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Oct
12
reviewed Approve suggested edit on Why does $M\otimes k(\mathfrak{m})=M_\mathfrak{m}/\mathfrak{m}M_\mathfrak{m}$? (From Matsumura, proof of Theorem 4.8.)
Oct
6
comment relationship between Exact sequences and Universal mapping property
@Jason Nope, in the first part I just showed $K\simeq K'$, since you asked if they were necessarily isomorphic.
Oct
6
answered relationship between Exact sequences and Universal mapping property
Oct
5
answered Tensor product of abelian groups
Oct
4
comment Proving if $a$ and $b$ are positive rational numbers and $\mathbb Q(\sqrt{a})=\mathbb Q(\sqrt{b})$ then $b=ac^2$ for some $c\in \mathbb Q$.
@Vincent Yes, that's it.
Oct
4
comment Proving if $a$ and $b$ are positive rational numbers and $\mathbb Q(\sqrt{a})=\mathbb Q(\sqrt{b})$ then $b=ac^2$ for some $c\in \mathbb Q$.
@Vincent If $d\neq 0$, you can isolate $\sqrt{a}$ in the displayed equation by subtracting $d^2$ and $c^2a$, and dividing by $2dc$, since $2dc\neq 0$. That expression for $\sqrt{a}$ will be rational.
Oct
4
answered Proving if $a$ and $b$ are positive rational numbers and $\mathbb Q(\sqrt{a})=\mathbb Q(\sqrt{b})$ then $b=ac^2$ for some $c\in \mathbb Q$.
Oct
4
revised Proving $(ab)^{-1}=a^{-1}b^{-1}$ where $F$ is a field and $a,b\in F$.
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Oct
4
comment Proving $(ab)^{-1}=a^{-1}b^{-1}$ where $F$ is a field and $a,b\in F$.
@Vincent Right, but $\frac{1}{[7]_{13}}=[2]_{13}$ as well.
Oct
4
comment Proving $(ab)^{-1}=a^{-1}b^{-1}$ where $F$ is a field and $a,b\in F$.
@Vincent What does $\frac{1}{[7]_{13}}$ mean if not the multiplicative inverse of $7$ modulo $13$?
Oct
4
answered Proving $(ab)^{-1}=a^{-1}b^{-1}$ where $F$ is a field and $a,b\in F$.
Oct
2
revised A relation involving an endomorphism of a finitely generated module
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Oct
2
answered A relation involving an endomorphism of a finitely generated module
Sep
30
awarded  Explainer
Sep
24
awarded  Autobiographer
Sep
23
comment A4 has no subgroup of order 6
@user156441 It's a well known fact that for any $n\geq 3$, $A_n$ is generated by $3$-cycles (here is a proof). So if $H$ contained all $3$-cycles, it would be $A_4$.
Aug
26
answered If $G$ is a simple $f$ an homomorphism, and $A\lhd H$ is such that $[H:A]=2$, show $f(G) \subset A$
Aug
25
comment Does $G$ always have a subgroup isomorphic to $G/N$?
@JorgeFernández Yes, take a look at this question.
Aug
25
answered Does $G$ always have a subgroup isomorphic to $G/N$?
Aug
25
comment Is any submanifold of $\mathbb{R}^{n}$ the zero set of some polynomial?
@QuangHoang Yes, open sets are (embedded) submanifolds.