Algebras with surjection and equal length are isomorphic rings

Question

Let $A, B$ be two $\mathcal{O}$-algebras with a surjective ring homomorphism $A \rightarrow B$, such that $\operatorname{length}_{\mathcal{O}}A = \operatorname{length}_{\mathcal{O}}B$. Can we conclude that $A, B$ are isomorphic as algebras, or at least isomorphic as rings?

Answer 1

No. Let $R$ and $S$ be two fields of different characteristic — say, $R=\mathbb{Z}/(2)$ and $S=\mathbb{Z}/(3)$.

Suppose $\mathcal{O}=R\times S$. For any two cardinals $\alpha$ and $\beta$, the direct sum ring $X_{\alpha,\beta}=R^{\oplus\alpha}\oplus S^{\oplus\beta}$ is an $\mathcal{O}$-algebra, where $(r,s)(\{v_j\}_{j\in\alpha},\{w_k\}_{k\in\beta})=(\{rv_j\}_{j\in\alpha},\{sw_k\}_{k\in\beta})$. Moreover, $\mathrm{length}(X_{\alpha,\beta})=\alpha+\beta$.

But there exists a surjection $X_{\alpha_1,\beta_1}\to X_{\alpha_2,\beta_2}$ iff $\alpha_1\leq\alpha_2$ and $\beta_1\leq\beta_2$. In particular, the two algebras are isomorphic iff $\alpha_1=\alpha_2$ and $\beta_1=\beta_2$, which does not follow from $\alpha_1+\beta_1=\alpha_2+\beta_2$.

For example, $R\times S^{\omega_0}\ncong S^{\omega_0}$, but both have countable length and the latter is a homomorphic image of the former (quotient by $R$).