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Oct
26
comment Every $R$-module is free $\implies$ $R$ is a division ring
@Pete: I second Amitesh's comment, for sure, and I thank you for a wonderful set of notes! I didn't know the noncommutative algebra notes; I looked up the argument there and expanded the answer (which, in fact, was a bit confusing, as I myself was confused by the matter of sidedness). Thank you.
Oct
26
revised Every $R$-module is free $\implies$ $R$ is a division ring
edited to clarify
Oct
26
answered Every $R$-module is free $\implies$ $R$ is a division ring
Oct
26
accepted Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution?
Oct
26
comment Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution?
Thank you very much, it's crystal clear. In fact, after the first simple manipulation, the proof is of the general fact that a function with a constant limit in infinity must have vanishing derivative in infinity, which is geometrically obvious. I will accept, though, the other answer as it adresses the original question, I hope you will understand.
Oct
23
accepted Particular case of Nakayama's lemma
Oct
23
answered Particular case of Nakayama's lemma
Oct
23
accepted Stability of autonomous linear systems of ODEs
Oct
23
accepted Square roots of $-1$ in quaternion ring
Oct
23
revised Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution?
edited to clarify hypothesis, and add particular case I'm interested in
Oct
23
suggested approved edit on Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution?
Oct
23
comment Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution?
@JuliánAguirre Would you please elaborate? I'm very interested in the autonomous case.
Oct
23
asked Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution?
Oct
22
revised Stability of autonomous linear systems of ODEs
added 481 characters in body
Oct
21
asked Stability of autonomous linear systems of ODEs
Oct
18
awarded  Yearling
Oct
12
comment Particular case of Nakayama's lemma
@Jack Ah, I see. Using elements as I was doing, it's $m=an_1\in M \Rightarrow m=a^2n_2\in M \Rightarrow \dots \Rightarrow m=a^kn_k=0$. Thank you.
Oct
12
asked Particular case of Nakayama's lemma
Oct
10
comment Isn't it cheating to consider an $ \mathbb{R}^3 $ vector as a “pure quaternion”?
@Mariano: what is MU?
Oct
1
comment Square roots of $-1$ in quaternion ring
Ouch, I found my mistake. I must remember that the binomial theorem only works in commutative rings. Thank you!