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I am being asked to find the following:

Let $F$ denote the field $\dfrac{\mathbb{F}_2[\alpha]}{(\alpha^3 + \alpha + 1)}$. Simplify $\alpha(\alpha + 1)(\alpha + 1)$ in $F$ and calculate $\alpha^{-1}$ in $F$.

I am trying to understand what exactly $\dfrac{\mathbb{F}_2[\alpha]}{(\alpha^3 + \alpha + 1)}$ means. Does it mean the set of all polynomials with variable $\alpha$ of the form $(\alpha^3 + \alpha + 1)$? If so, would it be right that I simply have $\alpha(\alpha + 1)(\alpha + 1) = \alpha^3 + 2\alpha^2 + \alpha = \alpha^3 + \alpha$ since $2\alpha^2 = 0$ since we are working in $\mod 2$? I am also unsure as to how to even begin calculating $\alpha^{-1}$.

Thank you in advance for your help.

letsmakemuffinstogether
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    You should observe that $\alpha^3$ and $\alpha+1$ are in the same coset, because their difference $$\alpha^3-(\alpha+1)=\alpha^3+\alpha+1$$ is in the ideal that you mod out. Therefore in this field $\alpha^3=\alpha+1$ (by a slight abuse of notation). Therefore the polynomial $\alpha^3+\alpha$ simplifies further. The answer to the latter question becomes apparent, if you carry out this simplification. – Jyrki Lahtonen Apr 07 '15 at 07:30
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    I'm not sure why you're being asked this question if you don't know about the quotient operation. The structure you're working on is that of polynomials in $\alpha$ over the binary field, mod $\alpha^3 + \alpha + 1$. – G. H. Faust Apr 07 '15 at 07:32
  • @ Jyrki - So then is it true that I have: $\alpha(\alpha + 1)(\alpha + 1) = \alpha + 1 + \alpha = 2\alpha + 1 = 1$? – letsmakemuffinstogether Apr 07 '15 at 18:45
  • Yes, that is indeed true. – Lubin Apr 08 '15 at 03:41

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