Question

Consider the ideal I<x³-1,2x⁴+2x³+7x²+5x+5> in Q[x]. Find p(x) ∈Q[x] such that I =<p(x)> . Is Q[x]/I a field? Give reasons for your answer.

Answer 1

$p(x)=(x^3-1,2x^4+2x^3+7x^2+5x+5)$  is the greatest common divisor of $x^3-1$ and $2x^4+2x^3+7x^2+5x+5$ .

Find $p(x)$ :

$2x^4+2x^3+7x^2+5x+5:x^3-1$

$\begin{matrix} 2x^4+2x^3+7x^2+5x+5 &&&& |x^3-1 \\ 2x^4-2x &&&& 2x+2\\ &&2x^3+7x^2+5x+5\\ &&2x^3-2&\\ &&&7x^2+7x+7\\ &&&x^2+x+1 \end{matrix}$

$x^3-1:x^2+x+1$

$\begin{matrix} x^3-1 &&& |x^2+x+1 \\ x^3+x^2+x &&&x-1 \\ & -x^2-x-1\\ &-x^2-x-1\\ &&0 \end{matrix}$

Then $p(x)=x^2+x+1$ .

Then

$Q[x]/p(x)=Q[x] /<x^2+x+1>=\\ =Q[x]/Q[x]=\{0\}$

It is not a field, because there $x^2+x+1=0$

Answer: $p(x)=x^2+x+1$, $Q[x]/I$ is not a field