Computation in a Quotient Ring

Question

I have $f(x)=x^3 +x -2 \in \mathbb{Q[x]}$ and R is the ring $\mathbb{Q[x]}/\langle f \rangle$.

$\alpha = x +\langle f \rangle$ is the image of $x$ in R. Note R is not a field as $f(x)=(x-1)(x^2 +x+2)$.

I need to calculate

$(\alpha ^2 - 2)^2$

and

$(\alpha ^2 - 2)^{-1}$ which is the inverse.

Answers needs to be given in the form of $a\alpha^2 +b\alpha+c$ where $a,b,c \in \mathbb{Q}$.

For the first one, I used the ideal addition and multiplication laws, and reached $(x^4 -4x^2 +4) + \langle f \rangle$ which is basically $(\alpha^2 -2)(\alpha^2 -2)$ but I cannot write it in the form it is asking in terms of $a\alpha^2 +b\alpha+c$. If someone can give me a hint, and just make sure I have got my final calculations correct, and possibly verify the inverse for me as well, I really appreciate their kind favor and time.

Many Thanks

Answer 1

• $\alpha^3+\alpha-2=0$, so $\alpha^3=-\alpha+2$ and $\alpha^4=-\alpha^2+2\alpha$,

so $\alpha^4-4\alpha^2+4$ simplifies to $-5\alpha^2+2\alpha+4$.

• Now say $(a\alpha^2+b\alpha+c)(\alpha^2-2)=1$.

Thus $a\alpha^4+b\alpha^3+c\alpha^2-2a\alpha^2-2b\alpha-2c=1.$

Thus $a(-2\alpha^2+2\alpha)+b(-\alpha+2)+(c-2a)\alpha^2-2b\alpha-2c=1$.

Thus $(c-3a)\alpha^2+(2a-3b)\alpha+(2b-2c)=1.$

Therefore $c-3a=0, 2a-3b=0$, and $2b-2c=1$.

Can you take it from here?