555win cung cấp cho bạn một cách thuận tiện, an toàn và đáng tin cậy [cong chua ori p3]
Aug 9, 2025 · You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how …
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 a line …
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 …
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 …
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 …
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 …
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.
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 $2$.
In Dummit & Foote, it is an exercise to show that $\mathbb Q \otimes_\mathbb Z \mathbb Q$ is a $1$-dimensional $\mathbb Q$-vector space. This is fairly easy: a $\mathbb Q$-basis for …
Feb 23, 2023 · 1 Prove that Aut ($\mathbb Z \times \mathbb Z$) $\cong$ $\text {GL}_2 (\mathbb Z)$. This is a HW problem for an Algebra course, hints/suggestions welcome. I didn't find this …
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