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seen Nov 28 '11 at 5:19

Abstract Algebra is ruining my life!


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 Abstract algebra rings - How to prove it?
Oh thank you so much. That made sense.
Oct
22
comment Abstract algebra rings - How to prove it?
There's no typo. This exact practice problem that the professor printed for us.
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
comment Abstract algebra rings - How to prove it?
Is the statement that I'm trying to prove only true for primes p > 2?
Oct
22
comment Abstract algebra rings - How to prove it?
I suppose the extra condition would be that if one case doesn't work out go to the the next $p$ value. $3(x + 1)$ can be expressed as $(5x + 1)(3x + 3)$ in $Z_5[x]$
Oct
22
comment Abstract algebra rings - How to prove it?
But wait, 1 is a root for $(x + 1)$ in $Z_2[x]$ so wouldn't that mean that $(x + 1)$ is reducible in $Z_2[x]$?
Oct
22
comment Abstract algebra rings - How to prove it?
I already know that 3 is a non-unit and hence irreducible in $Z[x]$ and so is $(x+1)$. $3(x + 1)$ has to be reducible in $Z[x]$.
Oct
22
awarded  Commentator
Oct
22
comment Abstract algebra rings - How to prove it?
Now I understand you, you're saying that $(x+1)$ is irreducible in $Z_2[x]$ and it is. Hmm...tricky
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 $p(x)$?
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