TITLE: Quotient ring being isomorphic to the initial ring
QUESTION [11 upvotes]: This may be a very basic question. 

Let $R$ be a commutative ring with unity (may be assumed to be an integral domain, if necessary) and $I$ be an ideal of $R$. Assume that $R/I$ is isomorphic to $R$. Does this mean $I=(0)$ ?

REPLY [12 votes]: There's a subtlety here: what do you mean by "is isomorphic to"?
In a setting where the constructions of $A$ and/or $B$ yield a canonical map $f:A \to B$, the phrase "$A$ is isomorphic to $B$" is often used as shorthand for saying the specific map $f$ is an isomorphism, rather than merely asserting there exists some map that is an isomorphism.
And it is true that, if the quotient map $R \to R/I$ is an isomorphism, then $I = (0)$.
But if you simply assert there is some isomorphism between $R$ and $R/I$ without implicitly requiring it to be the quotient map, then it is indeed possible for $I$ to be nontrivial as demonstrated in the other answer. For variety, I will give another example.
Let $R = \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \times \cdots$ be the product of infinitely many copies of $\mathbb{Z}$ (let's say, countably many).
There is a surjective "forget the first component and shift left by one place" map $R \to R$ whose kernel is $I = \mathbb{Z} \times 0 \times 0 \times \cdots $. Then, you have $R \cong R/I$.