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I am a graduate student at UCLA.


2h
comment If R is $(a,b)R(c,d) \iff a+d =b+c$ show that R is an equivalence relation.
Yes, it's correct. The exercise hints at the construction of the integers from the natural numbers; if you didn't have the integers to begin with, you wouldn't be able to write down "$a-b$", and then you'd have to do a bit more work to verify it's an equivalence relation. If the exercise makes no mention of this (or some similar) caveat, though, then you're probably good with this solution.
2d
comment Please justify this statement: “[A] holomorphic function is (n+1)-to-1 near a zero of its derivative of order n”.
It's a standard fact of complex analysis, following from the Taylor expansion around the point.
Jul
29
comment Why does $x$ vanish in the group $\langle x,y\mid x^3, y^3, yxyxy\rangle$?
@B.S. You're welcome to if you want to. I was indeed looking for an alternative to TC in this question. :-)
Jul
28
comment Why does $x$ vanish in the group $\langle x,y\mid x^3, y^3, yxyxy\rangle$?
@B.S. I don't have my copy of Artin handy, but I believe when I asked this problem, it was because he reduced it explicitly using Todd-Coxeter, and not by directly manipulating the relations.
Jul
25
awarded  Nice Answer
Jul
25
awarded  Mortarboard
Jul
25
comment A question about algebraically closed fields
Wow, you might want to check your tone there.
Jul
25
comment A question about algebraically closed fields
Then I invite someone else to intercede, as I will be on the road for the next 7 hours. :-) Have you read rschwieb's answer?
Jul
25
comment A question about algebraically closed fields
If you take the union of all $\mathbb F_{p^{(N!)^i}}$, for $i>0$, then you get an algebraically closed in practice field which is not algebraically closed. It is algebraically closed in practice because every polynomial with coefficients in the field has coefficients in some fixed $\mathbb F_{p^{(N!)^i}}$, which then splits in $\mathbb F_{p^{(N!)^{i+1}}}$. It is not algebraically closed because it does not contain $\mathbb F_{p^q}$ for $q$ a prime bigger than $N$.
Jul
25
comment A question about algebraically closed fields
Oh, you're right, nevermind, I just understood why my example doesn't work. I suspect you can still make the idea behind it work, and it will have to be, as you say, an infinite field, but I'll have to think about it.
Jul
25
comment A question about algebraically closed fields
It'll be a little confusing if you haven't seen the theory of finite fields before. Finite fields are a nice go-to example for some of the simpler questions in Galois theory. Have a look at the wikipedia page on splitting fields and see if that helps. Also finite fields; assuming you're familiar with the field obtained by taking the integers modulo a prime number $p$, the general finite fields are obtained by "extending" this simplest case of a finite field, via splitting fields.
Jul
25
comment A question about algebraically closed fields
My point is that the splitting field is not algebraically closed, as it is a finite extension of $\mathbb F_p$, and the algebraic closure of $\mathbb F_p$ is infinite.
Jul
25
comment A question about algebraically closed fields
Easily. Take the splitting field over a finite field ($\mathbb F_p$) of the polynomial $f(x)$ obtained by multiplying all polynomials of degree $\le 10^{10}$ together.
Jul
25
answered In Cantor's Diagonalization Argument, why are you allowed to assume you have a bijection from naturals to rationals but not from naturals to reals?
Jul
25
comment What are some 'conceptualizations' that work in mathematics but are not strictly true?
In particular one could illuminate the ambiguity in the phrase "$\lim_{x \to a} f(x)$ is the value of $f(x)$ as $x$ gets closer and closer to $a$" by giving alternate formal interpretations, and giving examples to demonstrate that these alternate interpretations are wrong.
Jul
25
awarded  Good Answer
Jul
24
comment What are some 'conceptualizations' that work in mathematics but are not strictly true?
@StevenStadnicki I just realized that there's something confusing me on this point. I know there are non-measurable sets whose construction depends on AC; are there sets whose existence can be proven within ZF but which can only be proven immeasurable with choice?
Jul
24
comment how many words can be formed using all letters in the word EXAMINATION
@mistermarko not on math stackexchange, it isn't. ;-)
Jul
24
awarded  Nice Answer
Jul
24
comment What are some 'conceptualizations' that work in mathematics but are not strictly true?
@Brady what does $ f_x $ mean? There are two possible distinct answers.