# user1306

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 Feb24 answered Irreducible action Lie algebras Oct18 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. Sep3 revised Field with natural numbersmaking the notation consistent Sep3 suggested suggested edit on Field with natural numbers Sep3 comment Field with natural numbersBrian 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. Aug31 revised Standard parabolic Lie subalgebras and conjugacycorrected example Aug31 comment Standard parabolic Lie subalgebras and conjugacyOf course, I wrote the Levi subalgebras. Thanks for the correction. Aug31 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. Aug31 awarded Critic Aug31 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". Aug31 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. Aug31 comment Simple Trace QuestionThis 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. Aug31 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). Aug31 awarded Editor Aug31 revised Simple Trace Questionformatting Aug31 suggested suggested edit on Simple Trace Question Aug31 suggested suggested edit on Proof of the result below without the primitive element theorem? Aug31 asked Standard parabolic Lie subalgebras and conjugacy Aug22 awarded Yearling Aug22 answered $\mathbb{Z}/n\mathbb{Z}[T]$ and zero divisors