What does "$\cong$" sign represent? - Mathematics Stack Exchange
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 …
Proof of $ (\mathbb {Z}/m\mathbb {Z}) \otimes_\mathbb {Z} …
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 …
Difference between "≈", "≃", and "≅" - Mathematics Stack Exchange
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 …
Is $\mathcal {O}_K/\mathfrak {P}^e \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, …
abstract algebra - Prove that $\mathbb Z_ {m}\times\mathbb Z_ …
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. Any help would be …
Computing the Canonical bundle $K_ {\mathbb {P}^n} \cong …
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 …
Notations involving squiggly lines over horizontal lines
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 $e$ is an idempotent element of $R$ but not a central …
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 …
How do I show $(\\mathbb{Z} / 12 \\mathbb{Z})^{*} \\cong …
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 $ …
Clarifying what is meant by 'identified' for $\mathrm {SU}\left (2 ...
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 …