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Let $\alpha$ denote the image of $x$ in $\mathbb Q[x]/(x^3 + 3x + 3)$. Express each of $1/α,1/(1+ \alpha),1/(1+ \alpha^2)$ in the form $c_2\alpha^2+c_1\alpha+c_0$ with $c_0, c_1,c_2 \in \mathbb Q$.

I can prove $x^3 + 3x + 3$ is irreducible in $\mathbb Q[x]$ and $\mathbb Q[x]/(x^3 + 3x + 3)$ is a field. However, I don't know how to proceed from here.

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  • $\begingroup$ Do you know how to write down all elements of this field? if yes, then write, say, $1/\alpha$ in this form and compute. $\endgroup$ – Dietrich Burde Feb 11 at 15:51
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    $\begingroup$ Generally we can use EEA = Extended Euclidean Algorithm to compute inverses, e.g. see here. But the first is immediate by $\, x(x^2+3) = -3\ $ $\endgroup$ – Bill Dubuque Feb 11 at 16:08
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Let $P(x)=x^3+3x+3$, $Q_1(x)=x$, $Q_2(x)=1+x$ and $Q_3(x)=1+x^2.$

By Euclid's algorithm, we have that $$P(x)=(x^2+3)Q_1(x) + 3.$$

If we view this relation in $\mathbb{Q}[x]/(P(x))$, we get that $$0=P(\alpha)=(\alpha^2+3)Q_1(\alpha)+3,$$ hence $$3 = -(\alpha^2+3)\alpha,$$ and is $$\frac{1}{\alpha} = \frac{-(\alpha^2+3)}{3}=-\frac{1}{3}\alpha^2-1.$$

For $\frac{1}{1+\alpha}$, we solve in a similar way. We have that $$P(x)=(x^2-x+4)Q_2(x)-1,$$ then $$0=P(\alpha)=(\alpha^2-\alpha+4)Q_2(\alpha)-1.$$ So $$\frac{1}{1+\alpha} = \frac{1}{Q_2(\alpha)} = \alpha^2-\alpha+4.$$

For $\frac{1}{1+\alpha^2}$ we found it in two steps (I'm avoiding the use of the mentioned Extended Euclidean Algorithm). First, we know as above that $$P(x)=xQ_3(x)+2x+3,$$ and in $\mathbb{Q}[x]/(P(x))$ we have $$0 = P(\alpha)=\alpha Q_3(\alpha)+2\alpha+3,$$ so $$\frac{1}{1+\alpha^2}=\frac{1}{Q_3(\alpha)} = \frac{-\alpha}{2\alpha+3}.$$

Now, we compute $\frac{1}{2\alpha+3}$. Since $$P(x)=(\frac{1}{2}x^2-\frac{3}{4}x+\frac{21}{8})(2x+3)-\frac{39}{8},$$ is $$\frac{1}{2\alpha+3} = \frac{8}{39}(\frac{1}{2}\alpha^2-\frac{3}{4}\alpha+\frac{21}{8}).$$ You now only have to multiply $\frac{1}{2\alpha+3}$ by $-\alpha$ and simplify the expresion using the relation $\alpha^3+3\alpha+3=0.$

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  • $\begingroup$ Actually, I wrote that I'm avoiding the use of EEA but my two steps are equivalent to using directly EEA. $\endgroup$ – AlgebraicallyClosed Feb 12 at 17:57

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