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visits member for 2 years, 5 months
seen Feb 9 '13 at 12:37

Feb
5
accepted How do I define Polynomial ring $R[x]$ over a commutative ring $R$ with unity?
Feb
5
comment How do I define Polynomial ring $R[x]$ over a commutative ring $R$ with unity?
Aha! I wonder why author of my textbook didn't write that one simple word 'sequence'.. Thank you
Feb
5
asked How do I define Polynomial ring $R[x]$ over a commutative ring $R$ with unity?
Feb
4
comment How do i prove that any two distinct irreducible monic polynomials in a field are relatively prime?
@rschwieb Yes, I made that assumption. Thank you!
Feb
4
comment How do i prove that any two distinct irreducible monic polynomials in a field are relatively prime?
@rschwieb For example, $Z_2=\{0,1\}$ is a field if $1+1=0, 1+0=1, 0•0=0, 1•0=0$. Here, polynomials $f(x)=x^2 + x + 1$ and $g(x)=1$ are not equal polynomials, but they are equal functions.
Feb
4
comment How do i prove that any two distinct irreducible monic polynomials in a field are relatively prime?
@rschwieb I was confused with the notion "$=$". Is there a special notation to distinguish "Equivalence in $F[X]$" from "equivalence in $F$"? I don't understand why one does not use some equivalence notation such as $\equiv$ for polynomials..
Feb
4
accepted How do i prove that any two distinct irreducible monic polynomials in a field are relatively prime?
Feb
4
comment How do i prove that any two distinct irreducible monic polynomials in a field are relatively prime?
@Math This is an exercise in my book, so I'm trying to prove it myself. There is no reference.
Feb
4
asked How do i prove that any two distinct irreducible monic polynomials in a field are relatively prime?
Feb
4
accepted Is there a ring which doesn't satisfy Division Algorithm for polynomials?
Feb
4
asked Is there a ring which doesn't satisfy Division Algorithm for polynomials?
Jan
25
comment Lebesgue measure, Borel sets and Axiom of choice
It's very clear now. Thank you!
Jan
25
accepted Lebesgue measure, Borel sets and Axiom of choice
Jan
25
comment Lebesgue measure, Borel sets and Axiom of choice
Isn't Axiom of countable choice necessary to prove the 'Existence of a completion of Borel measure'? If we cannot prove the existence of the lebesgue measure, how does that make sense to say "There exists a non-Lebesgue measurable set is unprovable in ZF"?
Jan
25
comment Lebesgue measure, Borel sets and Axiom of choice
@Michael I have searched both here and mathoverflow before i post this question, but i couldn't find it. Would you please tell me the link?
Jan
25
revised Lebesgue measure, Borel sets and Axiom of choice
added 1 characters in body
Jan
25
asked Lebesgue measure, Borel sets and Axiom of choice
Jan
24
comment What is the usual construction of Lebesgue measure?
@Simen So far, I showed that there exists a sigma algebra $\mathfrak{M}$ on $\mathbb{R}^k$ and a measure $\mu$ such that (1)$\mathfrak{M}$ contains all the Borel sets and (2)$\mu$ is complete.
Jan
24
comment What is the usual construction of Lebesgue measure?
@Matt I'm sorry, but I don't understand what do you mean by 'domain of the Lebesgue measure is the sigma algebra of the Lebesgue measurable sets'. I asked the domain of the Lebesgue measure, in other words I asked what is that sigma algebra on $\mathbb{R}^k$. Of course it is the sigma algebra of Lebesgue measurable sets..
Jan
24
comment What is the usual construction of Lebesgue measure?
@Matt Is the domain of Lebesgue measure is exactly the completion of the set of all Borel sets of $\mathbb{R}^k$? And what is that uniqueness theorem called?