# Maximal Ideals

• Jan 27th 2011, 06:45 AM
mulaosmanovicben
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
• Jan 27th 2011, 07:25 AM
FernandoRevilla
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
• Jan 27th 2011, 09:34 AM
mulaosmanovicben
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 :)
• Jan 28th 2011, 05:12 PM
mathstew
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.