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Are there any intuitive reasons that can help us remember that $R/I$ is a field iff $I$ is a maximal ideal; $R/I$ is an integral domain iff $I$ is a prime ideal?

(I can understand the proof, but have problems with remembering the result correctly and intuitive understanding.)

I can roughly understand that if $I$ is maximal ideal, $R/I$ will have the minimal number of ideals namely zero ideal and itself, which is the criteria of being a field?

How about if $I$ is prime ideal? How do we intuitively see that $R/I$ is a domain?

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    $\begingroup$ If $I$ is a prime ideal of a commutative ring $R$, then we can say that $a\codt b\in I$ if and only if $a\in I$ or $b\in I$. When you take the quotient, the ideal $I$ becomes the zero of the ring $R/I$ and the condition that $I$ is prime becomes $(aI)\cdot(bI)=I$ if and only if $a\in I$ or $b\in I$ which implies there is no zero divisors. $\endgroup$ – Levent Jul 9 '16 at 9:46
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    $\begingroup$ Elements in the ideal $I$ are zero in $R/I$. Hence if you cannot multiply two elements not in $I$ and get something in $I$, you cannot multiply two non-zero elements in $R/I$ and get zero. There's not much intuition here I think, its literally the same condition phrased in different words. $\endgroup$ – user2520938 Jul 9 '16 at 9:47
  • $\begingroup$ @Levent Thanks. I like this way of thinking. $\endgroup$ – yoyostein Jul 9 '16 at 9:55
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The easiest way to remember which is which is probably to think that a field is also a domain and a maximal ideal is also prime. So when you quotient by a maximal ideal you get a field, but also a domain, because the ideal was also prime.

You should know that the, calling $\pi:R\to R/I$ projection map, then an ideal $J\subset R/I$ is prime (resp. maximal) if and only if $\pi^{-1}(J)$ is prime (resp. maximal) in $R$. In particular, $\{0\}$ is prime (resp. maximal) in $R/I$ if and only if $I$ is prime (resp. maximal) in $R$. So it is enough to consider the case $I=\{0\}$.

It can be useful to learn and understand many equivalent definitions of domain/field and maximal/prime ideal. For $R\neq \{0\}$ a commutative ring with $1$:

$R$ is a field $\iff$ every $0\neq a\in R$ is invertible $\iff$ the only ideals of $R$ are $\{0\}$ and $R$ $\iff \{0\}$ is maximal

and

$R$ is a domain $\iff$ for $a,b\in R$ if $ab=0$ then either $a=0$ or $b=0$ $\iff$ for $a,b\in R$ if $ab\in\{0\}$ then either $a\in\{0\}$ or $b\in\{0\}$ $\iff$ $\{0\}$ is prime

Remember that by definition the improper ideal $R$ is not maximal nor prime.

A little aside you didn't ask for, but it fits nice in this answer. A ring is called reduced if there are no nonzero nilpotents ($a\in R$ is called nilpotent if $a^n=0$ for some $n\in\mathbb{N}$). A proper ideal $I\subsetneq R$ is called radical if $\sqrt I = I$, that is if every $a\in R$ such that $a^n\in I$ for some $n\in \mathbb N$ is already in $I$. It follows that every prime ideal is radical and that a ring is reduced iff $\{0\}$ is radical. So:

$\{$maximal ideals$\}\subset\{$prime ideals$\}\subset\{$radical ideals$\}$

$\{0\}$ is maximal $\implies$ $\{0\}$ is prime $\implies$ $\{0\}$ is radical

$R$ is a field $\implies$ $R$ is a domain $\implies$ $R$ is reduced

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I had just the same questions when I first met those propositions, so I'd like to share my intuitions.

Before starting, let me say that I do think going through nilpotent elements - radical ideal - reduced ring first helps understanding prime and maximal ideas. At least, this was my learning process, and I found easier finding intuitive explanation of why a ring modulo a radical ideal gives a reduced ring. So maybe have a look at Relantionship between nilpotents - radical ideals - reduced rings. (If you understand that, I think you may even have all the elements to come up with the rest of the answer yourself.)

Now to prime and maximal ideals.

Prime ideals: a domain is a place where no two elements, when multiplied together, yield zero. We always need to keep this in the back of our minds, because this is what we would like to achieve. So how can we achieve that? Take whatever ring you want, how can we, in a way, build a domain ring out of it?

We say: okay, the problem is when $a*b = 0$ and $a, b \neq 0$. It seems stupid, but the whole solution is in the latter condition. If we could magically say that every time that $a*b = 0$, either $a$ or $b$ would become zero (whatever this means), then we would indeed have built a domain ring. But lo!, we do have a way of transforming elements to zero, and that is through a quotient space!

In fact, imagine to go through all possible products between all ring elements: every time one of those products would yield zero, you take one of the factors and put it in a set. At the end of the story, you have a set of most null-divisor elements (I say most, because in $\mathbb{Z}_{6}$, for example, $2*3 = 0$, and you can put in your set either $2$ or $3$, you don't need to take them both.) What it's true is that a such set is more than a set, it's indeed an ideal: what we call a prime ideal.

Now, what does it happen if we quotient the ring by a prime ideal? We are transforming to zero all elements which gave us problems in the ring (i.e. which gave rise to null-divisors), and lo!: if $a*b = 0 \mid a,b \neq 0$ in the ring, we are now sure that in the quotient, either $a$ or $b$ are indeed $0$! Thus we have a domain ring!

Maximal ideals: this gets way trickier, and I fear I don't have a ready intuition to share. I do have some hints, but I am not entirely sure they are correct, so take them as they are.

As above, we want to set to zero all elements which don't have multiplicative inverse. First of all, we know that if a maximal ideal contains an invertible element, than that ideal is actually the whole ring (i.e. it is not a proper ideal). So all invertible elements will for sure be out of a maximal ideal, which is made up entirely by non-invertible elements (the ones we want to get rid of).

However, things aren't so easy, because the strategy we used before doesn't work here: grouping together all non-invertible elements doesn't give us an ideal (think about the non-invertible of $\mathbb{Z}_{10}$, for example), so we can't expect to build a field just taking away all non-invertible elements.

For example, we know that for $\mathbb{Z}$, all ideals of the form $(p) \mid p $ prime are maximal, and indeed all $\mathbb{Z}_{p}$ are fields. But look at what happens! We are not taking away non-invertible elements (2 is non-invertible in $\mathbb{Z}$, and yet it's still there), but in that quotient elements manage to find a multiplicative inverse among the ones that are left! This fact (that inverses are created in the quotient) is what makes very difficult having an intuition for this fact, I believe.

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Understanding the various definitions helps a lot (for me at least).

A ring $A$ with $1 \neq 0$ is an integral domain if $A$ is commutative and has no (nonzero) zero divisors. That is $ab = 0$ implies $a = 0$ or $b =0$, for $a, b \in A$.

An ideal $P$ of a commutative ring $R$ is prime if $P \neq R$ and $ab \in P$ implies $a \in P$ or $b \in P$, for $a, b \in R$. If we pass to the quotient, then the condition is that $R/P \neq0$ and $\overline{ab} = \bar 0$ implies $\bar a = \bar 0$ or $\bar b = \bar 0$, where $\bar r = r+P$ is the coset of $P$ containing $r$. This is exactly the condition for $R/P$ to be an integral domain.


For maximal ideal question, your observation is right:

Suppose $R$ is commutative with $1\neq 0$, $M$ a maximal ideal. Then, as you said, $R/M$ will only have two ideals, the zero ideal $(\bar0)$ and itself $R/M = (\bar 1)$, and this does imply $R/M$ is a field. Here is the proof:

Proof: Suppose $A$ is a commutative ring with $1\neq 0$ whose ideals are $(0)$ and $A = (1)$. To show $A$ is a field, we just need to show that every $a \in A\smallsetminus 0$ has a multiplicative inverse. Take $a \in A \smallsetminus 0$; then $(0) \subsetneq (a)$ so $(a) = A = (1)$. Thus there exists $b \in A$ such that $ab =1$.

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  • $\begingroup$ I think the last statement is incorrect. $A$ must be a field for this to work, how do you know if $a$ even has an inverse? All you know is that $A$ contains units, no assumption that $a$ will have an inverse as this was chosen arbitrary. $\endgroup$ – jacob smith Sep 3 '16 at 21:05
  • $\begingroup$ I took an $a \in A\smallsetminus 0$ and by assumption (that $A$ has two ideals $(0)$ and $A)$), we must have that $(a) = (1)$; so $1$ multiple of $a$, that is $1 = ab$ for some $b \in B$. $\endgroup$ – cat Sep 10 '16 at 6:32

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