Krdytkyu

Let $I subset R$ be an ideal of $R$. Define $pi(a + I) = a$ with $a in R$, so $pi: R/I to R$

Then $ker pi = I = bar{0} in R/I$ and this is clearly onto so it is an isomorphism. So this map says $R$ is always isomorphic to its quotient. Isn't there something not right about this argument? Or is this just a simple trivialization of the correspondence property?

Suppose we have the containment of ideals $I subset A subset R$

Then the correspondence theorem say $I subset A subset R stackrel{pi}leftrightsquigarrow I subset A/I subset R/I$

So considering this, set $A = R$, then $R subset R$ is an ideal of itself and so is $R/I subset R/I$, we get what I wrote previously?

Why am I asking this? It is because I was working several problems in Dummit chapter 7.4 and at least half of them could be answered by this without writing much more, but I feel like I m cheating my way here.

Here is one that made me wrote this question

If $R$ is commutative (with $1$) then if $P$ is a prime ideal of $R$ with no zero divisors, then $R$ is integral domain.

$P$ is prime so $R/P$ is integral domain. By the correspondence $R/P approx R$. So $P$ having no zero divisors seem to play no role.

Your $pi$ is not a function in general. For example consider $R=mathbb{Z}$ and $I=2mathbb{Z}$. Then $0+2mathbb{Z}=2+2mathbb{Z}$. So what is $pi(0+2mathbb{Z})=pi(2+2mathbb{Z})$? Is it $0$ or $2$?

Of course you can arbitrarly choose for example $pi(0+2mathbb{Z}):= 0$ (and therefore $pi(2+2mathbb{Z})= 0$) and so on for every coset but then you will find out that $pi$ is no longer a homomorphism. Indeed, let's simplify notation, I will use $[x]$ instead of $x+I$. In our case $R/I={[0],[1]}$. Note that we have $[1]+[1]=[0]$. Put for example $pi([0])=0$ and $pi([1])=1$. Then

$$pi([1]+[1])=pi([0])=0$$
$$pi([1])+pi([1])=1+1=2$$

and so $pi$ is not a homomorphism. And in this particular case there is no way to fix that. The only choice of values making $phi$ a homomorphism is $pi([x])=0$. Not to mention that $R/I$ has two elements while $R$ is infinite so there's no chance for $pi$ to be surjective.

So as you can see there are no shortcuts. Also the existence of an isomorphism $R/Isimeq R$ for $Ineq 0$ is a very rare situation. First of all if $R$ is nontrivial then $R/R$ is trivial. If you are looking for an example of proper ideal then consider $R$ which is not simple (so it has a proper, nontrivial ideal). Then there's always a proper quotient $R/I$ not isomorphic to $R$, namely for any maximal ideal $I$.

Your function $pi$ is only well-defined if $I$ is the ideal ${0}$. E.g., with $R = Bbb{Z}$ and $I = 2Bbb{Z}$, we have $1 + I = 3 + I$, so if $pi(a + I) = a $, we would have to have $1 = 3$, which holds in $R/I$, but does not hold in $R$.

For each $ain R$ there are several $bin R$ such that $a+I=b+I$ (unless $I={0}$ in the first place). So it is unclear what $a$ you are referring to.