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Aug 9, 2025 · Are you sure that $\left\ {\pmatrix { a & b\\0&a^2}\mid a,b\in K, a\ne 0\right\}\cong (K,+)$?
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 well defined map $$\phi: \mathbb {Z} [i] …
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Nov 13, 2015 · A symbol I have in my math homework looks like a ~ above a =. (That is, $\\cong$.) What does this mean? I'm studying Congruency at the moment if that helps.
Aug 22, 2023 · Q1: Yes, this is the definition of the determinant of a one-dimensional vector space. Q2: Yes, the dual of the trivial line bundle is the trivial line bundle (for instance, use that a line bundle is trivial iff it has a non-vanishing global section). Your proof is correct. (In my opinion the hard part is the part where you go from the sequence to the fact about the determinant.)
In mathematical notation, what are the usage differences between the various approximately-equal signs '≈', '≃', and '≅'? The Unicode standard lists all of them inside the Mathematical Operators B...
Originally you asked for $\mathbb {Z}/ (m) \otimes \mathbb {Z}/ (n) \cong \mathbb {Z}/\text {gcd} (m,n)$, so any old isomorphism would do, but your proof above actually shows that $\mathbb {Z}/\text {gcd} (m,n)$ $\textit {is}$ the tensor product.
In geometry, $\cong$ means congruence of figures, which means the figures have the same shape and size. (In advanced geometry, it means one is the image of the other under a mapping known as an 'isometry', which provides a formal definition of what 'same shape and size' means) Two congruent triangles look exactly the same, but they are not the ...
Feb 10, 2025 · Let $R$ be a ring with unity, and let $e$ be an idempotent element of $R$ such that $e^2 = e$. If $e$ is a central idempotent of $R$, then we obtain the following ring isomorphism: $$ R/ReR \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 regard it as a trivial fact.
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