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Pierre-Yves Gaillard's user avatar
Pierre-Yves Gaillard's user avatar
Pierre-Yves Gaillard's user avatar
Pierre-Yves Gaillard
  • Member for 15 years, 2 months
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On a theorem of Jacobson
Dear Mariano: Thanks for mentioning this question. I think one should ask if $S\cup(-S)$ [and not $S$] is maximal for the property in question.
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Ranks of free submodules of free modules
If I understand correctly, this question will be deleted. In my opinion, this would be unfortunate because the answer to it are (I think) of at least as high quality as the answers to the "Atiyah-MacDonald, exercise 2.11" question (mathoverflow.net/questions/136/atiyah-macdonald-exercise-2-‌​11). The best would be of course to append the answers to this question to the answers to the other question. But if this is too complicated, it would be better to reopen this question.
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Which manifolds admit a diffeomorphism of order $n$?
Dear Andrey Rekalo: The definition given by Łukasz is used for any element of any group. (This has nothing to do with diffeomorphisms.)
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is there a solution for equation $\arcsin((1-x)^{1/2})=\arccos(x^{1/2})$ in which $x$ is rational number
Dear Jacques Carette: David Speyer wrote above: "I don't understand. For every value of $x$ between $0$ and $1$, we have $\sin^{-1}((1-x)^{1/2}) = \cos^{-1}(x^{1/2})$. So take $x$ to be any rational number in this range." What's wrong with this argument?
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Cardinality of maximal linearly independent subset
Dear kwan: Here is how I understand your answer. Lazarus proves the existence of non-equipotent maximal linearly independent subsets of modules over commutative rings. Is this interpretation of your wording correct?
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Abstract thought vs calculation
Jacobson's Theorem says that the subset $S$ of $\mathbb Z[X]$ formed by the polynomials $X^n-X$, $n > 1$, has the following property. If, for every $a$ in any given ring $A$, there is an $f$ in $S$ such that $f(a)=0$, then $A$ is commutative. Is $S\cup-S$ maximal for this property?
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Nontrivial theorems with trivial proofs
Replaced "associative" by "commutative"
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The isomorphism inference rule
Why don't you correct it?
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The isomorphism inference rule
You wrote "When a group A is isomorphic to a group B and the group G is simple, then we can infer that the group B is also simple." Don't you mean A instead of G?
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answered
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Unbounded operator bounded in a dense subset
It looks like the best possible answer!
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What interesting/nontrivial results in Algebraic geometry require the existence of universes?
Dear Harry: Thank you very much for your answer, which confirms my intuition. Would it be indiscreet to ask: What is your personal position on these questions?
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