安裝中文字典英文字典辭典工具!
安裝中文字典英文字典辭典工具!
|
- 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 mapping known as an "isometry", which provides a formal definition of what "same shape and size" means) Two congruent triangles look exactly the same, but they are not the
- 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
- abstract algebra - Prove that $\mathbb Z_ {m}\times\mathbb Z_ {n} \cong . . .
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 appreciated
- abstract algebra - On proving that $\operatorname {Aut} A_n \cong . . .
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 regard it as a trivial fact
- Is $\\operatorname{Hom}_k(R,k) \\cong E(k)$ for a local ring $R$?
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 rings this is true?
- Proof of $ (\mathbb {Z} m\mathbb {Z}) \otimes_\mathbb {Z} (\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 $\mathbb {Z} \text {gcd} (m,n)$ $\textit {is}$ the tensor product
- ring theory - Is $\mathbb {Z}_p \cong \mathbb {Z} [ [X]] (X-p . . .
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 it's not obvious that $\mathbb {Z} [ [X]] (X-p)$ should actually be the inverse limit of $\mathbb {Z} [X] (X^n,X-p)$
- $\\hom (M, \\coprod_i N_i) \\cong \\bigoplus_i \\hom (M, N_i)$ in . . .
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 that every mono is either an iso or zero 3 The nlab gives the dual definition (using epis rather than monos) whereas wikipedia uses your definition This certainly at least suggests that the
|
|
|