how to prove the opposite direction without simply repeating the same proof backwards.

Answer 1

Regarding your added part: Let $A$ be a commutative ring identity (this assumption is missing, you should add this one also) and let $I$ be an ideal of $A.$ Then the zero element (i.e. additive identity) of $A/I$ is $0 + I$ and $a + I = 0 + I \Leftrightarrow a \in I.$

Suppose $A/I$ is an integral domain. We want to show that $I$ is a prime ideal. For this we need to show that if $x, y \in A$ are such that $xy \in I,$ then either $x \in I$ or $y \in I.$ Let $x, y \in A$ and assume that $xy \in I.$ Now in the quotient ring $A/I,$ we have $(x+I)(y+I) = xy + I = 0 + I.$ But it is an integral domain. So either $x+I = 0 + I$ or $y+I = 0+I.$ Hence either $x \in I$ or $y \in I.$

Next assume that $I$ is a prime ideal. Let $x+I, y+I \in A/I$ be such that $(x+I)(y+I) = 0+I.$ Then $xy+I = 0+I \Rightarrow xy \in I \Rightarrow$ either $x \in I$ or $y \in I \Rightarrow$ either $x+I = 0+I$ or $y+I = 0+I.$ Hence $A/I$ is an integral.

Note: This proof is exactly same as the existing answer (by pre-kidney). The only difference is that I use the notation $x+I$ instead of $[x]$ (because you want to use cosets instead of equivalence classes, though these two are same).

Answer 2

$A/m$ is a field iff $m$ is maximal:

The Noether Correspondence states that the ideals containing $m$ are in bijection with the ideals of $A/m$. Thus $A/m$ is a field iff its only ideal is 0 iff the only ideal containing $m$ is $m$ itself, i.e. $m$ is maximal.

$A/p$ is an integral domain iff $p$ is prime:

Let $[x]$ denote the equivalence class in $A/p$ corresponding to an element $x\in A$. Then $ab\in p\iff [a][b]=[0]$. Hence the condition $[a][b]=[0]\implies [a]\text{ or }[b]=0$ is equivalent to the definition of primality.