query: Show that there exists a ring isomorphism $R/(I\cap J) \cong R/I \times R/J$
positive_passages.docid: 156910
positive_passages.text: Necessary and sufficient condition for $R/IJ \cong R/I \times R/J$
negative_passages.docid: 61491
negative_passages.text: Complex Ring Isomorphism
negative_passages.docid: 45627
negative_passages.text: Find a sub-ring of M(2, $\mathbb{Q}$) which is isomorphic to $\mathbb{Q}$
negative_passages.docid: 120035
negative_passages.text: Pattern in Isomorphisms between $\mathbb Z_m \times \mathbb Z_n$ and $\mathbb Z_{mn}$
negative_passages.docid: 55900
negative_passages.text: Is there a ring homomorphism $M_2(\mathbb Z)\to \mathbb Z$?
negative_passages.docid: 37393
negative_passages.text: $R$ is commutative, $I$,$J$ are ideals, $I+J=R$, then $IJ=I\cap J$
negative_passages.docid: 41398
negative_passages.text: For a ring $R$, does $\operatorname{End}_R(R)\cong R^{\mathrm{op}}$?
negative_passages.docid: 13606
negative_passages.text: Prove that $K$ is isomorphic to a subfield of the ring of $n\times n$ matrices over $F$.
negative_passages.docid: 87946
negative_passages.text: Proving that the unities of a ring form a group under multiplication
negative_passages.docid: 6896
negative_passages.text: If $A/I \cong A/J$ as rings and $I\subseteq J,$ then $I=J.$
negative_passages.docid: 57269
negative_passages.text: A special bijection between $\mathbb{R}\times \mathbb{R}$ and $\mathbb{R}$
negative_passages.docid: 2235
negative_passages.text: Showing $\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \cong S_3$
negative_passages.docid: 44983
negative_passages.text: Isomorphism of R-modules
negative_passages.docid: 149856
negative_passages.text: Image of ring homomorphism is a subring
negative_passages.docid: 123130
negative_passages.text: Showing an isomorphism between exterior products of a free module
negative_passages.docid: 55902
negative_passages.text: Show $R[x]/(x)\cong R$
negative_passages.docid: 137739
negative_passages.text: Prove that the group isomorphism $\mathbb{Z}^m \cong \mathbb{Z}^n$ implies that $m = n$