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23h
comment Structure of $\text{Aut}_R(R[X])$
Since Gilmer's paper is not easily available to everybody, let me just mention that it says that the $R$-automorphisms of $R[X]$ are indeed the $R$-endomorphisms mapping $X$ to a polynomial $\sum_{n\ge 0}a_nX^n$ with $a_1$ invertible and $a_n$ nilpotent for every $n\ge 2$ (as in Matthieu Romagny's comment).
1d
comment Structure of $\text{Aut}_R(R[X])$
If $R$ is not reduced, pick nonzero $b$ with $b^2=0$ in $R$, and let $P$ be any polynomial such that $bP\neq 0$, for instance $P(X)=X^2$. Define an $R$-algebra endomorphism by $f_b:X\mapsto X+bP(X)$. Then it is an automorphism with inverse $f_{-b}$. (Because $bP(X+bP(X))=bP(X)$.)
1d
answered If $R$ is generated by idempotents, then $\text{Ann}(R)=0$?
1d
comment If $R$ is generated by idempotents, then $\text{Ann}(R)=0$?
generated by idempotents in which sense? as an additive group?
2d
revised Minimal dimension of a Lie algebra of matrices, with a restrictive property
correct two last G into g
2d
revised Minimal dimension of a Lie algebra of matrices, with a restrictive property
Corrected English and typing
Mar
28
comment Maximal abelian subgroup of general linear groups
These are the intersections of the maximal abelian subalgebras of $M_n(F)$ with $GL_n(F)$. It is probably more natural to begin with the study of the latter question.
Mar
27
comment Subgroups generated by opposite root groups
If $G$ is not split, I think that $U_\alpha$ is not necessarily a subgroup; I'm not even sure how to define it, but say in char. 0, in the Lie algebra $\mathfrak{u}_\alpha$ makes sense but is not necessarily a subalgebra, if $2\alpha$ is a root, unless I misunderstand the context.
Mar
27
comment Simply connected Lie groups homeomorphic to R^n are solvable
See the book by Onishnik and Vinberg: Lie Groups and Lie Algebras III: Structure of Lie Groups and Lie Algebras. Encyclopaedia of Math. Sciences, Springer. p52 (theorem 3.2) books.google.fr/books?id=l8nJCNiIQAAC&pg=PA51&lpg=PA51
Mar
27
comment Extension of scalars and projective limits
It's just common practice. If you like to call the reals $\mathbf{C}$ and the complex numbers $\mathbf{R}$, it's fine too.
Mar
27
comment Simply connected Lie groups homeomorphic to R^n are solvable
A topological group is homeomorphic to a Euclidean space if and only if it is a semidirect product $S\ltimes R$ with $S\simeq\tilde{SL}_2(\mathbf{R})^k$ for some $k$ and $R$ a simply connected solvable Lie group.
Mar
26
comment Arbitrary chains of prime ideals in $R[X]$
It's that in the noetherian case the chains are well-ordered (for the reverse inclusion order) and then it's natural to consider the ordinal rather than the cardinal, to get a much more refined notion of Krull dimension. (Btw, are there examples in the non-noetherian case where $\dim(S[X])>\dim(S)+1$?)
Mar
26
comment Arbitrary chains of prime ideals in $R[X]$
Just to be sure: you don't assume noetherian, right?
Mar
25
comment Extension of scalars and projective limits
If you have $R\stackrel{h}\to S\stackrel{g}\to T$, then $(g\circ h)_*=g_*\circ h_*$. The notation upper star would better fit the contravariant case, i.e. when $(g\circ h)^*=h^*\circ g^*$.
Mar
25
comment Extension of scalars and projective limits
A necessary condition is that $S$ is a finitely generated $R$-module. Indeed, suppose that $h_*$ commutes with taking the $S$-fold product $M\mapsto M^S$ (where $S$ is just viewed as a set!). Then $S^S$ is generated by $h(R)^S$ as an $S$-module. In particular, we can write $\mathrm{id}_S=\sum_{i=1}^ks_if_i$ with $f_i\in h(R)^S$. Thus $s=\sum_{i=1}^kf_i(s)s_i$ for all $s\in S$. This means that $s_1,\dots,s_k$ generates $S$ as an $R$-module.
Mar
25
comment Extension of scalars and projective limits
$h^*$ is misleading because it's covariant, you should write it $h_*$
Mar
25
comment Decidable properties of the Cayley complex of a presentation
You have to say clearly what you call "Cayley complex"; I assume it includes multiple edges in case some generators are equal in the group, and self-loops in case some generators are trivial in the group. For instance, the Cayley complex of the presentation $\langle x\mid x\rangle$ of the trivial group should be one vertex, with a self-loop and a 1-gon filling this self-loop.
Mar
24
awarded  Enlightened
Mar
24
revised When do two non-degenerate quadratic forms give rise to isomorphic Lie algebras?
gave the picture for $n=2$
Mar
24
revised When do two non-degenerate quadratic forms give rise to isomorphic Lie algebras?
improved to all $n\neq 8$