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 February 26th, 2013, 02:20 PM #1 Member   Joined: Oct 2012 Posts: 36 Thanks: 0 maximal ideal Let F be the field and f(x)=x-1,g(x)=x^2-1 and F[x]/(f(x)) is isomorphism to F, is it g(x) maximal?? I will say no.Since g(x) is not 0, the dieal (x^2-1) in a prime idea domain F is maximal iff (x^2-1) is irreducible. And we say (x^2-1) is irreducible if it is not a unit, but x^2-1=(x+1)(x-1) implies that either (x+1) or (x-1) is a unit. but I can find a taylor expansion of 1/(x^2-1) which means (x^2-1) is a unit, contradicts irreducible is my idea right ???
 February 27th, 2013, 05:06 AM #2 Member   Joined: Jan 2013 Posts: 93 Thanks: 0 Re: Maximal ideal You’ve probably got the right idea though I’m not sure what the Taylor expansion of $x^2-1$ has to do here. First, the statement $F[x}/\langle f(x)\rangle\underline{\underline{\sim}}F$ tells you that $F$ is a two-element field since the elements of $F[x}/\langle f(x)\rangle$ are $\langle f(x)\rangle$ and $1+\langle f(x)\rangle$. So the elements of $F[x}/\langle g(x)\rangle$ are $\langle g(x)\rangle$, $1+\langle g(x)\rangle$, $x+\langle g(x)\rangle$, $x+1+\langle g(x)\rangle$. (NB: $1=-1$ in the field $F$.) Now $\langle g(x)\rangle$ is maximal in $F[x]$ if and only if $F[x}/\langle g(x)\rangle$ is a field. But it’s not a field as $x+1\notin\langle g(x)\rangle$ but $(x+1)^2\in\langle g(x)\rangle$.

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