In geometry, $\cong$ means congruence of figures, which means the figures have the same shape and size. (In advanced geometry, it means one is the image of the other under a mapping known as an …

3 days ago · I am currently studying the number of solutions to $x^2 \equiv y^2 \pmod {p^q}$ in the ring $\mathbb {Z}_ {p^q} [\sqrt \delta] \cong \mathbb {Z}_ {p^q} [t] / (t^2 - \delta)$, where $p>2$ is a prime …

Originally you asked for $\mathbb {Z}/ (m) \otimes \mathbb {Z}/ (n) \cong \mathbb {Z}/\text {gcd} (m,n)$, so any old isomorphism would do, but your proof above actually shows that $\mathbb {Z}/\text {gcd} …

In mathematical notation, what are the usage differences between the various approximately-equal signs "≈", "≃", and "≅"? The Unicode standard lists all of them inside the Mathematical Operators B...

Prove that $\mathbb Z_ {m}\times\mathbb Z_ {n} \cong \mathbb Z_ {mn}$ implies $\gcd (m,n)=1$. This is the converse of the Chinese remainder theorem in abstract algebra.

May 4, 2025 · Yes, this is correct. The isomorphism is not just of $\mathbb {Z}$ -modules (which would just be the same as additive groups), but of rings. However, as user Lullaby points out, your …

Jan 1, 2025 · I went through several pages on the web, each of which asserts that $\operatorname {Aut} A_n \cong \operatorname {Aut} S_n \; (n\geq 4)$ or an equivalent statement without proof, and many …

Feb 10, 2025 · Let $R$ be a ring with unity, and let $e$ be an idempotent element of $R$ such that $e^2 = e$. If $e$ is a central idempotent of $R$, then we obtain the following ring isomorphism: $$ R/ReR …

Sep 28, 2011 · The symbol $\cong$ can in principle be used to designate an isomorphism in any category (e.g., isometric, diffeomorphic, homeomorphic, linearly isomorphic, etc.).

This approach uses the chinese remainder lemma and it illustrates the "unique factorization of ideals" into products of powers of maximal ideals in Dedekind domains: It follows $-1 \cong 10-1 \cong 9$ …