Explain the proof behind maximal ideals $\iff $ quotient is a field.

Question

$R$ is a commutative ring with unity and $M$ is a maximal ideal $\iff$ $R/M$ is a field.

Proof is here

1. Can someone explain the inclusion $J \subset \ker \psi$?

2. Can someone also explain the hypothesis that

To see this note that if for all $r \in R$, there is some $r_1 \in J$ such that $$\psi(r) = \psi(r_1)$$

Sorry but where did this hypothesis come from and why are we assuming this to prove $J = R$?

Proof is here

3. Can someoene explain the statement

This ideal $\psi^{-1}(r + M)$ contains $M$ because $r \in \psi^{-1}(r+M)$.

Isn't the definition of the preimage of $r+M$ just $$\psi^{-1}(r+M) = \{r \in R : \psi(r) = r +M \}$$

I mean I could just prove the inclusion, but such an attempt won't really reveal what the statement's intent.

($x \in M$, $\psi(x) = x + M = M$)

Answer 1

$1.$ $0+M$ is the zero of $R/M$. So $\psi(J)=0+M$ implies $J\subset ker(\psi)$.

$2.$ Since $\psi(J)=R/M$, then for all $r\in R$, we have $$\psi(r)=r+M\in\psi(J),$$thus there exists $r_1\in J$ such that $$\psi(r_1)=r+M=\psi(r).$$ Here, $\psi(r-r_1)=0+M$ and so $r-r_1\in M$, i.e., $r-r_1=m$, where $m\in M$. Therefore $$r=r_1+m\in J+M\subseteq J$$ Since $r$ is arbitrary, it follows that $R\subseteq J$.

$3.$ $\psi^{-1}(r+M)=\{s\in R:\psi(s)=s+M=r+M\}$.

Answer 2

I'll try to go in order here.

The first point that's somewhat brushed under the rug here is that if $R, S$ are commutative rings, $\phi: R \to S$ is surjective, and $I$ is an ideal of $R$, then $\phi(I)$ is an ideal of $S$. To see this, note that if $s \in S$, then $s = \phi(r)$ for some $r \in R$, so \begin{equation} \phi(r')s = \phi(r')\phi(r) = \phi(rr') \subset \phi(R) \end{equation}

Then, one uses the fact that a field has only two ideals, and we examine each case individually; $\phi(J) = 0 \subset R/M$, and $\phi(J) = R/M$.

It seems like you feel comfortable with the first case, $\phi(J) = 0$.

For the second case, we $\textbf{assume}$ the map is surjective (as we have handled the other case in which it is not), and so the statement that $\phi(r_1) = r + M$ for some $r_1 \in J$ is precisely this surjectivity.

I think the proof of the converse is perhaps too complicated. One can show that if $R$ is a commutative ring with unity, $R$ is a field $\textbf{if and only if}$ $R$ has precisely two ideals. It turns out that the ideals of $R/M$ are precisely the images of the ideals of $R$ containing $M$ under the quotient map. To prove this, start with the fact that the pre-images of ideals are ideals and examine ideals of $R/M$.

Once you've established that, the result follows easily. Since $M$ is maximal, there are only two such ideals, $\phi(R)$ and $\phi(M)$.

Answer 3

Q1.

Let $j\in J$.

$\psi(j)=0+M$ since $\psi(J)=0+M$.

This means $j\in\ker\psi$.

So $J\subseteq\ker\psi$.

Q2. Note that $\psi(J)=R/M$ implies that $R/M\subseteq \psi(J)$.

So any $r+M=\psi(r_1)$ for some $r_1\in J$.

Q3. For Q3 the keyword is "properly".

To see the ideal $\psi^{-1}(r+M)$ contains $M$, take $m\in M$.

$\psi(m)=m+M=M\subseteq (r+M)$ since every ideal contains zero and zero in $R/M$ is $M$. (Do you know this? Comment below if you need details.)

As mentioned, this is not really the main point. The keyword is "properly". This follows from the fact that $r\notin M$ but $r\in\psi^{-1}(r+M)$.