Question

Consider the ideal I<x³-1.x⁴+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

$p(x)=(x^3-1,x^4+2x^3+7x^2+5x+5)$ is the greatest common divisor of $x^3-1$ and $x^4+2x^3+7x^2+5x+5$, because there are polynomials $P(x)$ and $Q(x)$ such that $(x^3-1,x^4+2x^3+7x^2+5x+5)=P(x)(x^3-1)+$

$+Q(x)(x^4+2x^3+7x^2+5x+5)$, so $\langle(x^3-1,x^4+2x^3+7x^2+5x+5)\rangle\subset$ $\subset\langle x^3-1,x^4+2x^3+7x^2+5x+5\rangle$ . But $x^3-1$ and $x^4+2x^3+7x^2+5x+5$ are divisible by $\langle(x^3-1,x^4+2x^3+7x^2+5x+5)\rangle$, so $\langle x^3-1,x^4+2x^3+7x^2+5x+5\rangle\subset$

$\subset\langle(x^3-1,x^4+2x^3+7x^2+5x+5)\rangle$ , so $\langle x^3-1,x^4+2x^3+7x^2+5x+5\rangle=$

$=\langle(x^3-1,x^4+2x^3+7x^2+5x+5)\rangle$ .

We have $x^3-1=(x-1)(x^2+x+1)$, where $x-1$ and $x^2+x+1$ are irreducible polynomials in $\mathbb Q[x]$. So we need to check if $x^4+2x^3+7x^2+5x+5$ divisible by $x-1$ or $x^2+x+1$

$1$ is not a solution of $x^4+2x^3+7x^2+5x+5=0$, so $x^4+2x^3+7x^2+5x+5$ is not divisible by$x-1$.

$x^4+2x^3+7x^2+5x+5$ also is not divisible by $x^2+x+1$, because $x^4+2x^3+7x^2+5x+5=(x^2+x+1)(x^2+x+5)+x$

So $p(x)=(x^3-1,x^4+2x^3+7x^2+5x+5)=1$

Then $\mathbb Q[x]/\langle1\rangle=\mathbb Q[x]/\mathbb Q[x]=\{0\}$ . It is not a field, because there $1=0$ .

Answer: $p(x)=1$ , $\mathbb Q[x]/I$ is not a field.