# 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 $$I=(0)$$ ?

• Just a side note, “initial ring” might be interpreted as “the initial object in the category of Rings with unity”, which is $\mathbb Z$. May 2, 2018 at 23:17

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

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 $$R \cong R/I$$.

• It is worth noting that if $R$ and $R/I$ are isomorphic as $R$-algebras, then $I$ must be trivial. Apr 29, 2018 at 19:52
• Why a non-specific isomorphism between $R$ and $R/I$ as $R$- algebras will make $I$ trivial? @tomasz
– Sam
Apr 30, 2018 at 6:42
• @Sam: To be slightly more specific, I guess it implies that $I$ is trivial in the sense that $RI=0$. This is easy to see, as $R/I\cdot I=0$. Apr 30, 2018 at 11:51
• My current interest is in the following for which I hope the answer to be true but unable to prove it. Let $D\subseteq\mathbb{C}^n$ be a domain and $\mathcal{O}_{D}$ be the sheaf of holomorphic functions on $D$. Define the ideal sheaf $\mathcal{I}_{D}:=\mathcal{O}_{D}f_{1}+\cdots +\mathcal{O}_{D}f_{k}$. Now, for each fixed $x\in D$, we have $\mathcal{O}_{D,x}/\mathcal{I}_{D,x}\simeq \mathcal{O}_{D,x}$ as $\mathbb{C}$-algebras. Does this mean that $\mathcal{I}_{D,x}\simeq {0}$ for each $x\in D$ or equivalently, $\mathcal{I}_{D}={0}?$ - Does this go along the lines of what you said? @tomasz
– Sam
May 1, 2018 at 4:54
• @Sam: An $R$-algebra structure on a ring $S$ is determined by the action on the identity: $r \cdot s = r \cdot (1_S s) = (r \cdot 1_S)s$, and $r \mapsto r \cdot 1_S$ gives a homomorphism $R \to S$ (conversely, every such homomorphism gives an $R$-algebra structure on $S$). If you compute $R/I$ as a quotient of $R$-algebras rather than as a quotient of rings, the $R$-algebra structure on $R/I$ is the one whose corresponding homomorphism is the quotient map.
– user14972
May 1, 2018 at 18:13