user1306

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	441 reputation	bio	website		visits	member for	2 years, 10 months
28 badges		location			seen	May 4 at 12:22

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Feb 24	answered	Irreducible action Lie algebras
Oct 18	comment	Transcendental extensions of $\mathbb{Q}$ containing algebraic elements. @KevinCarlson There is a trivial case that $a=b$, and then $a=\frac{av+b}{v+1}$. For a less trivial example, let $a=\sqrt{2}$ and $b=\sqrt{6}$. Then $(4v)^{-1}((av+b)^2-2v^2-6)=\sqrt{3}$. So it does happen.
Sep 3	revised	Field with natural numbers making the notation consistent
Sep 3	suggested	suggested edit on Field with natural numbers
Sep 3	comment	Field with natural numbers Brian and Gerry beat me to the punch, but to supplement their answers, the Wikipedia page quote probably means "the smallest field containing the integers as a subring (in the usual sense) is $\mathbb Q$"; that's because one way to define $\mathbb Q$ is as the "field of fractions" of $\mathbb Z$, so this is true essentially by definition.
Aug 31	revised	Standard parabolic Lie subalgebras and conjugacy corrected example
Aug 31	comment	Standard parabolic Lie subalgebras and conjugacy Of course, I wrote the Levi subalgebras. Thanks for the correction.
Aug 31	comment	Why do we need to learn integration techniques? In my opinion, saying that learning integration techniques involves no critical thinking is ridiculous. Even in the extreme situation that students are only learning methods and none of the thinking to get to them, you are still excluding the process of deciding the best method to use, which in many problems is nontrivial. If your complaint is that sometimes classes just teach the methods with no nontrivial examples, then it sounds like something is wrong with those classes, not the practice in general.
Aug 31	awarded	Critic
Aug 31	comment	Simple Trace Question @DonAntonio, that was my point: you showed the subspace GENERATED by matrices which are commutators is equal to the subspace of traceless matrices. In other words, "If the trace is zero, the matrix is a linear combination of commutators of matrices". The OP's question is how to prove the stronger statement "If the trace is 0, show it is the commutator of two matrices".
Aug 31	comment	Simple Trace Question @AsinglePANCAKE, let me call the basis elements $e,f,h$ respectively (so $h$ is the diagonal one). Then check that $[e,f]=h$, $[h,e]=2e$ and $[h,f]=-2f$, so those elements are indeed in M'. This is the standard basis for traceless 2 by 2 matrices viewed as a Lie algebra, by the way.
Aug 31	comment	Simple Trace Question This doesn't actually answer the question; the question is whether or not every traceless matrix is the commutator of two matrices. The missing ingredient is showing that the sum of commutators is a commutator (since scalar multiplication is trivial), and thus that the span of commutators = set of commutators.
Aug 31	comment	Simple Trace Question @joriki, it seems to me that the question you reference wags the dog on this; it (sort of) assumes the the sum of commutators is a commutator, which is far from obvious (at least to me).
Aug 31	awarded	Editor
Aug 31	revised	Simple Trace Question formatting
Aug 31	suggested	suggested edit on Simple Trace Question
Aug 31	suggested	suggested edit on Proof of the result below without the primitive element theorem?
Aug 31	asked	Standard parabolic Lie subalgebras and conjugacy
Aug 22	awarded	Yearling
Aug 22	answered	$\mathbb{Z}/n\mathbb{Z}[T]$ and zero divisors