What is wrong with this argument? $R/I \approx R$

Question

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.

Answer 1

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

Answer 2

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

Of course you can arbitrarly choose for example $\pi(0+2\mathbb{Z}):= 0$ (and therefore $\pi(2+2\mathbb{Z})= 0$) 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 $R/I\simeq R$ for $I\neq 0$ 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 $I$.

Answer 3

Your function $\pi$ is only well-defined if $I$ is the ideal $\{0\}$. E.g., with $R = \Bbb{Z}$ and $I = 2\Bbb{Z}$, 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 $R$.