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 …

5 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 …

I've just started to learn about the tensor product and I want to show: $$ (\mathbb {Z}/m\mathbb {Z}) \otimes_\mathbb {Z} (\mathbb {Z} / n \mathbb {Z}) \cong \mathbb ...

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.

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 …

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, …

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.).

If $Aut (G)\cong \mathbb {Z}_n$ then $Aut (G)$ is cyclic, which implies that $G$ is abelian. But if $G$ is abelian then the inversion map $x\mapsto x^ {-1}$ is an automorphism of order $2$.

Aug 22, 2023 · Q1: Yes, this is the definition of the determinant of a one-dimensional vector space. Q2: Yes, the dual of the trivial line bundle is the trivial line bundle (for instance, use that …

Sep 28, 2024 · Claim: $\operatorname {Hom}_ {G} (V,W) \cong \operatorname {Hom}_ {G} (\mathbf {1},V^ {*} \otimes W)$ I'm looking for hints as to how to approach the proof of this claim.