Quotient ring being isomorphic to the initial ring

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 $I=(0)$ ?

Solution 1:

Counterexample:

$R=\Bbb Q [X_1, X_2, \dots , X_n, \dots]$. Define the ring morphism $f: R \to R$ by $$\begin{cases}X_i \mapsto X_{i-1} & i>1 \\ X_1 \mapsto 0 \end{cases} $$

and call $I= \ker f$.

Then $$R/I \cong \mathrm{Im} f= R$$

Solution 2:

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 $R \cong R/I$.