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 …

Oct 11, 2014 · 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 ...

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, …

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 …

The most obvious way to show this is by using the Chinese Remainder Theorem to see that $ {\mathbb Z}_ {12} \cong {\mathbb Z}_4 \times {\mathbb Z}_3$ (as rings) and therefore also $ …

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 …

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 $2$.

Jan 26, 2021 · Hence $R^m \cong R^n$ for any natural numbers $m \neq n$? I could simply answer this question with " $R^m$ is not module isomorphic to $R^n$ for $n\neq m$ because …

Dec 5, 2025 · In the fifth example in page $19$ the author explains why $\mathrm {SU} \left (2 \right) \cong \mathbb {S}^ {3}$ as real Lie groups. There is a step I don't seem to understand …