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 $I=(0)$ ?
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$\begingroup$ Just a side note, “initial ring” might be interpreted as “the initial object in the category of Rings with unity”, which is $\mathbb Z$. $\endgroup$– Lukas JuhrichMay 2, 2018 at 23:17
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2 Answers
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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 $$R/I \cong \mathrm{Im} f= R$$
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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 $R \cong R/I$.
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2$\begingroup$ It is worth noting that if $R$ and $R/I$ are isomorphic as $R$-algebras, then $I$ must be trivial. $\endgroup$– tomaszApr 29, 2018 at 19:52
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$\begingroup$ Why a non-specific isomorphism between $R$ and $R/I$ as $R$- algebras will make $I$ trivial? @tomasz $\endgroup$– SamApr 30, 2018 at 6:42
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$\begingroup$ @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 $R/I\cdot I=0$. $\endgroup$– tomaszApr 30, 2018 at 11:51
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$\begingroup$ 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 $\endgroup$– SamMay 1, 2018 at 4:54
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1$\begingroup$ @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. $\endgroup$– user14972May 1, 2018 at 18:13