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 Sep24 awarded Autobiographer Nov3 asked Given fields $M/E/F$, why does $[M:F] = [M:E][E:F]$? Oct22 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. Oct22 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 Oct22 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. Oct22 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. Oct22 awarded Scholar Oct22 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 $p(x)$? Oct22 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. Oct22 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. Oct22 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] Oct22 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 Oct22 awarded Commentator Oct22 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 Oct22 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 $p(x)$? Sep30 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! Sep30 awarded Student Sep30 awarded Editor Sep30 revised Abstract Algebra: Why is the number of prime numbers to a set $\mathbb{Z}_n$ usually $\varphi(n)$ added 7 characters in body Sep30 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.