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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

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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

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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 
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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 
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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 
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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 
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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 
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Arbitrary chains of prime ideals in $R[X]$
It's that in the noetherian case the chains are wellordered (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 nonnoetherian case where $\dim(S[X])>\dim(S)+1$?) 
Mar 26 
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Arbitrary chains of prime ideals in $R[X]$
Just to be sure: you don't assume noetherian, right? 
Mar 25 
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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 
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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 
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Extension of scalars and projective limits
$h^*$ is misleading because it's covariant, you should write it $h_*$ 
Mar 25 
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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 selfloops 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 selfloop and a 1gon filling this selfloop. 
Mar 24 
awarded  Enlightened 
Mar 24 
revised 
When do two nondegenerate quadratic forms give rise to isomorphic Lie algebras?
gave the picture for $n=2$ 
Mar 24 
revised 
When do two nondegenerate quadratic forms give rise to isomorphic Lie algebras?
improved to all $n\neq 8$ 