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Let $\alpha \in \mathbb{C}$ be a root of the irreducible polynomial $f(x)=x^3+x+1$. Write $\alpha^{-1}$ and $(\alpha+1)^{-1}$ in terms of $\{1,\alpha,\alpha^2\}$.

So I was able to find a similar question here and was able to solve some of the problems i was given where $\alpha$ had a positive power but I am confused on how to divide the polynomials when it is instead raised to a negative power? Thanks in advance!

  • 0
    You can use the Euclidean algorithm to find these inverses. [See this related post](http://math.stackexchange.com/questions/1963526/expressing-alpha-1-1-where-alpha3-alpha2-alpha2-0). – 2017-01-26

4 Answers 4

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Let us assume that $\alpha^{-1} = a \cdot 1 + b \cdot \alpha + c \cdot \alpha^2$ and try and find the coefficients $a,b,c \in \mathbb{C}$. Multiplying both sides of the equation by $\alpha$ we get

$$ 1 = a \cdot \alpha + b \cdot \alpha^2 + c \cdot \alpha^3. $$

Since $\alpha^3 + \alpha + 1 = 0$ we have $\alpha^3 = -\alpha - 1$ and we get

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

or

$$ (-c -1) \cdot 1 + (a - c) \cdot \alpha + b \cdot \alpha^2 = 0. $$

Since $1,\alpha,\alpha^2$ are linearly independent, we must have $-c - 1 = a - c = b = 0$ so $a = c = -1$ and $b = 0$ leading to

$$ \alpha^{-1} = -1 - \alpha^2. $$

This can be verified directly by multiplying both sides by $\alpha$:

$$ 1 =^{?} (-1 - \alpha^2)(\alpha) = -\alpha -\alpha^3 = -(\alpha + \alpha^3) = -(-1) = 1. $$

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.So we have that $\alpha^3 + \alpha + 1 = 0$. Divide by $\alpha$ and transpose, you get $\frac{1}{\alpha} = - \alpha^2 - 1$.

On the other hand, note that $\alpha + 1 = -\alpha^3$, so that $\frac{1}{\alpha+1} = \alpha^{-3} = -(\alpha^2 - 1)^{3}$. Expanding this, and using $\alpha^3 = -\alpha-1$, gives you the answer.

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For the first part, we know that $\alpha^3+\alpha+1=0$. What happens when you multiply this by $\alpha^{-1}$?

The second part works in a similar way, starting with $1+\alpha=-\alpha^3$ and using the first part.

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$\alpha^3+ \alpha +1=0\Rightarrow \alpha (\alpha^2+1)=-1 \Rightarrow \alpha^{-1}=-\alpha^2-1$.

$\alpha^3+ \alpha +1=0\Rightarrow \alpha^3+1^3+\alpha=0 \Rightarrow (\alpha+1)(\alpha^2-\alpha+1)+\alpha=0 \Rightarrow (\alpha+1)(\alpha^2-\alpha+1)+\alpha+1=1 \Rightarrow (\alpha+1)(\alpha^2-\alpha+2)=1$

$\Rightarrow (1+\alpha)^{-1}=\alpha^2-\alpha+2 $