Let $R$ be an infinite commutative Artinian ring such that for any two distinct ideals $I, J$ of $R$, $R/I$ and $R/J$ has different cardinalities; then is it true that $R$ is a PIR (principal ideal ring) ? If this is true, then can we reduce the Artinian hypothesis to Noetherian ?



commutative, infinite, artinian ring (with unity) in which distinct ideals has distinct index
Let $R$ be an infinite, commutative, Artinian ring with the distinct-index property for distinct ideals. The claim is that $R$ is a field.


**Case 1.** $R$ has a nonzero ideal $I$ such that $I^2=0$.


Reasoning for this case: $I$ must be a finitely generated ideal, since Artinian rings are Noetherian. Hence $I$ is f.g. as an $R/I$-module. This implies that the cardinality of $I$ is dominated by the cardinality of some f.g. free $R/I$-module, which gives us the estimate $|I|\leq |R/I|^n$ for some finite $n$. Hence $|R|=|R/I|\cdot |I|\leq |R/I|^{n+1}.$ If $R/I$ were finite, this bound would force $R$ to be finite, which is contrary to assumption. Hence $|R/I|=\kappa$ is infinite. But now 
$$\kappa=|R/I|\leq |R|\leq |R/I|^{n+1} = \kappa^{n+1}=\kappa,$$
so the ideals $I$ and $(0)$ both have index $\kappa$. This contradicts either the assumption $I\neq 0$ or the distinct-index property. \\\


THUS, we must be in


**Case 2.** $R$ has no nonzero ideal that squares to $0$. 


Reasoning for this case: By the structure theorem for commutative Artinian rings, $R$ must be a finite product of fields, say $R\cong \mathbb F\_1\times \cdots\times \mathbb F\_k$, where $|\mathbb F\_i|=\kappa\_i$ and $\kappa\_1\leq \cdots\leq \kappa\_k$. 
Since the finite product $\prod \kappa\_i=|R|$ is infinite, it follows that this product of cardinals equals its largest factor, i.e. $|R|=\kappa\_k=|\mathbb F\_k|$. This implies that the ideals $(0)$ and 
$\mathbb F\_1\times \cdots\times \mathbb F\_{k-1}\times \{0\}$
have the same index (= $\kappa\_k$). This contradicts the distinct-index property, unless $k=1$ and $R=\mathbb F\_k$. \\\




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**Edit, to answer this question:**
*what if I only wanted that distinct **non-zero** ideals have distinct index?*


You can modify the above argument to fit this case. The result is that either 


**(i)** $R$ has at most one nonzero proper ideal, $I$, $R/I$ is infinite, and $I^2=0$, or 


**(ii)** $R$ is a product of at most $2$ fields, and the factor fields have unequal cardinality.


To see this, suppose that $R$ has a minimal ideal $I$ such that $I^2=0$. Apply the Case 1 estimate $|R|\leq |R/I|^{n+1}$ to deduce that $R/I$ is infinite and satisfies the distinct-index property for distinct ideals, even for its zero ideal. Hence by the answer to the original question $R/I$ is a field. Given this and $I^2=0$, it follows that the only ideals of $R$ are $0, I, R$ and **(i)** holds. (There are rings satisfying these conditions, like $\mathbb F[x]/(x^2)$ for $\mathbb F$ an infinite field.)


Now if there are no minimal ideals $I$ such that $I^2=0$, then as in the Case 2 argument above $R$ must be a finite product of fields, $R\cong \mathbb F\_1\times \cdots\times \mathbb F\_k$, where $|\mathbb F\_i|=\kappa\_i$ and $\kappa\_1\leq \cdots\leq \kappa\_k$. Since $R$ is infinite, $\kappa\_k$ is infinite. If there are at least $3$ factors, then the nonzero ideals $\mathbb F\_1\times \cdots\times \mathbb F\_{k-1}\times \{0\}$ and 
$\mathbb F\_1\times \cdots\times \mathbb F\_{k-2}\times \{0\}\times \{0\}$
have the same index (= $\kappa\_k$), a contradiction. If there are exactly two factors, and they have the same cardinality, then $\mathbb F\_1\times \{0\}$ and $\{0\}\times \mathbb F\_2$ are nonzero ideals that have the same index. Hence, when $R$ has no nonzero ideals that square to zero $R$ must be a product of at most two fields. If there are two factors, then they must be of different cardinalities and at least one of them must be infinite. (It is easy to produce rings like this.)


