Question

Find a maximal ideal of R [x] containing 

the ideal <x²— 1, x³— 1>

Answer 1

$Solution: ~As~ x²-1 = (x-1)(x+1) ~and ~x³-1=(x-1)(x²+x+1), ~\\the ~ideal ~J = <x-1> ~contains ~every ~element ~of~ the ~given~ ideal~ which~ are ~\\of~ the~ form \\~~~~~~~~~~~~~~~~~~~~~~~~~~f(x)(x²-1)+g(x)(x³-1), \\where ~f(x) ~and ~g(x)~ are~ polynomials ~in ~R[x],~ because ~the ~above ~can ~be ~written~ as~ \\(x-1)[(x+1)f(x)+(x²+x+1)g(x)]. \\Also~ the~ ideal~ <x-1>~ is~ a~ maximal~ ideal~ because~ if~ any~ ideal~ I ~properly~ \\contains~ J,~ then~ take~ any ~polynomial ~p(x)~ in ~I \notin~ J. \\By ~the~ Division~ algorithm ~in~ R[x], ~we ~can~ find~ polynomials~ q(x)~ and~ r(x)~ such~that \\p(x)=q(x-1)+r(x), ~with ~deg(r(x))<deg(x-1)=1 \\ \Rightarrow ~ r(x)=0,~ or~ r(x)~ is ~a ~non~zero~ constant. \\But~ r(x) = p(x) - q(x)(x-1)~ is~ in~ I,~ and ~if ~the~ ideal ~I~ contains~ a~ nonzero~ \\constant~ r, ~then~ as ~the~ inverse ~of~ r~ belongs~ to ~the~ field ~R ~and~ hence ~\\1 ~will~ belong~ to~ I ~\\ \Rightarrow I=R. \\ \therefore~ J = <x-1>~ is~a ~maximal ~ideal ~and ~hence~ it~ is~ the~ maximal ~ideal ~containing ~\\<x²-1, x³-1>.$