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What does "$\cong||quot; 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 mapping known as an "isometry", which provi...
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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 Operators B...
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$\Bbb Z [i]/ (a+bi)\cong \Bbb Z/ (a^2+b^2)$ if $ (a,b)=1$....
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 $-1 \cong 10-1 \cong 9$ hence you get a ...
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abstract algebra - On proving that $\operatorname {Aut} A_n \cong...
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 of them seem to...
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Is $\\operatorname{Hom}_k(R,k) \\cong E(k)$ for a local ring $R$?
Jun 16, 2024 · Another injective cogenerator is the injective hull of the residue field $E (k)$. How this two injective cogenerators are related in general? Is it true that $M \cong E (k)$? If not in general for what ...
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abstract algebra - Prove that $\mathbb Z_ {m}\times\mathbb Z_ {n}...
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.
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ring theory - Is $\mathbb {Z}_p \cong \mathbb {Z} [ [X]]/ (X-p...
Nov 5, 2023 · A tempting cleaner-looking way would be to observe that $\mathbb {Z} [X]/ (X^n,X-p)\cong \mathbb {Z}/ (p^n)$ and take the inverse limit. I don't see a way to make that work purely formally, though, since...
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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 $\mathbb {Z}/\text {gcd} (m,n)$ $\tex...
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$\\hom (M, \\coprod_i N_i) \\cong \\bigoplus_i \\hom (M, N_i)$ in...
Feb 15, 2015 · A few points on the definition of "simple": 1. Every mono (and epi) is regular in an abelian category, so no need to worry about that. 2. The zero morphism $0 \to X$ is always mono, so you need to say t...
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Proving the isomorphism $Pic (X) \cong H^1 (X,\mathcal {O}^*_X) $...
Feb 6, 2025 · The proof given for the theorem is the following: "The description of line bundles in terms of their cocycles provides us with an isomorphism $\operatorname {Pic} (X) \cong \check {H^1} (X,\mathcal {O}^*_X) $ ".