
Maximal Ideals
Prove <x^2+1> is maximal in R[x]
R[x] = {all polynomials with real coefficients}
<x^2+1> = {f(x)*(x^2+1) ; f(x) is from R[x]}
Could you guys give me some hints? I am not really sure how to even start this. This exact problem is actually an example in my book but I could not even follow it! (Gallian)
Thank you

One way (there is an immediate characterization in terms of irredutible polynomials):
$\displaystyle <x^2+1>\;\textrm{maximal}\;\Leftrightarrow \mathbb{R}[x]/<x^2+1>\;\textrm{is\:a\;field}$
and if $\displaystyle I\subset \mathbb{R}[x]$ is an ideal, then $\displaystyle \mathbb{R}/I$ is a commutative ring.
So, you only have to prove that
$\displaystyle ax+b+<x^2+1>\;\;(ax+b\neq 0)$
is a unit in $\displaystyle \mathbb{R}[x]/<x^2+1>$ .
Fernando Revilla

Thanks I was actually trying to prove that R[x]/<x^2+1> was a field by showing <x^2+1> was maximal but I guess showing that its all the elements are units would be easier :)

yes that is the easiest way to show that something is maximal. Think about what happens when you take the quotient R[x]/(x^2+1).
Remember when you mod out by an ideal you partion the ring into cosets: a+(x^2+1) for each a in R[x]. Something gets "squashed" to 0 iff it is in the ideal (x^2+1).
So what type of elements are left? What can you decide this quotient ring is isomorphic to? Think about this for a bit, you should be able to show it's a field without too much trouble.