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I'm studying for my algebra exam, and came across the following problem, which I'm not sure how to solve

Let $f = X^2 - 1 \in \mathbb{F}_3[X]$ and $\alpha = X + \langle f \rangle \in \mathbb{F}_3[X]/\langle f \rangle$.

Prove the following properties:

$(\alpha - 1)^2 = \alpha - 1$

$(-\alpha - 1)^2 = -\alpha - 1$

$(\alpha - 1)(-\alpha - 1) = 0$

$(\alpha - 1) +(-\alpha - 1) = 1$

I've shown that $\alpha^2 = 1$ and tried to use that in proving the first property: $(\alpha - 1)^2 = \alpha^2 - 2\alpha + 1 = 1 - 2\alpha + 1 = 2 - 2\alpha$

But I'm unsure of how to proceed from here

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    $\begingroup$ Remember, we are working in $\mathbb F_3$. $2$ and $-1$ are the same thing! $\endgroup$ – RKD Dec 30 '13 at 16:21
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Note that in $\mathbb{F}_3$ you have $2 = -1$ and $1 = -2$.

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Look, the right way to think about it is that

  • each element of $K = \mathbb{F}_3[X]/\langle f \rangle$ can be written uniquely as $a + b \alpha$, for $a, b \in \mathbb{F}_3$, and
  • $\alpha^2 = 1$.

You need no more to be able to calculate in $K$.

Also, note that $2 = -1$ in $\mathbb{F}_3$, and thus in $K$.

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