Thread: 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.