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I need a hint to solve a problem in the algebra book of Dummit and Foote. It is what is the inverse of $\dfrac{1+\theta}{1+\theta+\theta^{2}}$ in which $\theta$ is the root of $x^3-2x-2=0$.

I think there are two ways to solve this problem:

  1. Every element in $\mathbb{Q}(\theta)$ is of the form $a+b\theta+c\theta^{2}$ so we have $(1+\theta+\theta^{2})(a+b\theta+c\theta^{2})=1+\theta$, then we can multiply out but it is getting complicated.
  2. $\mathbb{Q}(\theta)$ is nothing but $\mathbb{Q}[x]/(x^3-2x-2)$, then there exist two polynomial $a(x), b(x)$ such that $a(x)(1+x+x^2)+b(x)(x^3-2x-2)=1$. If we can figure out what is $a(x)$, then its image in the quotient $\mathbb{Q}[x]/(x^3-2x-2)$ is exactly the inverse of $1+\theta+\theta^2$. Then we get the result by multiplying that image with $1+\theta$. But how can we compute $a(x)$ and $b(x)$?

Please help me.

Thanks.

user80756
  • 83

1 Answers1

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You're on the right track. Just multiply it out and equate coefficients. Since the 2nd, 3rd, and 4th power coefficients are zero on the right-hand side, things should simplify. I think $a=1$ and $c=0$, and you can use the fact that $\theta^3 -2 \theta -2 = 0$ to help simplify further (e.g. to eliminate factors of $\theta^3$).

  • I do not think it is that simply. Multiplying out we will get a polynomial of degree 4, and it is may be the result by multiplying $\theta^3-2\theta-2$ with a monomial. – user80756 Nov 21 '13 at 03:33
  • It is that simple. –  Nov 21 '13 at 07:15