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seen Nov 28 '11 at 5:19

Abstract Algebra is ruining my life!


Sep
24
awarded  Autobiographer
Nov
3
asked Given fields $M/E/F$, why does $[M:F] = [M:E][E:F]$?
Oct
22
comment Might $a(x)$ be irreducible in $F[x]$ but reducible in $F[x]/\langle p(x)\rangle$, where $a(x)$ is not a multiple of $p(x)$?
Thanks your comments were very helpful to my understanding.
Oct
22
revised Might $a(x)$ be irreducible in $F[x]$ but reducible in $F[x]/\langle p(x)\rangle$, where $a(x)$ is not a multiple of $p(x)$?
added 302 characters in body
Oct
22
comment Might $a(x)$ be irreducible in $F[x]$ but reducible in $F[x]/\langle p(x)\rangle$, where $a(x)$ is not a multiple of $p(x)$?
I'm talking about irreducibility of polynomials.
Oct
22
comment Might $a(x)$ be irreducible in $F[x]$ but reducible in $F[x]/\langle p(x)\rangle$, where $a(x)$ is not a multiple of $p(x)$?
Maybe I didn't get the memo here but it seems like everyone is saying x^4 - 5= (x^2 - sqrt(5))(x^2 + sqrt(5)) without mentioning that x^2 = 5 (mod p(x)). That is the only part that doesn't make sense to me. It's like everyone is ignoring that fact.
Oct
22
awarded  Scholar
Oct
22
accepted Might $a(x)$ be irreducible in $F[x]$ but reducible in $F[x]/\langle p(x)\rangle$, where $a(x)$ is not a multiple of $p(x)$?
Oct
22
comment Might $a(x)$ be irreducible in $F[x]$ but reducible in $F[x]/\langle p(x)\rangle$, where $a(x)$ is not a multiple of $p(x)$?
I have a question about your a(x) = x^4 + 1 and p(x) = x^2 + 1. Aren't we modding out by p(x) so x^2 is really congruent to -1 (mod p(x))? So x^4 + 1 would be (x^2)(x^2) + 1 = 1 + 1 = 2? What's the y supposed to be, I'm kind of confused about that notation.
Oct
22
comment Might $a(x)$ be irreducible in $F[x]$ but reducible in $F[x]/\langle p(x)\rangle$, where $a(x)$ is not a multiple of $p(x)$?
I understand your first comment. It was helpful.
Oct
22
comment Might $a(x)$ be irreducible in $F[x]$ but reducible in $F[x]/\langle p(x)\rangle$, where $a(x)$ is not a multiple of $p(x)$?
I just want a real example like functions a(x), p(x) and any ring F[x]
Oct
22
comment Might $a(x)$ be irreducible in $F[x]$ but reducible in $F[x]/\langle p(x)\rangle$, where $a(x)$ is not a multiple of $p(x)$?
Chinese remaindering? I don't follow you when you say that Bill. I'm only taking an introductory abstract algebra course
Oct
22
awarded  Commentator
Oct
22
revised Might $a(x)$ be irreducible in $F[x]$ but reducible in $F[x]/\langle p(x)\rangle$, where $a(x)$ is not a multiple of $p(x)$?
added 2 characters in body
Oct
22
asked Might $a(x)$ be irreducible in $F[x]$ but reducible in $F[x]/\langle p(x)\rangle$, where $a(x)$ is not a multiple of $p(x)$?
Sep
30
comment Abstract Algebra: Why is the number of prime numbers to a set $\mathbb{Z}_n$ usually $\varphi(n)$
Thank you Arturo, you are so understandable!
Sep
30
awarded  Student
Sep
30
awarded  Editor
Sep
30
revised Abstract Algebra: Why is the number of prime numbers to a set $\mathbb{Z}_n$ usually $\varphi(n)$
added 7 characters in body
Sep
30
comment Abstract Algebra: Why is the number of prime numbers to a set $\mathbb{Z}_n$ usually $\varphi(n)$
I actually meant n^p - n^(p - 1). Sorry.