Dec
18
reviewed Leave Closed How to select the right books?
Dec
18
reviewed No Action Needed Group theory applications along with a solved example
Dec
18
comment Blackboard bold, Bold, Fraktur, and Reserved Variable.
It is user preference based on discipline and context. Notation nazis who are convinced there is a "right notation" exist but are usually politely ignored. Just do your best to be clear to your intended audience.
Dec
17
reviewed Looks OK Blocks of Pyramid Pattern Expression
Dec
17
reviewed Approve For any $p,q\in\mathbb{Z}[i]$, $\mathrm{N}(\gcd(p, q))$ must divide $\gcd(\mathrm{N}(p), \mathrm{N}(q))$
Dec
17
comment What's the algebraic closure of the quaternions?
@MannyReyes I for one don't have any idea what the proposed limit looks like. Maybe the quotients are all domains and the inductive limit is too? It would seem like a bit of a miracle if the whole thing was a division ring...
Dec
17
comment What's the algebraic closure of the quaternions?
@AdamHughes Then maybe this really is the brick wall, since $R\langle x\rangle/(xi+ix-j)$ does not make $x$ algebraic, as was hoped.
Dec
17
comment Abstract Linear Transformation Question
Do you know what $T^k$ means? Isn't it clear what happens each time you apply $T$?
Dec
17
comment Gallian: is it true that the well-ordering principle can't be proven from properties of arithmetic?
If by "that" you mean the link, then yes, that is a proof of the well-ordering theorem that only relies on set theory.
Dec
17
comment Gallian: is it true that the well-ordering principle can't be proven from properties of arithmetic?
@AnalysisIncarnate I can't offer any disproof of the existence of a proof without knowing what the axioms are that he is talking about. To successfully prove there is no proof, you'd have to give an example which isn't well-ordered by nevertheless satisfies the "arithmetic axioms" that he is talking about.
Dec
17
comment Gallian: is it true that the well-ordering principle can't be proven from properties of arithmetic?
OK, must be a typo on his part then. The integers are of course not well ordered.
Dec
17
comment Gallian: is it true that the well-ordering principle can't be proven from properties of arithmetic?
@AnalysisIncarnate The ring axioms for $\Bbb Z$ do not even mention order, so I can't imagine how one would talk about well ordering in any case.
Dec
17
comment Prerequisites for Hartshorne: Euclid and beyond?
Hartshorne is really a wonderful author. Based on my reading, I thought it would be accessible to anyone comfortable with proof classes, and not much other prerequisite would be necessary. It probably differs from person to person, but I won't let that stand in the way of my recommendation of this book :)
Dec
17
comment What's the algebraic closure of the quaternions?
Sebastien, you want to read this paper by Lam on the quaternions for that theorem. It's an awesome paper for anyone interested in the quaternions :)
Dec
17
comment What's the algebraic closure of the quaternions?
I don't even know that it exists, and I have much less chance of knowing if it's finite dimensional over whatever field/division ring you are imagining.
Dec
17
revised Gallian: is it true that the well-ordering principle can't be proven from properties of arithmetic?
better title
Dec
17
comment Gallian: is it true that the well-ordering principle can't be proven from properties of arithmetic?
I sure hope he didn't claim the integers are well ordered...
Dec
17
answered Gallian: is it true that the well-ordering principle can't be proven from properties of arithmetic?
Dec
17
comment What's the algebraic closure of the quaternions?
The "best" theorem I know for any notion of algebraic closure of $\Bbb H$ in terms of free-polynomials $\Bbb H\langle x\rangle$ is this: every polynomial in $\Bbb H\langle x\rangle$ whose highest degree term is a single monomial has a root in $\Bbb H$. The polynomial $xi+ix-j$ fails this because the part of degree $1$ has two pieces. But, for example, $xkx+xi+ix-j$ would have a root, since its highest degree is $2$ and it only has $xkx$ in that degree.
Dec
17
comment What's the algebraic closure of the quaternions?
Still, I guess the inductive system still exists, and indeed $p(x)$ has a root in the quotient ring, so maybe we get a remnant of what we wanted? It's been a while for me too, so I can't remember if the finiteness of the extensions in the inductive system is vital.